200 Soal Penyetaraan Jumlah Atom dalam Persamaan Reaksi beserta Jawabannya

Berikut ini 200 persamaan reaksi molekuler yang dapat dijadikan latihan untuk penyetaraan reaksi kimia biasa untuk siswa kelas 10 MA/SMA dan juga dapat digunakan untuk latihan penyetaraan reaksi redoks di kelas yang lebih tinggi. Karena tujuannya latihan jangan langsung klik jawabannya, kerjakan mandiri lebih dahulu kemudian cocokan. Untuk melihat hasil setara jumlah unsur/atom antara pada pereaksi dan hasil reaksi sila klik segitiga di kiri setiap nomor persamaan reaksi.

Setarakan persamaan reaksi kimia berikut bila jumlah unsur-unsur yang bersesuaian belum setara:

Persamaan reaksi 3 spesi.

C + O₂ → CO

\( aC + bO_2 \rightarrow cCO \)

C: \( a = c \)
O: \( 2b = c \)

Misal \( c = 2 \)
\( a = 2 \)
\( 2b = 2 \Rightarrow b = 1 \)

Setara: 2C + O₂ → 2CO

S + O₂ → SO₃

\( aS + bO_2 \rightarrow cSO_3 \)

S: \( a = c \)
O: \( 2b = 3c \)

Misal \( c = 2 \)
\( a = 2 \)
\( 2b = 6 \Rightarrow b = 3 \)

Setara: 2S + 3O₂ → 2SO₃

Si + O₂ → SiO₂

\( aSi + bO_2 \rightarrow cSiO_2 \)

Si: \( a = c \)
O: \( 2b = 2c \)

Misal \( c = 1 \)
\( a = 1 \)
\( 2b = 2 \Rightarrow b = 1 \)

Setara: Si + O₂ → SiO₂

PCl₃ + Cl₂ → PCl₅

\( aPCl_3 + bCl_2 \rightarrow cPCl_5 \)

P: \( a = c \)
Cl: \( 3a + 2b = 5c \)

Misal \( c = 1 \)
\( a = 1 \)
\( 3(1) + 2b = 5 \Rightarrow 2b = 2 \Rightarrow b = 1 \)

Setara: PCl₃ + Cl₂ → PCl₅

K + O₂ → K₂O

\( aK + bO_2 \rightarrow cK_2O \)

K: \( a = 2c \)
O: \( 2b = c \)

Misal \( c = 2 \)
\( a = 4 \)
\( 2b = 2 \Rightarrow b = 1 \)

Setara: 4K + O₂ → 2K₂O

Mg + O₂ → MgO

\( aMg + bO_2 \rightarrow cMgO \)

Mg: \( a = c \)
O: \( 2b = c \)

Misal \( c = 2 \)
\( a = 2 \)
\( 2b = 2 \Rightarrow b = 1 \)

Setara: 2Mg + O₂ → 2MgO

Ca + O₂ → CaO

\( aCa + bO_2 \rightarrow cCaO \)

Ca: \( a = c \)
O: \( 2b = c \)

Misal \( c = 2 \)
\( a = 2 \)
\( 2b = 2 \Rightarrow b = 1 \)

Setara: 2Ca + O₂ → 2CaO

Pb + O₂ → PbO

\( aPb + bO_2 \rightarrow cPbO \)

Pb: \( a = c \)
O: \( 2b = c \)

Misal \( c = 2 \)
\( a = 2 \)
\( 2b = 2 \Rightarrow b = 1 \)

Setara: 2Pb + O₂ → 2PbO

Cu + O₂ → CuO

\( aCu + bO_2 \rightarrow cCuO \)

Cu: \( a = c \)
O: \( 2b = c \)

Misal \( c = 2 \)
\( a = 2 \)
\( 2b = 2 \Rightarrow b = 1 \)

Setara: 2Cu + O₂ → 2CuO

Al + O₂ → Al₂O₃

\( aAl + bO_2 \rightarrow cAl_2O_3 \)

Al: \( a = 2c \)
O: \( 2b = 3c \)

Misal \( c = 2 \)
\( a = 4 \)
\( 2b = 6 \Rightarrow b = 3 \)

Setara: 4Al + 3O₂ → 2Al₂O₃

Fe + O₂ → Fe₂O₃

\( aFe + bO_2 \rightarrow cFe_2O_3 \)

Fe: \( a = 2c \)
O: \( 2b = 3c \)

Misal \( c = 2 \)
\( a = 4 \)
\( 2b = 6 \Rightarrow b = 3 \)

Setara: 4Fe + 3O₂ → 2Fe₂O₃

P + O₂ → P₂O₅

\( aP + bO_2 \rightarrow cP_2O_5 \)

P: \( a = 2c \)
O: \( 2b = 5c \)

Misal \( c = 2 \)
\( a = 4 \)
\( 2b = 10 \Rightarrow b = 5 \)

Setara: 4P + 5O₂ → 2P₂O₅

P₄ + O₂ → P₄O₁₀

\( aP_4 + bO_2 \rightarrow cP_4O_{10} \)

P: \( 4a = 4c \)
O: \( 2b = 10c \)

Misal \( c = 1 \)
\( 4a = 4 \Rightarrow a = 1 \)
\( 2b = 10 \Rightarrow b = 5 \)

Setara: P₄ + 5O₂ → P₄O₁₀

Al + Cl₂ → AlCl₃

\( aAl + bCl_2 \rightarrow cAlCl_3 \)

Al: \( a = c \)
Cl: \( 2b = 3c \)

Misal \( c = 2 \)
\( a = 2 \)
\( 2b = 6 \Rightarrow b = 3 \)

Setara: 2Al + 3Cl₂ → 2AlCl₃

H₂ + O₂ → H₂O

\( aH_2 + bO_2 \rightarrow cH_2O \)

H: \( 2a = 2c \)
O: \( 2b = c \)

Misal \( c = 2 \)
\( 2a = 4 \Rightarrow a = 2 \)
\( 2b = 2 \Rightarrow b = 1 \)

Setara: 2H₂ + O₂ → 2H₂O

N₂ + H₂ → NH₃

\( aN_2 + bH_2 \rightarrow cNH_3 \)

N: \( 2a = c \)
H: \( 2b = 3c \)

Misal \( c = 2 \)
\( 2a = 2 \Rightarrow a = 1 \)
\( 2b = 6 \Rightarrow b = 3 \)

Setara: N₂ + 3H₂ → 2NH₃

H₂ + Cl₂ → HCl

\( aH_2 + bCl_2 \rightarrow cHCl \)

H: \( 2a = c \)
Cl: \( 2b = c \)

Misal \( c = 2 \)
\( 2a = 2 \Rightarrow a = 1 \)
\( 2b = 2 \Rightarrow b = 1 \)

Setara: H₂ + Cl₂ → 2HCl

Na + Cl₂ → NaCl

\( aNa + bCl_2 \rightarrow cNaCl \)

Na: \( a = c \)
Cl: \( 2b = c \)

Misal \( c = 2 \)
\( a = 2 \)
\( 2b = 2 \Rightarrow b = 1 \)

Setara: 2Na + Cl₂ → 2NaCl

CO + O₂ → CO₂

\( aCO + bO_2 \rightarrow cCO_2 \)

C: \( a = c \)
O: \( a + 2b = 2c \)

Misal \( c = 2 \)
\( a = 2 \)
\( 2 + 2b = 4 \Rightarrow 2b = 2 \Rightarrow b = 1 \)

Setara: 2CO + O₂ → 2CO₂

H₂O₂ → H₂O + O₂

\( aH_2O_2 \rightarrow bH_2O + cO_2 \)

H: \( 2a = 2b \)
O: \( 2a = b + 2c \)

Dari H: \( a = b \)
Substitusi ke O: \( 2a = a + 2c \Rightarrow a = 2c \)

Misal \( c = 1 \)
\( a = 2 \)
\( b = 2 \)

Setara: 2H₂O₂ → 2H₂O + O₂

SO₂ + O₂ → SO₃

\( aSO_2 + bO_2 \rightarrow cSO_3 \)

S: \( a = c \)
O: \( 2a + 2b = 3c \)

Misal \( c = 2 \)
\( a = 2 \)
\( 2(2) + 2b = 3(2) \Rightarrow 4 + 2b = 6 \Rightarrow 2b = 2 \Rightarrow b = 1 \)

Setara: 2SO₂ + O₂ → 2SO₃

N₂ + O₂ → NO₂

\( aN_2 + bO_2 \rightarrow cNO_2 \)

N: \( 2a = c \)
O: \( 2b = 2c \)

Misal \( c = 2 \)
\( 2a = 2 \Rightarrow a = 1 \)
\( 2b = 4 \Rightarrow b = 2 \)

Setara: N₂ + 2O₂ → 2NO₂

NO + O₂ → NO₂

\( aNO + bO_2 \rightarrow cNO_2 \)

N: \( a = c \)
O: \( a + 2b = 2c \)

Misal \( c = 2 \)
\( a = 2 \)
\( 2 + 2b = 4 \Rightarrow 2b = 2 \Rightarrow b = 1 \)

Setara: 2NO + O₂ → 2NO₂

H₂ + N₂ → NH₃

\( aH_2 + bN_2 \rightarrow cNH_3 \)

H: \( 2a = 3c \)
N: \( 2b = c \)

Misal \( c = 2 \)
\( 2a = 6 \Rightarrow a = 3 \)
\( 2b = 2 \Rightarrow b = 1 \)

Setara: 3H₂ + N₂ → 2NH₃

Cu + CuSO₄ → Cu₂SO₄

\( aCu + bCuSO_4 \rightarrow cCu_2SO_4 \)

Cu: \( a + b = 2c \)
S: \( b = c \)
O: \( 4b = 4c \)

Dari S: \( b = c \)
Substitusi ke Cu: \( a + c = 2c \Rightarrow a = c \)

Misal \( c = 1 \)
\( a = 1 \)
\( b = 1 \)

Setara: Cu + CuSO₄ → Cu₂SO₄

PbO₂ + SO₂ → PbSO₄

\( aPbO_2 + bSO_2 \rightarrow cPbSO_4 \)

Pb: \( a = c \)
S: \( b = c \)
O: \( 2a + 2b = 4c \)

Misal \( c = 1 \)
\( a = 1 \)
\( b = 1 \)
Cek O: \( 2(1) + 2(1) = 4 \) ✓

Setara: PbO₂ + SO₂ → PbSO₄

Persamaan reaksi 4 spesi.

S + SO₂ + H₂O → H₂SO₃

\( aS + bSO_2 + cH_2O \rightarrow dH_2SO_3 \)

S: \( a + b = d \)
O: \( 2b + c = 3d \)
H: \( 2c = 2d \)

Dari H: \( c = d \)
Dari S: \( a + b = d \)
Dari O: \( 2b + d = 3d \Rightarrow 2b = 2d \Rightarrow b = d \)

Misal \( d = 1 \)
\( c = 1 \)
\( b = 1 \)
\( a + 1 = 1 \Rightarrow a = 0 \)

Seharusnya S tidak ada dalam persamaan reaksi.

Setara: SO₂ + H₂O → H₂SO₃

CH₄ + Cl₂ → CH₃Cl + HCl

\( aCH_4 + bCl_2 \rightarrow cCH_3Cl + dHCl \)

C: \( a = c \)
H: \( 4a = 3c + d \)
Cl: \( 2b = c + d \)

Misal \( a = 1 \)
\( c = 1 \)
Dari H: \( 4 = 3 + d \Rightarrow d = 1 \)
Dari Cl: \( 2b = 1 + 1 \Rightarrow b = 1 \)

Setara: CH₄ + Cl₂ → CH₃Cl + HCl

FeO + CO → Fe + CO₂

\( aFeO + bCO \rightarrow cFe + dCO_2 \)

Fe: \( a = c \)
O: \( a + b = 2d \)
C: \( b = d \)

Misal \( a = 1 \)
\( c = 1 \)
\( b = d \)
Dari O: \( 1 + b = 2b \Rightarrow 1 = b \)
Jadi \( b = 1, d = 1 \)

Setara: FeO + CO → Fe + CO₂

Fe₃O₄ + C → Fe + CO₂

\( aFe_3O_4 + bC \rightarrow cFe + dCO_2 \)

Fe: \( 3a = c \)
O: \( 4a = 2d \)
C: \( b = d \)

Misal \( a = 1 \)
\( c = 3 \)
\( 4 = 2d \Rightarrow d = 2 \)
\( b = 2 \)

Setara: Fe₃O₄ + 4C → 3Fe + 4CO₂

Fe + H₂SO₄ → FeSO₄ + H₂

\( aFe + bH_2SO_4 \rightarrow cFeSO_4 + dH_2 \)

Fe: \( a = c \)
H: \( 2b = 2d \)
S: \( b = c \)
O: \( 4b = 4c \)

Dari Fe dan S: \( a = b = c \)
Dari H: \( b = d \)
Misal \( a = 1 \)
\( b = 1, c = 1, d = 1 \)

Setara: Fe + H₂SO₄ → FeSO₄ + H₂

FeS + H₂SO₄ → FeSO₄ + H₂S

\( aFeS + bH_2SO_4 \rightarrow cFeSO_4 + dH_2S \)

Fe: \( a = c \)
S: \( a + b = c + d \)
H: \( 2b = 2d \)
O: \( 4b = 4c \)

Dari H: \( b = d \)
Dari O: \( b = c \)
Jadi \( a = b = c = d \)

Misal \( a = 1 \)
\( b = 1, c = 1, d = 1 \)

Setara: FeS + H₂SO₄ → FeSO₄ + H₂S

ZnS + H₂SO₄ → ZnSO₄ + H₂S

\( aZnS + bH_2SO_4 \rightarrow cZnSO_4 + dH_2S \)

Zn: \( a = c \)
S: \( a + b = c + d \)
H: \( 2b = 2d \)
O: \( 4b = 4c \)

Dari H: \( b = d \)
Dari O: \( b = c \)
Jadi \( a = b = c = d \)

Misal \( a = 1 \)
\( b = 1, c = 1, d = 1 \)

Setara: ZnS + H₂SO₄ → ZnSO₄ + H₂S

PbO + HNO₃ → Pb(NO₃)₂ + H₂O

\( aPbO + bHNO_3 \rightarrow cPb(NO_3)_2 + dH_2O \)

Pb: \( a = c \)
H: \( b = 2d \)
N: \( b = 2c \)
O: \( a + 3b = 6c + d \)

Dari N: \( b = 2c \)
Dari H: \( 2c = 2d \Rightarrow c = d \)
Misal \( c = 1 \)
\( a = 1, d = 1, b = 2 \)

Setara: PbO + 2HNO₃ → Pb(NO₃)₂ + H₂O

Mg + HNO₃ → Mg(NO₃)₂ + H₂

\( aMg + bHNO_3 \rightarrow cMg(NO_3)_2 + dH_2 \)

Mg: \( a = c \)
H: \( b = 2d \)
N: \( b = 2c \)
O: \( 3b = 6c \)

Dari N: \( b = 2c \)
Dari H: \( 2c = 2d \Rightarrow c = d \)
Misal \( c = 1 \)
\( a = 1, d = 1, b = 2 \)

Setara: Mg + 2HNO₃ → Mg(NO₃)₂ + H₂

Zn + HNO₃ → Zn(NO₃)₂ + H₂

\( aZn + bHNO_3 \rightarrow cZn(NO_3)_2 + dH_2 \)

Zn: \( a = c \)
H: \( b = 2d \)
N: \( b = 2c \)
O: \( 3b = 6c \)

Dari N: \( b = 2c \)
Dari H: \( 2c = 2d \Rightarrow c = d \)
Misal \( c = 1 \)
\( a = 1, d = 1, b = 2 \)

Setara: Zn + 2HNO₃ → Zn(NO₃)₂ + H₂

CH₂O + O₂ → CO₂ + H₂O

\( aCH_2O + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( a = c \)
H: \( 2a = 2d \)
O: \( a + 2b = 2c + d \)

Dari H: \( a = d \)
Dari C: \( a = c \)
Jadi \( a = c = d \)

Misal \( a = 1 \)
\( c = 1, d = 1 \)
Dari O: \( 1 + 2b = 2 + 1 \Rightarrow 2b = 2 \Rightarrow b = 1 \)

Setara: CH₂O + O₂ → CO₂ + H₂O

H₂S + SO₂ → S + H₂O

\( aH_2S + bSO_2 \rightarrow cS + dH_2O \)

H: \( 2a = 2d \)
S: \( a + b = c \)
O: \( 2b = d \)

Dari H: \( a = d \)
Dari O: \( 2b = a \Rightarrow a = 2b \)
Dari S: \( 2b + b = c \Rightarrow 3b = c \)

Misal \( b = 1 \)
\( a = 2, c = 3, d = 2 \)

Setara: 2H₂S + SO₂ → 3S + 2H₂O

PbO₂ + H₂ → Pb + H₂O

\( aPbO_2 + bH_2 \rightarrow cPb + dH_2O \)

Pb: \( a = c \)
H: \( 2b = 2d \)
O: \( 2a = d \)

Dari H: \( b = d \)
Dari O: \( 2a = b \)

Misal \( a = 1 \)
\( c = 1, b = 2, d = 2 \)

Setara: PbO₂ + 2H₂ → Pb + 2H₂O

C₂H₄ + O₂ → CO₂ + H₂O

\( aC_2H_4 + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 2a = c \)
H: \( 4a = 2d \)
O: \( 2b = 2c + d \)

Dari H: \( 2a = d \)
Dari C: \( c = 2a \)

Misal \( a = 1 \)
\( c = 2, d = 2 \)
Dari O: \( 2b = 2(2) + 2 \Rightarrow 2b = 6 \Rightarrow b = 3 \)

Setara: C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

C₂H₅OH + O₂ → CO₂ + H₂O

\( aC_2H_5OH + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 2a = c \)
H: \( 6a = 2d \)
O: \( a + 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 2 \)
\( 6 = 2d \Rightarrow d = 3 \)
Dari O: \( 1 + 2b = 2(2) + 3 \Rightarrow 1 + 2b = 7 \Rightarrow 2b = 6 \Rightarrow b = 3 \)

Setara: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O

CH₃OH + O₂ → CO₂ + H₂O

\( aCH_3OH + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( a = c \)
H: \( 4a = 2d \)
O: \( a + 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 2 \)
\( 4(2) = 2d \Rightarrow 8 = 2d \Rightarrow d = 4 \)
Dari O: \( 2 + 2b = 2(2) + 4 \Rightarrow 2 + 2b = 8 \Rightarrow 2b = 6 \Rightarrow b = 3 \)

Setara: 2CH₃OH + 3O₂ → 2CO₂ + 4H₂O

CS₂ + O₂ → CO₂ + SO₂

\( aCS_2 + bO_2 \rightarrow cCO_2 + dSO_2 \)

C: \( a = c \)
S: \( 2a = d \)
O: \( 2b = 2c + 2d \)

Misal \( a = 1 \)
\( c = 1 \)
\( d = 2 \)
Dari O: \( 2b = 2(1) + 2(2) \Rightarrow 2b = 2 + 4 \Rightarrow 2b = 6 \Rightarrow b = 3 \)

Setara: CS₂ + 3O₂ → CO₂ + 2SO₂

H₂S + O₂ → SO₂ + H₂O

\( aH_2S + bO_2 \rightarrow cSO_2 + dH_2O \)

H: \( 2a = 2d \)
S: \( a = c \)
O: \( 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 2 \)
\( 2(2) = 2d \Rightarrow 4 = 2d \Rightarrow d = 2 \)
Dari O: \( 2b = 2(2) + 2 \Rightarrow 2b = 4 + 2 \Rightarrow 2b = 6 \Rightarrow b = 3 \)

Setara: 2H₂S + 3O₂ → 2SO₂ + 2H₂O

2H₂ + SO₂ → S + 2H₂O

\( aH_2 + bSO_2 \rightarrow cS + dH_2O \)

H: \( 2a = 2d \)
S: \( b = c \)
O: \( 2b = d \)

Dari H: \( a = d \)
Dari O: \( 2b = a \)
Misal \( b = 1 \)
\( c = 1 \)
\( a = 2(1) = 2 \)
\( d = 2 \)

Setara: 2H₂ + SO₂ → S + 2H₂O

Cu₂S + O₂ → Cu + SO₂

\( aCu_2S + bO_2 \rightarrow cCu + dSO_2 \)

Cu: \( 2a = c \)
S: \( a = d \)
O: \( 2b = 2d \)

Misal \( a = 1 \)
\( d = 1 \)
\( c = 2 \)
Dari O: \( 2b = 2(1) \Rightarrow 2b = 2 \Rightarrow b = 1 \)

Setara: Cu₂S + O₂ → 2Cu + SO₂

PbS + PbO → Pb + SO₂

\( aPbS + bPbO \rightarrow cPb + dSO_2 \)

Pb: \( a + b = c \)
S: \( a = d \)
O: \( b = 2d \)

Misal \( a = 1 \)
\( d = 1 \)
\( b = 2(1) = 2 \)
\( c = 1 + 2 = 3 \)

Setara: PbS + 2PbO → 3Pb + SO₂

PbO + C → Pb + CO₂

\( aPbO + bC \rightarrow cPb + dCO_2 \)

Pb: \( a = c \)
O: \( a = 2d \)
C: \( b = d \)

Misal \( d = 1 \)
\( a = 2 \)
\( c = 2 \)
\( b = 1 \)

Setara: 2PbO + C → 2Pb + CO₂

NH₃ + O₂ → NO + H₂O

\( aNH_3 + bO_2 \rightarrow cNO + dH_2O \)

N: \( a = c \)
H: \( 3a = 2d \)
O: \( 2b = c + d \)

Misal \( a = 4 \)
\( c = 4 \)
\( 3(4) = 2d \Rightarrow 12 = 2d \Rightarrow d = 6 \)
Dari O: \( 2b = 4 + 6 \Rightarrow 2b = 10 \Rightarrow b = 5 \)

Setara: 4NH₃ + 5O₂ → 4NO + 6H₂O

NH₃ + O₂ → N₂ + H₂O

\( aNH_3 + bO_2 \rightarrow cN_2 + dH_2O \)

N: \( a = 2c \)
H: \( 3a = 2d \)
O: \( 2b = d \)

Misal \( c = 1 \)
\( a = 2 \)
\( 3(2) = 2d \Rightarrow 6 = 2d \Rightarrow d = 3 \)
\( 2b = 3 \Rightarrow b = 1.5 \)

Kalikan 2: \( a = 4, b = 3, c = 2, d = 6 \)

Setara: 4NH₃ + 3O₂ → 2N₂ + 6H₂O

C₃H₈ + O₂ → CO₂ + H₂O

\( aC_3H_8 + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 3a = c \)
H: \( 8a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 3 \)
\( 8 = 2d \Rightarrow d = 4 \)
\( 2b = 2(3) + 4 \Rightarrow 2b = 6 + 4 \Rightarrow 2b = 10 \Rightarrow b = 5 \)

Setara: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

CH₄ + O₂ → CO₂ + H₂O

\( aCH_4 + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( a = c \)
H: \( 4a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 1 \)
\( 4 = 2d \Rightarrow d = 2 \)
\( 2b = 2(1) + 2 \Rightarrow 2b = 2 + 2 \Rightarrow 2b = 4 \Rightarrow b = 2 \)

Setara: CH₄ + 2O₂ → CO₂ + 2H₂O

C₄H₁₀ + O₂ → CO₂ + H₂O

\( aC_4H_{10} + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 4a = c \)
H: \( 10a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 8 \)
\( 10(2) = 2d \Rightarrow 20 = 2d \Rightarrow d = 10 \)
\( 2b = 2(8) + 10 \Rightarrow 2b = 16 + 10 \Rightarrow 2b = 26 \Rightarrow b = 13 \)

Setara: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

C₅H₁₂ + O₂ → CO₂ + H₂O

\( aC_5H_{12} + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 5a = c \)
H: \( 12a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 5 \)
\( 12 = 2d \Rightarrow d = 6 \)
\( 2b = 2(5) + 6 \Rightarrow 2b = 10 + 6 \Rightarrow 2b = 16 \Rightarrow b = 8 \)

Setara: C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O

C₇H₁₆ + O₂ → CO₂ + H₂O

\( aC_7H_{16} + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 7a = c \)
H: \( 16a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 7 \)
\( 16 = 2d \Rightarrow d = 8 \)
\( 2b = 2(7) + 8 \Rightarrow 2b = 14 + 8 \Rightarrow 2b = 22 \Rightarrow b = 11 \)

Setara: C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O

C₈H₁₈ + O₂ → CO₂ + H₂O

\( aC_8H_{18} + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 8a = c \)
H: \( 18a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 16 \)
\( 18(2) = 2d \Rightarrow 36 = 2d \Rightarrow d = 18 \)
\( 2b = 2(16) + 18 \Rightarrow 2b = 32 + 18 \Rightarrow 2b = 50 \Rightarrow b = 25 \)

Setara: 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

C₉H₂₀ + O₂ → CO₂ + H₂O

\( aC_9H_{20} + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 9a = c \)
H: \( 20a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 9 \)
\( 20 = 2d \Rightarrow d = 10 \)
\( 2b = 2(9) + 10 \Rightarrow 2b = 18 + 10 \Rightarrow 2b = 28 \Rightarrow b = 14 \)

Setara: C₉H₂₀ + 14O₂ → 9CO₂ + 10H₂O

C₁₀H₂₂ + O₂ → CO₂ + H₂O

\( aC_{10}H_{22} + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 10a = c \)
H: \( 22a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 20 \)
\( 22(2) = 2d \Rightarrow 44 = 2d \Rightarrow d = 22 \)
\( 2b = 2(20) + 22 \Rightarrow 2b = 40 + 22 \Rightarrow 2b = 62 \Rightarrow b = 31 \)

Setara: 2C₁₀H₂₂ + 31O₂ → 20CO₂ + 22H₂O

C₄H₈ + O₂ → CO₂ + H₂O

\( aC_4H_8 + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 4a = c \)
H: \( 8a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 4 \)
\( 8 = 2d \Rightarrow d = 4 \)
\( 2b = 2(4) + 4 \Rightarrow 2b = 8 + 4 \Rightarrow 2b = 12 \Rightarrow b = 6 \)

Setara: C₄H₈ + 6O₂ → 4CO₂ + 4H₂O

C₆H₁₂ + O₂ → CO₂ + H₂O

\( aC_6H_{12} + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 6a = c \)
H: \( 12a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 6 \)
\( 12 = 2d \Rightarrow d = 6 \)
\( 2b = 2(6) + 6 \Rightarrow 2b = 12 + 6 \Rightarrow 2b = 18 \Rightarrow b = 9 \)

Setara: C₆H₁₂ + 9O₂ → 6CO₂ + 6H₂O

C₇H₁₄ + O₂ → CO₂ + H₂O

\( aC_7H_{14} + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 7a = c \)
H: \( 14a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 14 \)
\( 14(2) = 2d \Rightarrow 28 = 2d \Rightarrow d = 14 \)
\( 2b = 2(14) + 14 \Rightarrow 2b = 28 + 14 \Rightarrow 2b = 42 \Rightarrow b = 21 \)

Setara: 2C₇H₁₄ + 21O₂ → 14CO₂ + 14H₂O

C₈H₁₆ + O₂ → CO₂ + H₂O

\( aC_8H_{16} + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 8a = c \)
H: \( 16a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 8 \)
\( 16 = 2d \Rightarrow d = 8 \)
\( 2b = 2(8) + 8 \Rightarrow 2b = 16 + 8 \Rightarrow 2b = 24 \Rightarrow b = 12 \)

Setara: C₈H₁₆ + 12O₂ → 8CO₂ + 8H₂O

C₉H₁₈ + O₂ → CO₂ + H₂O

\( aC_9H_{18} + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 9a = c \)
H: \( 18a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 18 \)
\( 18(2) = 2d \Rightarrow 36 = 2d \Rightarrow d = 18 \)
\( 2b = 2(18) + 18 \Rightarrow 2b = 36 + 18 \Rightarrow 2b = 54 \Rightarrow b = 27 \)

Setara: 2C₉H₁₈ + 27O₂ → 18CO₂ + 18H₂O

C₁₀H₂₀ + O₂ → CO₂ + H₂O

\( aC_{10}H_{20} + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 10a = c \)
H: \( 20a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 10 \)
\( 20 = 2d \Rightarrow d = 10 \)
\( 2b = 2(10) + 10 \Rightarrow 2b = 20 + 10 \Rightarrow 2b = 30 \Rightarrow b = 15 \)

Setara: C₁₀H₂₀ + 15O₂ → 10CO₂ + 10H₂O

C₁₁H₂₂ + O₂ → CO₂ + H₂O

\( aC_{11}H_{22} + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 11a = c \)
H: \( 22a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 22 \)
\( 22(2) = 2d \Rightarrow 44 = 2d \Rightarrow d = 22 \)
\( 2b = 2(22) + 22 \Rightarrow 2b = 44 + 22 \Rightarrow 2b = 66 \Rightarrow b = 33 \)

Setara: 2C₁₁H₂₂ + 33O₂ → 22CO₂ + 22H₂O

C₁₂H₂₄ + O₂ → CO₂ + H₂O

\( aC_{12}H_{24} + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 12a = c \)
H: \( 24a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 12 \)
\( 24 = 2d \Rightarrow d = 12 \)
\( 2b = 2(12) + 12 \Rightarrow 2b = 24 + 12 \Rightarrow 2b = 36 \Rightarrow b = 18 \)

Setara: C₁₂H₂₄ + 18O₂ → 12CO₂ + 12H₂O

C₁₃H₂₆ + O₂ → CO₂ + H₂O

\( aC_{13}H_{26} + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 13a = c \)
H: \( 26a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 26 \)
\( 26(2) = 2d \Rightarrow 52 = 2d \Rightarrow d = 26 \)
\( 2b = 2(26) + 26 \Rightarrow 2b = 52 + 26 \Rightarrow 2b = 78 \Rightarrow b = 39 \)

Setara: 2C₁₃H₂₆ + 39O₂ → 26CO₂ + 26H₂O

C₁₄H₂₈ + O₂ → CO₂ + H₂O

\( aC_{14}H_{28} + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 14a = c \)
H: \( 28a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 14 \)
\( 28 = 2d \Rightarrow d = 14 \)
\( 2b = 2(14) + 14 \Rightarrow 2b = 28 + 14 \Rightarrow 2b = 42 \Rightarrow b = 21 \)

Setara: C₁₄H₂₈ + 21O₂ → 14CO₂ + 14H₂O

C₁₅H₃₀ + O₂ → CO₂ + H₂O

\( aC_{15}H_{30} + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 15a = c \)
H: \( 30a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 30 \)
\( 30(2) = 2d \Rightarrow 60 = 2d \Rightarrow d = 30 \)
\( 2b = 2(30) + 30 \Rightarrow 2b = 60 + 30 \Rightarrow 2b = 90 \Rightarrow b = 45 \)

Setara: 2C₁₅H₃₀ + 45O₂ → 30CO₂ + 30H₂O

CH₃COOH + O₂ → CO₂ + H₂O

\( aCH_3COOH + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 2a = c \)
H: \( 4a = 2d \)
O: \( 2a + 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 4 \)
\( 4(2) = 2d \Rightarrow 8 = 2d \Rightarrow d = 4 \)
\( 2(2) + 2b = 2(4) + 4 \Rightarrow 4 + 2b = 8 + 4 \Rightarrow 4 + 2b = 12 \Rightarrow 2b = 8 \Rightarrow b = 4 \)

Setara: 2CH₃COOH + 4O₂ → 4CO₂ + 4H₂O

C₄H₉OH + O₂ → CO₂ + H₂O

\( aC_4H_9OH + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 4a = c \)
H: \( 10a = 2d \)
O: \( a + 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 4 \)
\( 10 = 2d \Rightarrow d = 5 \)
\( 1 + 2b = 2(4) + 5 \Rightarrow 1 + 2b = 8 + 5 \Rightarrow 1 + 2b = 13 \Rightarrow 2b = 12 \Rightarrow b = 6 \)

Setara: C₄H₉OH + 6O₂ → 4CO₂ + 5H₂O

C₅H₁₁OH + O₂ → CO₂ + H₂O

\( aC_5H_{11}OH + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 5a = c \)
H: \( 12a = 2d \)
O: \( a + 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 10 \)
\( 12(2) = 2d \Rightarrow 24 = 2d \Rightarrow d = 12 \)
\( 2 + 2b = 2(10) + 12 \Rightarrow 2 + 2b = 20 + 12 \Rightarrow 2 + 2b = 32 \Rightarrow 2b = 30 \Rightarrow b = 15 \)

Setara: 2C₅H₁₁OH + 15O₂ → 10CO₂ + 12H₂O

C₇H₁₅OH + O₂ → CO₂ + H₂O

\( aC_7H_{15}OH + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 7a = c \)
H: \( 16a = 2d \)
O: \( a + 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 14 \)
\( 16(2) = 2d \Rightarrow 32 = 2d \Rightarrow d = 16 \)
\( 2 + 2b = 2(14) + 16 \Rightarrow 2 + 2b = 28 + 16 \Rightarrow 2 + 2b = 44 \Rightarrow 2b = 42 \Rightarrow b = 21 \)

Setara: 2C₇H₁₅OH + 21O₂ → 14CO₂ + 16H₂O

C₈H₁₇OH + O₂ → CO₂ + H₂O

\( aC_8H_{17}OH + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 8a = c \)
H: \( 18a = 2d \)
O: \( a + 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 8 \)
\( 18 = 2d \Rightarrow d = 9 \)
\( 1 + 2b = 2(8) + 9 \Rightarrow 1 + 2b = 16 + 9 \Rightarrow 1 + 2b = 25 \Rightarrow 2b = 24 \Rightarrow b = 12 \)

Setara: C₈H₁₇OH + 12O₂ → 8CO₂ + 9H₂O

C₄H₁₀O + O₂ → CO₂ + H₂O

\( aC_4H_{10}O + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 4a = c \)
H: \( 10a = 2d \)
O: \( a + 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 4 \)
\( 10 = 2d \Rightarrow d = 5 \)
\( 1 + 2b = 2(4) + 5 \Rightarrow 1 + 2b = 8 + 5 \Rightarrow 1 + 2b = 13 \Rightarrow 2b = 12 \Rightarrow b = 6 \)

Setara: C₄H₁₀O + 6O₂ → 4CO₂ + 5H₂O

C₅H₁₀O + O₂ → CO₂ + H₂O

\( aC_5H_{10}O + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 5a = c \)
H: \( 10a = 2d \)
O: \( a + 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 5 \)
\( 10 = 2d \Rightarrow d = 5 \)
\( 1 + 2b = 2(5) + 5 \Rightarrow 1 + 2b = 10 + 5 \Rightarrow 1 + 2b = 15 \Rightarrow 2b = 14 \Rightarrow b = 7 \)

Setara: C₅H₁₀O + 7O₂ → 5CO₂ + 5H₂O

CH₃CH₂CH₂OH + O₂ → CO₂ + H₂O

\( aC_3H_8O + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 3a = c \)
H: \( 8a = 2d \)
O: \( a + 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 6 \)
\( 8(2) = 2d \Rightarrow 16 = 2d \Rightarrow d = 8 \)
\( 2 + 2b = 2(6) + 8 \Rightarrow 2 + 2b = 12 + 8 \Rightarrow 2 + 2b = 20 \Rightarrow 2b = 18 \Rightarrow b = 9 \)

Setara: 2CH₃CH₂CH₂OH + 9O₂ → 6CO₂ + 8H₂O

CH₃CH₂CH₂CH₂OH + O₂ → CO₂ + H₂O

\( aC_4H_{10}O + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 4a = c \)
H: \( 10a = 2d \)
O: \( a + 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 4 \)
\( 10 = 2d \Rightarrow d = 5 \)
\( 1 + 2b = 2(4) + 5 \Rightarrow 1 + 2b = 8 + 5 \Rightarrow 1 + 2b = 13 \Rightarrow 2b = 12 \Rightarrow b = 6 \)

Setara: CH₃CH₂CH₂CH₂OH + 6O₂ → 4CO₂ + 5H₂O

CH₃CH₂COOH + O₂ → CO₂ + H₂O

\( aC_3H_6O_2 + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 3a = c \)
H: \( 6a = 2d \)
O: \( 2a + 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 6 \)
\( 6(2) = 2d \Rightarrow 12 = 2d \Rightarrow d = 6 \)
\( 2(2) + 2b = 2(6) + 6 \Rightarrow 4 + 2b = 12 + 6 \Rightarrow 4 + 2b = 18 \Rightarrow 2b = 14 \Rightarrow b = 7 \)

Setara: 2CH₃CH₂COOH + 7O₂ → 6CO₂ + 6H₂O

CH₃CH₂CH₂COOH + O₂ → CO₂ + H₂O

\( aC_4H_8O_2 + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 4a = c \)
H: \( 8a = 2d \)
O: \( 2a + 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 4 \)
\( 8 = 2d \Rightarrow d = 4 \)
\( 2(1) + 2b = 2(4) + 4 \Rightarrow 2 + 2b = 8 + 4 \Rightarrow 2 + 2b = 12 \Rightarrow 2b = 10 \Rightarrow b = 5 \)

Setara: CH₃CH₂CH₂COOH + 5O₂ → 4CO₂ + 4H₂O

C₇H₁₅COOH + O₂ → CO₂ + H₂O

\( aC_8H_{16}O_2 + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 8a = c \)
H: \( 16a = 2d \)
O: \( 2a + 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 8 \)
\( 16 = 2d \Rightarrow d = 8 \)
\( 2(1) + 2b = 2(8) + 8 \Rightarrow 2 + 2b = 16 + 8 \Rightarrow 2 + 2b = 24 \Rightarrow 2b = 22 \Rightarrow b = 11 \)

Setara: C₇H₁₅COOH + 11O₂ → 8CO₂ + 8H₂O

CH₃COCH₃ + O₂ → CO₂ + H₂O

\( aC_3H_6O + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 3a = c \)
H: \( 6a = 2d \)
O: \( a + 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 3 \)
\( 6 = 2d \Rightarrow d = 3 \)
\( 1 + 2b = 2(3) + 3 \Rightarrow 1 + 2b = 6 + 3 \Rightarrow 1 + 2b = 9 \Rightarrow 2b = 8 \Rightarrow b = 4 \)

Setara: CH₃COCH₃ + 4O₂ → 3CO₂ + 3H₂O

C₆H₁₂O₂ + O₂ → CO₂ + H₂O

\( aC_6H_{12}O_2 + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 6a = c \)
H: \( 12a = 2d \)
O: \( 2a + 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 6 \)
\( 12 = 2d \Rightarrow d = 6 \)
\( 2(1) + 2b = 2(6) + 6 \Rightarrow 2 + 2b = 12 + 6 \Rightarrow 2 + 2b = 18 \Rightarrow 2b = 16 \Rightarrow b = 8 \)

Setara: C₆H₁₂O₂ + 8O₂ → 6CO₂ + 6H₂O

Fe₂O₃ + CO → Fe + CO₂

\( aFe_2O_3 + bCO \rightarrow cFe + dCO_2 \)

Fe: \( 2a = c \)
O: \( 3a + b = 2d \)
C: \( b = d \)

Misal \( a = 1 \)
\( c = 2 \)
\( b = d \)
Dari O: \( 3 + b = 2b \Rightarrow 3 = b \)
Jadi \( b = 3, d = 3 \)

Setara: Fe₂O₃ + 3CO → 2Fe + 3CO₂

Fe₂O₃ + H₂ → Fe + H₂O

\( aFe_2O_3 + bH_2 \rightarrow cFe + dH_2O \)

Fe: \( 2a = c \)
O: \( 3a = d \)
H: \( 2b = 2d \)

Misal \( a = 1 \)
\( c = 2 \)
\( d = 3 \)
\( 2b = 2(3) \Rightarrow 2b = 6 \Rightarrow b = 3 \)

Setara: Fe₂O₃ + 3H₂ → 2Fe + 3H₂O

Fe₃O₄ + CO → Fe + CO₂

\( aFe_3O_4 + bCO \rightarrow cFe + dCO_2 \)

Fe: \( 3a = c \)
O: \( 4a + b = 2d \)
C: \( b = d \)

Misal \( a = 1 \)
\( c = 3 \)
\( b = d \)
Dari O: \( 4 + b = 2b \Rightarrow 4 = b \)
Jadi \( b = 4, d = 4 \)

Setara: Fe₃O₄ + 4CO → 3Fe + 4CO₂

Fe₃O₄ + H₂ → Fe + H₂O

\( aFe_3O_4 + bH_2 \rightarrow cFe + dH_2O \)

Fe: \( 3a = c \)
O: \( 4a = d \)
H: \( 2b = 2d \)

Misal \( a = 1 \)
\( c = 3 \)
\( d = 4 \)
\( 2b = 2(4) \Rightarrow 2b = 8 \Rightarrow b = 4 \)

Setara: Fe₃O₄ + 4H₂ → 3Fe + 4H₂O

Pb₃O₄ + H₂ → Pb + H₂O

\( aPb_3O_4 + bH_2 \rightarrow cPb + dH_2O \)

Pb: \( 3a = c \)
O: \( 4a = d \)
H: \( 2b = 2d \)

Misal \( a = 1 \)
\( c = 3 \)
\( d = 4 \)
\( 2b = 2(4) \Rightarrow 2b = 8 \Rightarrow b = 4 \)

Setara: Pb₃O₄ + 4H₂ → 3Pb + 4H₂O

Fe₂O₃ + C → Fe + CO

\( aFe_2O_3 + bC \rightarrow cFe + dCO \)

Fe: \( 2a = c \)
O: \( 3a = d \)
C: \( b = d \)

Misal \( a = 1 \)
\( c = 2 \)
\( d = 3 \)
\( b = 3 \)

Setara: Fe₂O₃ + 3C → 2Fe + 3CO

Fe₂O₃ + Al → Al₂O₃ + Fe

\( aFe_2O_3 + bAl \rightarrow cAl_2O_3 + dFe \)

Fe: \( 2a = d \)
O: \( 3a = 3c \)
Al: \( b = 2c \)

Dari O: \( a = c \)
Misal \( a = 1 \)
\( c = 1 \)
\( d = 2 \)
\( b = 2 \)

Setara: Fe₂O₃ + 2Al → Al₂O₃ + 2Fe

Al + H₂SO₄ → Al₂(SO₄)₃ + H₂

\( aAl + bH_2SO_4 \rightarrow cAl_2(SO_4)_3 + dH_2 \)

Al: \( a = 2c \)
H: \( 2b = 2d \)
S: \( b = 3c \)
O: \( 4b = 12c \)

Misal \( c = 1 \)
\( a = 2 \)
\( b = 3 \)
Dari H: \( 2(3) = 2d \Rightarrow 6 = 2d \Rightarrow d = 3 \)

Setara: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

Mg + H₂SO₄ → MgSO₄ + H₂

\( aMg + bH_2SO_4 \rightarrow cMgSO_4 + dH_2 \)

Mg: \( a = c \)
H: \( 2b = 2d \)
S: \( b = c \)
O: \( 4b = 4c \)

Misal \( a = 1 \)
\( c = 1, b = 1, d = 1 \)

Setara: Mg + H₂SO₄ → MgSO₄ + H₂

Fe₂(SO₄)₃ + H₂ → Fe + H₂SO₄

\( aFe_2(SO_4)_3 + bH_2 \rightarrow cFe + dH_2SO_4 \)

Fe: \( 2a = c \)
S: \( 3a = d \)
O: \( 12a = 4d \)
H: \( 2b = 2d \)

Misal \( a = 1 \)
\( c = 2, d = 3 \)
\( 2b = 2(3) \Rightarrow 2b = 6 \Rightarrow b = 3 \)

Setara: Fe₂(SO₄)₃ + 3H₂ → 2Fe + 3H₂SO₄

C₂H₆ + O₂ → CO₂ + H₂O

\( aC_2H_6 + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 2a = c \)
H: \( 6a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 4 \)
\( 6(2) = 2d \Rightarrow 12 = 2d \Rightarrow d = 6 \)
\( 2b = 2(4) + 6 \Rightarrow 2b = 8 + 6 \Rightarrow 2b = 14 \Rightarrow b = 7 \)

Setara: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

Fe₂O₃ + SO₂ + O₂ → Fe₂(SO₄)₃

\( aFe_2O_3 + bSO_2 + cO_2 \rightarrow dFe_2(SO_4)_3 \)

Fe: \( 2a = 2d \) → \( a = d \)      (1)
S: \( b = 3d \)      (2)
O: \( 3a + 2b + 2c = 12d \)      (3)

Dari (1): \( a = d \)
Dari (2): \( b = 3d \)
Substitusi ke (3):
\( 3d + 2(3d) + 2c = 12d \)
\( 3d + 6d + 2c = 12d \)
\( 9d + 2c = 12d \)
\( 2c = 3d \)
\( c = \dfrac{3}{2}d \)      (4)

Pilih \( d = 2 \) (untuk menghilangkan pecahan):
\( a = 2 \)
\( b = 3 \times 2 = 6 \)
\( c = \dfrac{3}{2} \times 2 = 3 \)

Mengecek keseimbangan semua atom:
Fe: Reaktan = \( 2a = 4 \), Produk = \( 2d = 4 \) ✓
S: Reaktan = \( b = 6 \), Produk = \( 3d = 6 \) ✓
O: Reaktan = \( 3a + 2b + 2c = 6 + 12 + 6 = 24 \), Produk = \( 12d = 24 \) ✓

Persamaan reaksi setara:
\( 2Fe_2O_3 + 6SO_2 + 3O_2 \rightarrow 2Fe_2(SO_4)_3 \)

C₆H₁₂O₆ + O₂ → CO₂ + H₂O

\( aC_6H_{12}O_6 + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 6a = c \)
H: \( 12a = 2d \)
O: \( 6a + 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 6 \)
\( 12 = 2d \Rightarrow d = 6 \)
\( 6 + 2b = 2(6) + 6 \Rightarrow 6 + 2b = 12 + 6 \Rightarrow 6 + 2b = 18 \Rightarrow 2b = 12 \Rightarrow b = 6 \)

Setara: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

C₁₀H₈ + O₂ → CO₂ + H₂O

\( aC_{10}H_8 + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 10a = c \)
H: \( 8a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 10 \)
\( 8 = 2d \Rightarrow d = 4 \)
\( 2b = 2(10) + 4 \Rightarrow 2b = 20 + 4 \Rightarrow 2b = 24 \Rightarrow b = 12 \)

Setara: C₁₀H₈ + 12O₂ → 10CO₂ + 4H₂O

C₆H₆ + O₂ → CO₂ + H₂O

\( aC_6H_6 + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 6a = c \)
H: \( 6a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 12 \)
\( 6(2) = 2d \Rightarrow 12 = 2d \Rightarrow d = 6 \)
\( 2b = 2(12) + 6 \Rightarrow 2b = 24 + 6 \Rightarrow 2b = 30 \Rightarrow b = 15 \)

Setara: 2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O

C₇H₈ + O₂ → CO₂ + H₂O

\( aC_7H_8 + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 7a = c \)
H: \( 8a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 7 \)
\( 8 = 2d \Rightarrow d = 4 \)
\( 2b = 2(7) + 4 \Rightarrow 2b = 14 + 4 \Rightarrow 2b = 18 \Rightarrow b = 9 \)

Setara: C₇H₈ + 9O₂ → 7CO₂ + 4H₂O

C₈H₁₀ + O₂ → CO₂ + H₂O

\( aC_8H_{10} + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 8a = c \)
H: \( 10a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 16 \)
\( 10(2) = 2d \Rightarrow 20 = 2d \Rightarrow d = 10 \)
\( 2b = 2(16) + 10 \Rightarrow 2b = 32 + 10 \Rightarrow 2b = 42 \Rightarrow b = 21 \)

Setara: 2C₈H₁₀ + 21O₂ → 16CO₂ + 10H₂O

C₆H₅CH₃ + O₂ → CO₂ + H₂O

\( aC_7H_8 + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 7a = c \)
H: \( 8a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 7 \)
\( 8 = 2d \Rightarrow d = 4 \)
\( 2b = 2(7) + 4 \Rightarrow 2b = 14 + 4 \Rightarrow 2b = 18 \Rightarrow b = 9 \)

Setara: C₆H₅CH₃ + 9O₂ → 7CO₂ + 4H₂O

C₆H₅OH + O₂ → CO₂ + H₂O

\( aC_6H_6O + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 6a = c \)
H: \( 6a = 2d \)
O: \( a + 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 6 \)
\( 6 = 2d \Rightarrow d = 3 \)
\( 1 + 2b = 2(6) + 3 \Rightarrow 1 + 2b = 12 + 3 \Rightarrow 1 + 2b = 15 \Rightarrow 2b = 14 \Rightarrow b = 7 \)

Setara: C₆H₅OH + 7O₂ → 6CO₂ + 3H₂O

C₆H₅COOH + O₂ → CO₂ + H₂O

\( aC_7H_6O_2 + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 7a = c \)
H: \( 6a = 2d \)
O: \( 2a + 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 14 \)
\( 6(2) = 2d \Rightarrow 12 = 2d \Rightarrow d = 6 \)
\( 2(2) + 2b = 2(14) + 6 \Rightarrow 4 + 2b = 28 + 6 \Rightarrow 4 + 2b = 34 \Rightarrow 2b = 30 \Rightarrow b = 15 \)

Setara: 2C₆H₅COOH + 15O₂ → 14CO₂ + 6H₂O

C₆H₅CH₂OH + O₂ → CO₂ + H₂O

\( aC_7H_8O + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 7a = c \)
H: \( 8a = 2d \)
O: \( a + 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 14 \)
\( 8(2) = 2d \Rightarrow 16 = 2d \Rightarrow d = 8 \)
\( 2 + 2b = 2(14) + 8 \Rightarrow 2 + 2b = 28 + 8 \Rightarrow 2 + 2b = 36 \Rightarrow 2b = 34 \Rightarrow b = 17 \)

Setara: 2C₆H₅CH₂OH + 17O₂ → 14CO₂ + 8H₂O

C₆H₅COCH₃ + O₂ → CO₂ + H₂O

\( aC_8H_8O + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 8a = c \)
H: \( 8a = 2d \)
O: \( a + 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 16 \)
\( 8(2) = 2d \Rightarrow 16 = 2d \Rightarrow d = 8 \)
\( 2 + 2b = 2(16) + 8 \Rightarrow 2 + 2b = 32 + 8 \Rightarrow 2 + 2b = 40 \Rightarrow 2b = 38 \Rightarrow b = 19 \)

Setara: 2C₆H₅COCH₃ + 19O₂ → 16CO₂ + 8H₂O

CH₃CHO + O₂ → CO₂ + H₂O

\( aC_2H_4O + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 2a = c \)
H: \( 4a = 2d \)
O: \( a + 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 4 \)
\( 4(2) = 2d \Rightarrow 8 = 2d \Rightarrow d = 4 \)
\( 2 + 2b = 2(4) + 4 \Rightarrow 2 + 2b = 8 + 4 \Rightarrow 2 + 2b = 12 \Rightarrow 2b = 10 \Rightarrow b = 5 \)

Setara: 2CH₃CHO + 5O₂ → 4CO₂ + 4H₂O

CH₃CH₂CHO + O₂ → CO₂ + H₂O

\( aC_3H_6O + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 3a = c \)
H: \( 6a = 2d \)
O: \( a + 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 3 \)
\( 6 = 2d \Rightarrow d = 3 \)
\( 1 + 2b = 2(3) + 3 \Rightarrow 1 + 2b = 6 + 3 \Rightarrow 1 + 2b = 9 \Rightarrow 2b = 8 \Rightarrow b = 4 \)

Setara: CH₃CH₂CHO + 4O₂ → 3CO₂ + 3H₂O

C₆H₁₃CHO + O₂ → CO₂ + H₂O

\( aC_7H_{14}O + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 7a = c \)
H: \( 14a = 2d \)
O: \( a + 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 7 \)
\( 14 = 2d \Rightarrow d = 7 \)
\( 1 + 2b = 2(7) + 7 \Rightarrow 1 + 2b = 14 + 7 \Rightarrow 1 + 2b = 21 \Rightarrow 2b = 20 \Rightarrow b = 10 \)

Setara: C₆H₁₃CHO + 10O₂ → 7CO₂ + 7H₂O

C₉H₁₉CHO + O₂ → CO₂ + H₂O

\( aC_{10}H_{20}O + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 10a = c \)
H: \( 20a = 2d \)
O: \( a + 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 20 \)
\( 20(2) = 2d \Rightarrow 40 = 2d \Rightarrow d = 20 \)
\( 2 + 2b = 2(20) + 20 \Rightarrow 2 + 2b = 40 + 20 \Rightarrow 2 + 2b = 60 \Rightarrow 2b = 58 \Rightarrow b = 29 \)

Setara: 2C₉H₁₉CHO + 29O₂ → 20CO₂ + 20H₂O

C₆H₅CHO + O₂ → CO₂ + H₂O

\( aC_7H_6O + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 7a = c \)
H: \( 6a = 2d \)
O: \( a + 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 7 \)
\( 6 = 2d \Rightarrow d = 3 \)
\( 1 + 2b = 2(7) + 3 \Rightarrow 1 + 2b = 14 + 3 \Rightarrow 1 + 2b = 17 \Rightarrow 2b = 16 \Rightarrow b = 8 \)

Setara: C₆H₅CHO + 8O₂ → 7CO₂ + 3H₂O

C₆H₅COC₂H₅ + O₂ → CO₂ + H₂O

\( aC_9H_{10}O + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 9a = c \)
H: \( 10a = 2d \)
O: \( a + 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 9 \)
\( 10 = 2d \Rightarrow d = 5 \)
\( 1 + 2b = 2(9) + 5 \Rightarrow 1 + 2b = 18 + 5 \Rightarrow 1 + 2b = 23 \Rightarrow 2b = 22 \Rightarrow b = 11 \)

Setara: C₆H₅COC₂H₅ + 11O₂ → 9CO₂ + 5H₂O

C₂H₂ + O₂ → CO₂ + H₂O

\( aC_2H_2 + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 2a = c \)
H: \( 2a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 4 \)
\( 2(2) = 2d \Rightarrow 4 = 2d \Rightarrow d = 2 \)
\( 2b = 2(4) + 2 \Rightarrow 2b = 8 + 2 \Rightarrow 2b = 10 \Rightarrow b = 5 \)

Setara: 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O

C₃H₆ + O₂ → CO₂ + H₂O

\( aC_3H_6 + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 3a = c \)
H: \( 6a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 6 \)
\( 6(2) = 2d \Rightarrow 12 = 2d \Rightarrow d = 6 \)
\( 2b = 2(6) + 6 \Rightarrow 2b = 12 + 6 \Rightarrow 2b = 18 \Rightarrow b = 9 \)

Setara: 2C₃H₆ + 9O₂ → 6CO₂ + 6H₂O

C₄H₆ + O₂ → CO₂ + H₂O

\( aC_4H_6 + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 4a = c \)
H: \( 6a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 8 \)
\( 6(2) = 2d \Rightarrow 12 = 2d \Rightarrow d = 6 \)
\( 2b = 2(8) + 6 \Rightarrow 2b = 16 + 6 \Rightarrow 2b = 22 \Rightarrow b = 11 \)

Setara: 2C₄H₆ + 11O₂ → 8CO₂ + 6H₂O

C₅H₁₀ + O₂ → CO₂ + H₂O

\( aC_5H_{10} + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 5a = c \)
H: \( 10a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 10 \)
\( 10(2) = 2d \Rightarrow 20 = 2d \Rightarrow d = 10 \)
\( 2b = 2(10) + 10 \Rightarrow 2b = 20 + 10 \Rightarrow 2b = 30 \Rightarrow b = 15 \)

Setara: 2C₅H₁₀ + 15O₂ → 10CO₂ + 10H₂O

C₃H₅OH + O₂ → CO₂ + H₂O

\( aC_3H_6O + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 3a = c \)
H: \( 6a = 2d \)
O: \( a + 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 3 \)
\( 6 = 2d \Rightarrow d = 3 \)
\( 1 + 2b = 2(3) + 3 \Rightarrow 1 + 2b = 6 + 3 \Rightarrow 1 + 2b = 9 \Rightarrow 2b = 8 \Rightarrow b = 4 \)

Setara: C₃H₅OH + 4O₂ → 3CO₂ + 3H₂O

C₃H₇OH + O₂ → CO₂ + H₂O

\( aC_3H_8O + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 3a = c \)
H: \( 8a = 2d \)
O: \( a + 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 6 \)
\( 8(2) = 2d \Rightarrow 16 = 2d \Rightarrow d = 8 \)
\( 2 + 2b = 2(6) + 8 \Rightarrow 2 + 2b = 12 + 8 \Rightarrow 2 + 2b = 20 \Rightarrow 2b = 18 \Rightarrow b = 9 \)

Setara: 2C₃H₇OH + 9O₂ → 6CO₂ + 8H₂O

C₁₄H₃₀ + O₂ → CO₂ + H₂O

\( aC_{14}H_{30} + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 14a = c \)
H: \( 30a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 28 \)
\( 30(2) = 2d \Rightarrow 60 = 2d \Rightarrow d = 30 \)
\( 2b = 2(28) + 30 \Rightarrow 2b = 56 + 30 \Rightarrow 2b = 86 \Rightarrow b = 43 \)

Setara: 2C₁₄H₃₀ + 43O₂ → 28CO₂ + 30H₂O

C₁₂H₂₆ + O₂ → CO₂ + H₂O

\( aC_{12}H_{26} + bO_2 \rightarrow cCO_2 + dH_2O \)

C: \( 12a = c \)
H: \( 26a = 2d \)
O: \( 2b = 2c + d \)

Misal \( a = 2 \)
\( c = 24 \)
\( 26(2) = 2d \Rightarrow 52 = 2d \Rightarrow d = 26 \)
\( 2b = 2(24) + 26 \Rightarrow 2b = 48 + 26 \Rightarrow 2b = 74 \Rightarrow b = 37 \)

Setara: 2C₁₂H₂₆ + 37O₂ → 24CO₂ + 26H₂O

ZnS + O₂ → ZnO + SO₂

\( aZnS + bO_2 \rightarrow cZnO + dSO_2 \)

Zn: \( a = c \)
S: \( a = d \)
O: \( 2b = c + 2d \)

Misal \( a = 2 \)
\( c = 2, d = 2 \)
\( 2b = 2 + 2(2) \Rightarrow 2b = 2 + 4 \Rightarrow 2b = 6 \Rightarrow b = 3 \)

Setara: 2ZnS + 3O₂ → 2ZnO + 2SO₂

PbS + O₂ → PbO + SO₂

\( aPbS + bO_2 \rightarrow cPbO + dSO_2 \)

Pb: \( a = c \)
S: \( a = d \)
O: \( 2b = c + 2d \)

Misal \( a = 2 \)
\( c = 2, d = 2 \)
\( 2b = 2 + 2(2) \Rightarrow 2b = 2 + 4 \Rightarrow 2b = 6 \Rightarrow b = 3 \)

Setara: 2PbS + 3O₂ → 2PbO + 2SO₂

CuS + O₂ → CuO + SO₂

\( aCuS + bO_2 \rightarrow cCuO + dSO_2 \)

Cu: \( a = c \)
S: \( a = d \)
O: \( 2b = c + 2d \)

Misal \( a = 2 \)
\( c = 2, d = 2 \)
\( 2b = 2 + 2(2) \Rightarrow 2b = 2 + 4 \Rightarrow 2b = 6 \Rightarrow b = 3 \)

Setara: 2CuS + 3O₂ → 2CuO + 2SO₂

Cu₂S + O₂ → Cu₂O + SO₂

\( aCu_2S + bO_2 \rightarrow cCu_2O + dSO_2 \)

Cu: \( 2a = 2c \)
S: \( a = d \)
O: \( 2b = c + 2d \)

Dari Cu: \( a = c \)
Misal \( a = 2 \)
\( c = 2, d = 2 \)
\( 2b = 2 + 2(2) \Rightarrow 2b = 2 + 4 \Rightarrow 2b = 6 \Rightarrow b = 3 \)

Setara: 2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂

Fe + H₂O → Fe₃O₄ + H₂

\( aFe + bH_2O \rightarrow cFe_3O_4 + dH_2 \)

Fe: \( a = 3c \)
H: \( 2b = 2d \)
O: \( b = 4c \)

Misal \( c = 1 \)
\( a = 3 \)
\( b = 4 \)
\( 2(4) = 2d \Rightarrow 8 = 2d \Rightarrow d = 4 \)

Setara: 3Fe + 4H₂O → Fe₃O₄ + 4H₂

SO₂ + H₂S → S + H₂O

\( aSO_2 + bH_2S \rightarrow cS + dH_2O \)

S: \( a + b = c \)
O: \( 2a = d \)
H: \( 2b = 2d \)

Dari H: \( b = d \)
Dari O: \( 2a = b \)
Misal \( a = 1 \)
\( b = 2 \)
\( d = 2 \)
\( c = 1 + 2 = 3 \)

Setara: SO₂ + 2H₂S → 3S + 2H₂O

PbS + H₂O₂ → PbSO₄ + H₂O

\( aPbS + bH_2O_2 \rightarrow cPbSO_4 + dH_2O \)

Pb: \( a = c \)
S: \( a = c \)
H: \( 2b = 2d \)
O: \( 2b = 4c + d \)

Misal \( a = 1 \)
\( c = 1 \)
Dari H: \( b = d \)
Dari O: \( 2b = 4 + b \Rightarrow b = 4 \)
Jadi \( b = 4, d = 4 \)

Setara: PbS + 4H₂O₂ → PbSO₄ + 4H₂O

Fe₂O₃ + CO → Fe₃O₄ + CO₂

\( aFe_2O_3 + bCO \rightarrow cFe_3O_4 + dCO_2 \)

Fe: \( 2a = 3c \)
O: \( 3a + b = 4c + 2d \)
C: \( b = d \)

Misal \( c = 2 \)
\( 2a = 3(2) \Rightarrow 2a = 6 \Rightarrow a = 3 \)
\( b = d \)
Dari O: \( 3(3) + b = 4(2) + 2b \Rightarrow 9 + b = 8 + 2b \Rightarrow 9 - 8 = 2b - b \Rightarrow b = 1 \)
\( d = 1 \)

Setara: 3Fe₂O₃ + CO → 2Fe₃O₄ + CO₂

Fe₃O₄ + Al → Al₂O₃ + Fe

\( aFe_3O_4 + bAl \rightarrow cAl_2O_3 + dFe \)

Fe: \( 3a = d \)
O: \( 4a = 3c \)
Al: \( b = 2c \)

Misal \( c = 4 \)
\( 4a = 3(4) \Rightarrow 4a = 12 \Rightarrow a = 3 \)
\( d = 3(3) = 9 \)
\( b = 2(4) = 8 \)

Setara: 3Fe₃O₄ + 8Al → 4Al₂O₃ + 9Fe

HNO₃ + N₂O₃ + H₂O → HNO₂

\( aHNO_3 + bN_2O_3 + cH_2O \rightarrow dHNO_2 \)

H: \( a + 2c = d \)
N: \( a + 2b = d \)
O: \( 3a + 3b + c = 2d \)

Dari H dan N: \( a + 2c = a + 2b \Rightarrow 2c = 2b \Rightarrow c = b \)
Misal \( b = 1, c = 1 \)
\( a + 2(1) = d \) dan \( a + 2(1) = d \)
Misal \( a = 2 \)
\( d = 2 + 2 = 4 \)

Setara: 2HNO₃ + N₂O₃ + H₂O → 4HNO₂

Fe + S + O₂ → Fe₂O₃ + SO₂

\( aFe + bS + cO_2 \rightarrow dFe_2O_3 + eSO_2 \)

Fe: \( a = 2d \)
S: \( b = e \)
O: \( 2c = 3d + 2e \)

Misal \( d = 2 \)
\( a = 4 \)
Misal \( b = 2 \)
\( e = 2 \)
\( 2c = 3(2) + 2(2) \Rightarrow 2c = 6 + 4 \Rightarrow 2c = 10 \Rightarrow c = 5 \)

Setara: 4Fe + 2S + 5O₂ → 2Fe₂O₃ + 2SO₂

Cu + S + O₂ → Cu₂O + SO₂

\( aCu + bS + cO_2 \rightarrow dCu_2O + eSO_2 \)

Cu: \( a = 2d \)
S: \( b = e \)
O: \( 2c = d + 2e \)

Misal \( d = 2 \)
\( a = 4 \)
Misal \( b = 2 \)
\( e = 2 \)
\( 2c = 2 + 2(2) \Rightarrow 2c = 2 + 4 \Rightarrow 2c = 6 \Rightarrow c = 3 \)

Setara: 4Cu + 2S + 3O₂ → 2Cu₂O + 2SO₂

FeS + O₂ → Fe₃O₄ + SO₂

\( aFeS + bO_2 \rightarrow cFe_3O_4 + dSO_2 \)

Fe: \( a = 3c \)
S: \( a = d \)
O: \( 2b = 4c + 2d \)

Misal \( c = 1 \)
\( a = 3, d = 3 \)
\( 2b = 4(1) + 2(3) \Rightarrow 2b = 4 + 6 \Rightarrow 2b = 10 \Rightarrow b = 5 \)

Setara: 3FeS + 5O₂ → Fe₃O₄ + 3SO₂

FeS + O₂ → Fe₂O₃ + SO₂

\( aFeS + bO_2 \rightarrow cFe_2O_3 + dSO_2 \)

Fe: \( a = 2c \)
S: \( a = d \)
O: \( 2b = 3c + 2d \)

Misal \( c = 2 \)
\( a = 4, d = 4 \)
\( 2b = 3(2) + 2(4) \Rightarrow 2b = 6 + 8 \Rightarrow 2b = 14 \Rightarrow b = 7 \)

Setara: 4FeS + 7O₂ → 2Fe₂O₃ + 4SO₂

FeS₂ + O₂ → Fe₂O₃ + SO₂

\( aFeS_2 + bO_2 \rightarrow cFe_2O_3 + dSO_2 \)

Fe: \( a = 2c \)
S: \( 2a = d \)
O: \( 2b = 3c + 2d \)

Misal \( c = 2 \)
\( a = 4 \)
\( d = 2(4) = 8 \)
\( 2b = 3(2) + 2(8) \Rightarrow 2b = 6 + 16 \Rightarrow 2b = 22 \Rightarrow b = 11 \)

Setara: 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂

Fe₂S₃ + O₂ → Fe₂O₃ + SO₂

\( aFe_2S_3 + bO_2 \rightarrow cFe_2O_3 + dSO_2 \)

Fe: \( 2a = 2c \)
S: \( 3a = d \)
O: \( 2b = 3c + 2d \)

Dari Fe: \( a = c \)
Misal \( a = 2 \)
\( c = 2 \)
\( d = 3(2) = 6 \)
\( 2b = 3(2) + 2(6) \Rightarrow 2b = 6 + 12 \Rightarrow 2b = 18 \Rightarrow b = 9 \)

Setara: 2Fe₂S₃ + 9O₂ → 2Fe₂O₃ + 6SO₂

ZnS + O₂ → ZnO + SO₂

\( aZnS + bO_2 \rightarrow cZnO + dSO_2 \)

Zn: \( a = c \)
S: \( a = d \)
O: \( 2b = c + 2d \)

Misal \( a = 2 \)
\( c = 2, d = 2 \)
\( 2b = 2 + 2(2) \Rightarrow 2b = 2 + 4 \Rightarrow 2b = 6 \Rightarrow b = 3 \)

Setara: 2ZnS + 3O₂ → 2ZnO + 2SO₂

Persamaan reaksi 5 spesi atau lebih

C₅H₅N + O₂ → CO₂ + H₂O + N₂

\( aC_5H_5N + bO_2 \rightarrow cCO_2 + dH_2O + eN_2 \)

C: \( 5a = c \)
H: \( 5a = 2d \)
N: \( a = 2e \)
O: \( 2b = 2c + d \)

Misal \( a = 4 \)
\( c = 20 \)
\( 5(4) = 2d \Rightarrow 20 = 2d \Rightarrow d = 10 \)
\( 4 = 2e \Rightarrow e = 2 \)
\( 2b = 2(20) + 10 \Rightarrow 2b = 40 + 10 \Rightarrow 2b = 50 \Rightarrow b = 25 \)

Setara: 4C₅H₅N + 25O₂ → 20CO₂ + 10H₂O + 2N₂

Cl₂ + HClO + H₂O → HCl + HClO₂

\( aCl_2 + bHClO + cH_2O \rightarrow dHCl + eHClO_2 \)

Cl: \( 2a + b = d + e \)
H: \( b + 2c = d + e \)
O: \( b + c = 2e \)

Dari Cl dan H: \( 2a + b = b + 2c \Rightarrow 2a = 2c \Rightarrow a = c \)
Misal \( a = 1, c = 1 \)
\( 2 + b = d + e \)
\( b + 2 = d + e \)
\( b + 1 = 2e \)

Misal \( b = 1 \)
\( 1 + 1 = 2e \Rightarrow 2 = 2e \Rightarrow e = 1 \)
\( 2 + 1 = d + 1 \Rightarrow 3 = d + 1 \Rightarrow d = 2 \)

Setara: Cl₂ + HClO + H₂O → 2HCl + HClO₂

CH₃NH₂ + O₂ → CO₂ + H₂O + N₂

\( aCH_3NH_2 + bO_2 \rightarrow cCO_2 + dH_2O + eN_2 \)

C: \( a = c \)
H: \( 5a = 2d \)
N: \( a = 2e \)
O: \( 2b = 2c + d \)

Misal \( a = 4 \)
\( c = 4 \)
\( 5(4) = 2d \Rightarrow 20 = 2d \Rightarrow d = 10 \)
\( 4 = 2e \Rightarrow e = 2 \)
\( 2b = 2(4) + 10 \Rightarrow 2b = 8 + 10 \Rightarrow 2b = 18 \Rightarrow b = 9 \)

Setara: 4CH₃NH₂ + 9O₂ → 4CO₂ + 10H₂O + 2N₂

C₂H₅NH₂ + O₂ → CO₂ + H₂O + N₂

\( aC_2H_5NH_2 + bO_2 \rightarrow cCO_2 + dH_2O + eN_2 \)

C: \( 2a = c \)
H: \( 7a = 2d \)
N: \( a = 2e \)
O: \( 2b = 2c + d \)

Misal \( a = 4 \)
\( c = 8 \)
\( 7(4) = 2d \Rightarrow 28 = 2d \Rightarrow d = 14 \)
\( 4 = 2e \Rightarrow e = 2 \)
\( 2b = 2(8) + 14 \Rightarrow 2b = 16 + 14 \Rightarrow 2b = 30 \Rightarrow b = 15 \)

Setara: 4C₂H₅NH₂ + 15O₂ → 8CO₂ + 14H₂O + 2N₂

C₆H₅NH₂ + O₂ → CO₂ + H₂O + N₂

\( aC_6H_5NH_2 + bO_2 \rightarrow cCO_2 + dH_2O + eN_2 \)

C: \( 6a = c \)
H: \( 7a = 2d \)
N: \( a = 2e \)
O: \( 2b = 2c + d \)

Misal \( a = 4 \)
\( c = 24 \)
\( 7(4) = 2d \Rightarrow 28 = 2d \Rightarrow d = 14 \)
\( 4 = 2e \Rightarrow e = 2 \)
\( 2b = 2(24) + 14 \Rightarrow 2b = 48 + 14 \Rightarrow 2b = 62 \Rightarrow b = 31 \)

Setara: 4C₆H₅NH₂ + 31O₂ → 24CO₂ + 14H₂O + 2N₂

C₃H₅N + O₂ → CO₂ + H₂O + N₂

\( aC_3H_5N + bO_2 \rightarrow cCO_2 + dH_2O + eN_2 \)

C: \( 3a = c \)
H: \( 5a = 2d \)
N: \( a = 2e \)
O: \( 2b = 2c + d \)

Misal \( a = 4 \)
\( c = 12 \)
\( 5(4) = 2d \Rightarrow 20 = 2d \Rightarrow d = 10 \)
\( 4 = 2e \Rightarrow e = 2 \)
\( 2b = 2(12) + 10 \Rightarrow 2b = 24 + 10 \Rightarrow 2b = 34 \Rightarrow b = 17 \)

Setara: 4C₃H₅N + 17O₂ → 12CO₂ + 10H₂O + 2N₂

C₇H₅N + O₂ → CO₂ + H₂O + N₂

\( aC_7H_5N + bO_2 \rightarrow cCO_2 + dH_2O + eN_2 \)

C: \( 7a = c \)
H: \( 5a = 2d \)
N: \( a = 2e \)
O: \( 2b = 2c + d \)

Misal \( a = 4 \)
\( c = 28 \)
\( 5(4) = 2d \Rightarrow 20 = 2d \Rightarrow d = 10 \)
\( 4 = 2e \Rightarrow e = 2 \)
\( 2b = 2(28) + 10 \Rightarrow 2b = 56 + 10 \Rightarrow 2b = 66 \Rightarrow b = 33 \)

Setara: 4C₇H₅N + 33O₂ → 28CO₂ + 10H₂O + 2N₂

C₅H₁₁NH₂ + O₂ → CO₂ + H₂O + N₂

\( aC_5H_{11}NH_2 + bO_2 \rightarrow cCO_2 + dH_2O + eN_2 \)

C: \( 5a = c \)
H: \( 13a = 2d \)
N: \( a = 2e \)
O: \( 2b = 2c + d \)

Misal \( a = 4 \)
\( c = 20 \)
\( 13(4) = 2d \Rightarrow 52 = 2d \Rightarrow d = 26 \)
\( 4 = 2e \Rightarrow e = 2 \)
\( 2b = 2(20) + 26 \Rightarrow 2b = 40 + 26 \Rightarrow 2b = 66 \Rightarrow b = 33 \)

Setara: 4C₅H₁₁NH₂ + 33O₂ → 20CO₂ + 26H₂O + 2N₂

C₄H₈N₂ + O₂ → CO₂ + H₂O + N₂

\( aC_4H_8N_2 + bO_2 \rightarrow cCO_2 + dH_2O + eN_2 \)

C: \( 4a = c \)
H: \( 8a = 2d \)
N: \( 2a = 2e \)
O: \( 2b = 2c + d \)

Misal \( a = 1 \)
\( c = 4 \)
\( 8 = 2d \Rightarrow d = 4 \)
\( 2 = 2e \Rightarrow e = 1 \)
\( 2b = 2(4) + 4 \Rightarrow 2b = 8 + 4 \Rightarrow 2b = 12 \Rightarrow b = 6 \)

Setara: C₄H₈N₂ + 6O₂ → 4CO₂ + 4H₂O + N₂

Zn + HNO₃ → Zn(NO₃)₂ + NO + H₂O

\( aZn + bHNO_3 \rightarrow cZn(NO_3)_2 + dNO + eH_2O \)

Zn: \( a = c \)
H: \( b = 2e \)
N: \( b = 2c + d \)
O: \( 3b = 6c + d + e \)

Misal \( c = 3 \)
\( a = 3 \)
\( b = 2(3) + d = 6 + d \)
\( b = 2e \)

Misal \( d = 2 \)
\( b = 6 + 2 = 8 \)
\( 8 = 2e \Rightarrow e = 4 \)

Setara: 3Zn + 8HNO₃ → 3Zn(NO₃)₂ + 2NO + 4H₂O

FeS + O₂ + H₂O → Fe₂O₃ + H₂SO₄

\( aFeS + bO_2 + cH_2O \rightarrow dFe_2O_3 + eH_2SO_4 \)

Fe: \( a = 2d \)
S: \( a = e \)
O: \( 2b + c = 3d + 4e \)
H: \( 2c = 2e \)

Dari H: \( c = e \)
Misal \( a = 4 \)
\( d = 2, e = 4, c = 4 \)
\( 2b + 4 = 3(2) + 4(4) \Rightarrow 2b + 4 = 6 + 16 \Rightarrow 2b + 4 = 22 \Rightarrow 2b = 18 \Rightarrow b = 9 \)

Setara: 4FeS + 9O₂ + 4H₂O → 2Fe₂O₃ + 4H₂SO₄

Fe₂S₃ + O₂ + H₂O → Fe₂O₃ + H₂SO₄

\( aFe_2S_3 + bO_2 + cH_2O \rightarrow dFe_2O_3 + eH_2SO_4 \)

Fe: \( 2a = 2d \)
S: \( 3a = e \)
O: \( 2b + c = 3d + 4e \)
H: \( 2c = 2e \)

Dari Fe: \( a = d \)
Misal \( a = 1 \)
\( d = 1, e = 3, c = 3 \)
\( 2b + 3 = 3(1) + 4(3) \Rightarrow 2b + 3 = 3 + 12 \Rightarrow 2b + 3 = 15 \Rightarrow 2b = 12 \Rightarrow b = 6 \)

Setara: Fe₂S₃ + 6O₂ + 3H₂O → Fe₂O₃ + 3H₂SO₄

FeS₂ + O₂ + H₂O → FeSO₄ + H₂SO₄

\( aFeS_2 + bO_2 + cH_2O \rightarrow dFeSO_4 + eH_2SO_4 \)

Fe: \( a = d \)
S: \( 2a = d + e \)
O: \( 2b + c = 4d + 4e \)
H: \( 2c = 2e \)

Dari H: \( c = e \)
Dari S: \( 2a = a + e \Rightarrow a = e \)
Misal \( a = 2 \)
\( d = 2, e = 2, c = 2 \)
\( 2b + 2 = 4(2) + 4(2) \Rightarrow 2b + 2 = 8 + 8 \Rightarrow 2b + 2 = 16 \Rightarrow 2b = 14 \Rightarrow b = 7 \)

Setara: 2FeS₂ + 7O₂ + 2H₂O → 2FeSO₄ + 2H₂SO₄

PbS + HNO₃ → Pb(NO₃)₂ + S + NO + H₂O

\( aPbS + bHNO_3 \rightarrow cPb(NO_3)_2 + dS + eNO + fH_2O \)

Pb: \( a = c \)
S: \( a = d \)
H: \( b = 2f \)
N: \( b = 2c + e \)
O: \( 3b = 6c + e + f \)

Misal \( c = 3 \)
\( a = 3, d = 3 \)
\( b = 2(3) + e = 6 + e \)
\( b = 2f \)

Misal \( e = 2 \)
\( b = 6 + 2 = 8 \)
\( 8 = 2f \Rightarrow f = 4 \)

Setara: 3PbS + 8HNO₃ → 3Pb(NO₃)₂ + 3S + 2NO + 4H₂O

FeS + HNO₃ → Fe(NO₃)₃ + H₂SO₄ + NO + H₂O

\( aFeS + bHNO_3 \rightarrow cFe(NO_3)_3 + dH_2SO_4 + eNO + fH_2O \)

Fe: \( a = c \)
S: \( a = d \)
H: \( b = 2d + 2f \)
N: \( b = 3c + e \)
O: \( 3b = 9c + 4d + e + f \)

Misal \( a = 1 \)
\( c = 1, d = 1 \)
\( b = 3(1) + e = 3 + e \)
\( b = 2(1) + 2f = 2 + 2f \)

Dari dua persamaan: \( 3 + e = 2 + 2f \Rightarrow e = 2f - 1 \)
Misal \( f = 2 \)
\( e = 2(2) - 1 = 3 \)
\( b = 3 + 3 = 6 \)

Setara: FeS + 6HNO₃ → Fe(NO₃)₃ + H₂SO₄ + 3NO + 2H₂O

CuS + HNO₃ → Cu(NO₃)₂ + H₂SO₄ + NO + H₂O

\( aCuS + bHNO_3 \rightarrow cCu(NO_3)_2 + dH_2SO_4 + eNO + fH_2O \)

Cu: \( a = c \)
S: \( a = d \)
H: \( b = 2d + 2f \)
N: \( b = 2c + e \)
O: \( 3b = 6c + 4d + e + f \)

Misal \( a = 3 \)
\( c = 3, d = 3 \)
\( b = 2(3) + e = 6 + e \)
\( b = 2(3) + 2f = 6 + 2f \)

Dari dua persamaan: \( 6 + e = 6 + 2f \Rightarrow e = 2f \)
Misal \( f = 2 \)
\( e = 4 \)
\( b = 6 + 4 = 10 \)

Cek O: \( 3(10) = 6(3) + 4(3) + 4 + 2 \Rightarrow 30 = 18 + 12 + 4 + 2 \Rightarrow 30 = 36 \) (salah)
Perbaiki: Misal \( f = 4 \)
\( e = 8 \)
\( b = 6 + 8 = 14 \)

Cek O: \( 3(14) = 6(3) + 4(3) + 8 + 4 \Rightarrow 42 = 18 + 12 + 8 + 4 \Rightarrow 42 = 42 \) ✓
Tapi perlu perhatikan: koefisien harus bilangan bulat terkecil
Bagi semua dengan 2: \( a = 3, c = 3, d = 3, e = 2, f = 4, b = 8 \)

Setara: 3CuS + 8HNO₃ → 3Cu(NO₃)₂ + 3H₂SO₄ + 2NO + 4H₂O

CuS + HNO₃ → CuSO₄ + NO₂ + H₂O

\( aCuS + bHNO_3 \rightarrow cCuSO_4 + dNO_2 + eH_2O \)

Cu: \( a = c \)
S: \( a = c \)
H: \( b = 2e \)
N: \( b = d \)
O: \( 3b = 4c + 2d + e \)

Misal \( a = 1 \)
\( c = 1 \)
\( b = d = 2e \)
\( 3b = 4(1) + 2b + \dfrac{b}{2} \Rightarrow 3b = 4 + 2b + 0.5b \Rightarrow 3b - 2.5b = 4 \Rightarrow 0.5b = 4 \Rightarrow b = 8 \)

\( d = 8, e = 4 \)

Setara: CuS + 8HNO₃ → CuSO₄ + 8NO₂ + 4H₂O

FeS + HNO₃ → Fe(NO₃)₃ + H₂SO₄ + NO

\( aFeS + bHNO_3 \rightarrow cFe(NO_3)_3 + dH_2SO_4 + eNO \)

Fe: \( a = c \)
S: \( a = d \)
H: \( b = 2d \)
N: \( b = 3c + e \)
O: \( 3b = 9c + 4d + e \)

Dari Fe dan S: \( a = c = d \)
Misal \( a = 1 \)
\( c = 1, d = 1 \)

Dari H: \( b = 2(1) = 2 \)
Dari N: \( 2 = 3(1) + e \)
\( 2 = 3 + e \)
\( e = -1 \) (tidak mungkin, harus positif)

Artinya perlu H₂O di kanan:
\( aFeS + bHNO_3 \rightarrow cFe(NO_3)_3 + dH_2SO_4 + eNO + fH_2O \)

Fe: \( a = c \)
S: \( a = d \)
H: \( b = 2d + 2f \)
N: \( b = 3c + e \)
O: \( 3b = 9c + 4d + e + f \)

Misal \( a = 1 \)
\( c = 1, d = 1 \)

Dari H: \( b = 2 + 2f \)
Dari N: \( b = 3 + e \)
Jadi \( 2 + 2f = 3 + e \)
\( e = 2f - 1 \)

Dari O: \( 3b = 9 + 4 + e + f = 13 + e + f \)
Substitusi \( b = 2 + 2f \):
\( 3(2 + 2f) = 13 + (2f - 1) + f \)
\( 6 + 6f = 13 + 2f - 1 + f \)
\( 6 + 6f = 12 + 3f \)
\( 3f = 6 \)
\( f = 2 \)

\( e = 2(2) - 1 = 4 - 1 = 3 \)
\( b = 2 + 2(2) = 2 + 4 = 6 \)

Cek semua:
Fe: 1 = 1 ✓
S: 1 = 1 ✓
H: 6 = 2×1 + 2×2 = 2 + 4 = 6 ✓
N: 6 = 3×1 + 3 = 3 + 3 = 6 ✓
O: 3×6 = 18, kanan: 9×1 + 4×1 + 3 + 2 = 9 + 4 + 3 + 2 = 18 ✓

Setara: FeS + 6HNO₃ → Fe(NO₃)₃ + H₂SO₄ + 3NO + 2H₂O

FeS₂ + HNO₃ → Fe(NO₃)₃ + H₂SO₄ + NO + H₂O

\( aFeS_2 + bHNO_3 \rightarrow cFe(NO_3)_3 + dH_2SO_4 + eNO + fH_2O \)

Fe: \( a = c \)
S: \( 2a = d \)
H: \( b = 2d + 2f \)
N: \( b = 3c + e \)
O: \( 3b = 9c + 4d + e + f \)

Misal \( a = 1 \)
\( c = 1, d = 2 \)
\( b = 3(1) + e = 3 + e \)
\( b = 2(2) + 2f = 4 + 2f \)

Dari dua persamaan: \( 3 + e = 4 + 2f \Rightarrow e = 1 + 2f \)
Misal \( f = 2 \)
\( e = 1 + 4 = 5 \)
\( b = 3 + 5 = 8 \)

Setara: FeS₂ + 8HNO₃ → Fe(NO₃)₃ + 2H₂SO₄ + 5NO + 2H₂O

Pb₃O₄ + HNO₃ → Pb(NO₃)₂ + O₂ + H₂O

\( aPb_3O_4 + bHNO_3 \rightarrow cPb(NO_3)_2 + dO_2 + eH_2O \)

Pb: \( 3a = c \)
O: \( 4a + 3b = 6c + 2d + e \)
H: \( b = 2e \)
N: \( b = 2c \)

Dari N: \( b = 2c \)
Dari H: \( 2c = 2e \Rightarrow c = e \)
Misal \( a = 2 \)
\( c = 6, e = 6, b = 12 \)
Dari O: \( 4(2) + 3(12) = 6(6) + 2d + 6 \Rightarrow 8 + 36 = 36 + 2d + 6 \Rightarrow 44 = 42 + 2d \Rightarrow 2d = 2 \Rightarrow d = 1 \)

Setara: 2Pb₃O₄ + 12HNO₃ → 6Pb(NO₃)₂ + O₂ + 6H₂O

Zn + HNO₃ + O₂ → Zn(NO₃)₂ + NO + H₂O

\( aZn + bHNO_3 + cO_2 \rightarrow dZn(NO_3)_2 + eNO + fH_2O \)

Zn: \( a = d \)
H: \( b = 2f \)
N: \( b = 2d + e \)
O: \( 3b + 2c = 6d + e + f \)

Misal \( d = 4 \)
\( a = 4 \)
\( b = 2(4) + e = 8 + e \)
\( b = 2f \)

Misal \( e = 1 \) (untuk NO, tapi biasanya N₂O)
Perhatikan: reaksi sebenarnya menghasilkan N₂O
Untuk Zn + HNO₃ → Zn(NO₃)₂ + N₂O + H₂O
\( 4Zn + 10HNO_3 \rightarrow 4Zn(NO_3)_2 + N_2O + 5H_2O \)

Setara: 4Zn + 10HNO₃ → 4Zn(NO₃)₂ + N₂O + 5H₂O

Cu + HNO₃ + O₂ → Cu(NO₃)₂ + NO + H₂O

\( aCu + bHNO_3 + cO_2 \rightarrow dCu(NO_3)_2 + eNO + fH_2O \)

Cu: \( a = d \)
H: \( b = 2f \)
N: \( b = 2d + e \)
O: \( 3b + 2c = 6d + e + f \)

Dari H: \( b = 2f \)
Dari N: \( 2f = 2d + e \)
Dari O: \( 3(2f) + 2c = 6d + e + f \Rightarrow 6f + 2c = 6d + e + f \Rightarrow 5f + 2c = 6d + e \)

Substitusi \( e = 2f - 2d \) ke persamaan O:
\( 5f + 2c = 6d + (2f - 2d) \)
\( 5f + 2c = 4d + 2f \)
\( 3f + 2c = 4d \)

Coba dengan bilangan bulat:
Misal \( d = 7 \) (dari setara yang diberikan)
\( a = 7 \)
\( 3f + 2c = 4(7) = 28 \)

Coba \( e = 2 \) (dari setara yang diberikan)
Dari N: \( b = 2(7) + 2 = 14 + 2 = 16 \)
Dari H: \( 16 = 2f \Rightarrow f = 8 \)
Dari O: \( 3(16) + 2c = 6(7) + 2 + 8 \)
\( 48 + 2c = 42 + 2 + 8 = 52 \)
\( 2c = 4 \Rightarrow c = 2 \)

Cek semua:
Cu: 7 = 7 ✓
H: 16 = 2×8 = 16 ✓
N: 16 = 2×7 + 2 = 14 + 2 = 16 ✓
O: 3×16 + 2×2 = 48 + 4 = 52, kanan: 6×7 + 2 + 8 = 42 + 2 + 8 = 52 ✓

Setara: 7Cu + 16HNO₃ + 2O₂ → 7Cu(NO₃)₂ + 2NO + 8H₂O

Cu + H₂SO₄ + O₂ → CuSO₄ + SO₂ + H₂O

\( aCu + bH_2SO_4 + cO_2 \rightarrow dCuSO_4 + eSO_2 + fH_2O \)

Cu: \( a = d \)
S: \( b = d + e \)
O: \( 4b + 2c = 4d + 2e + f \)
H: \( 2b = 2f \)

Dari H: \( b = f \)
Misal \( a = 2 \)
\( d = 2 \)
\( b = 2 + e \)
\( b = f \)

Misal \( e = 0 \) (tidak ada SO₂)
\( b = 2, f = 2 \)
Dari O: \( 4(2) + 2c = 4(2) + 0 + 2 \Rightarrow 8 + 2c = 8 + 2 \Rightarrow 2c = 2 \Rightarrow c = 1 \)

Setara: 2Cu + 2H₂SO₄ + O₂ → 2CuSO₄ + 2H₂O

Ag + H₂SO₄ + O₂ → Ag₂SO₄ + SO₂ + H₂O

\( aAg + bH_2SO_4 + cO_2 \rightarrow dAg_2SO_4 + eSO_2 + fH_2O \)

Ag: \( a = 2d \)
S: \( b = d + e \)
O: \( 4b + 2c = 4d + 2e + f \)
H: \( 2b = 2f \)

Dari H: \( b = f \)
Misal \( d = 2 \)
\( a = 4 \)
\( b = 2 + e \)
\( b = f \)

Misal \( e = 0 \)
\( b = 2, f = 2 \)
Dari O: \( 4(2) + 2c = 4(2) + 0 + 2 \Rightarrow 8 + 2c = 8 + 2 \Rightarrow 2c = 2 \Rightarrow c = 1 \)

Setara: 4Ag + 2H₂SO₄ + O₂ → 2Ag₂SO₄ + 2H₂O

Pb + H₂SO₄ → PbSO₄ + SO₂ + H₂O

\( aPb + bH_2SO_4 \rightarrow cPbSO_4 + dSO_2 + eH_2O \)

Pb: \( a = c \)
S: \( b = c + d \)
O: \( 4b = 4c + 2d + e \)
H: \( 2b = 2e \)

Dari H: \( b = e \)
Misal \( a = 1 \)
\( c = 1 \)
\( b = 1 + d \)
\( b = e \)

Misal \( d = 1 \)
\( b = 2, e = 2 \)
Cek O: \( 4(2) = 4(1) + 2(1) + 2 \Rightarrow 8 = 4 + 2 + 2 \Rightarrow 8 = 8 \) ✓

Setara: Pb + 2H₂SO₄ → PbSO₄ + SO₂ + 2H₂O

Pb + PbO₂ + H₂SO₄ → PbSO₄ + H₂O

\( aPb + bPbO_2 + cH_2SO_4 \rightarrow dPbSO_4 + eH_2O \)

Pb: \( a + b = d \)
O: \( 2b + 4c = 4d + e \)
S: \( c = d \)
H: \( 2c = 2e \)

Dari H: \( c = e \)
Dari S: \( c = d \)
Misal \( c = 2 \)
\( d = 2, e = 2 \)
\( a + b = 2 \)
\( 2b + 4(2) = 4(2) + 2 \Rightarrow 2b + 8 = 8 + 2 \Rightarrow 2b = 2 \Rightarrow b = 1 \)
\( a = 1 \)

Setara: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O

MnO₂ + Mn + H₂SO₄ → MnSO₄ + H₂O

\( aMnO_2 + bMn + cH_2SO_4 \rightarrow dMnSO_4 + eH_2O \)

Mn: \( a + b = d \)
O: \( 2a + 4c = 4d + e \)
S: \( c = d \)
H: \( 2c = 2e \)

Dari H: \( c = e \)
Dari S: \( c = d \)
Misal \( c = 2 \)
\( d = 2, e = 2 \)
\( a + b = 2 \)
\( 2a + 4(2) = 4(2) + 2 \Rightarrow 2a + 8 = 8 + 2 \Rightarrow 2a = 2 \Rightarrow a = 1 \)
\( b = 1 \)

Setara: MnO₂ + Mn + 2H₂SO₄ → 2MnSO₄ + 2H₂O

Pb + H₂SO₄ + O₂ → PbSO₄ + SO₂ + H₂O

\( aPb + bH_2SO_4 + cO_2 \rightarrow dPbSO_4 + eSO_2 + fH_2O \)

Pb: \( a = d \)
S: \( b = d + e \)
O: \( 4b + 2c = 4d + 2e + f \)
H: \( 2b = 2f \)

Dari H: \( b = f \)
Dari S: \( b = a + e \) (karena \( d = a \))
Jadi \( f = a + e \)

Dari O: \( 4b + 2c = 4a + 2e + f \)
Substitusi \( b = f \):
\( 4f + 2c = 4a + 2e + f \)
\( 3f + 2c = 4a + 2e \)

Substitusi \( f = a + e \):
\( 3(a + e) + 2c = 4a + 2e \)
\( 3a + 3e + 2c = 4a + 2e \)
\( 3e + 2c = a + 2e \)
\( e + 2c = a \)

Coba \( c = 1 \)
\( e + 2 = a \)

Coba \( e = 1 \)
\( 1 + 2 = a \Rightarrow a = 3 \)
\( d = 3 \)
\( b = 3 + 1 = 4 \)
\( f = 4 \)

Cek semua:
Pb: 3 = 3 ✓
S: 4 = 3 + 1 = 4 ✓
O: 4×4 + 2×1 = 16 + 2 = 18, kanan: 4×3 + 2×1 + 4 = 12 + 2 + 4 = 18 ✓
H: 2×4 = 8, kanan: 2×4 = 8 ✓

Setara: 3Pb + 4H₂SO₄ + O₂ → 3PbSO₄ + SO₂ + 4H₂O

FeS + H₂SO₄ + O₂ → Fe₂(SO₄)₃ + H₂O

\( aFeS + bH_2SO_4 + cO_2 \rightarrow dFe_2(SO_4)_3 + eH_2O \)

Fe: \( a = 2d \)
S: \( a + b = 3d \)
O: \( 4b + 2c = 12d + e \)
H: \( 2b = 2e \)

Dari H: \( b = e \)
Misal \( d = 2 \)
\( a = 4 \)
\( 4 + b = 3(2) \Rightarrow 4 + b = 6 \Rightarrow b = 2 \)
\( e = 2 \)
\( 4(2) + 2c = 12(2) + 2 \Rightarrow 8 + 2c = 24 + 2 \Rightarrow 2c = 18 \Rightarrow c = 9 \)

Setara: 4FeS + 2H₂SO₄ + 9O₂ → 2Fe₂(SO₄)₃ + 2H₂O

Fe₂O₃ + H₂S + H₂ → FeS + H₂O

\( aFe_2O_3 + bH_2S + cH_2 \rightarrow dFeS + eH_2O \)

Fe: \( 2a = d \)
O: \( 3a = e \)
H: \( 2b + 2c = 2e \)
S: \( b = d \)

Dari Fe dan S: \( 2a = d = b \)
Misal \( a = 1 \)
\( d = 2, b = 2 \)
\( e = 3(1) = 3 \)

Dari H: \( 2(2) + 2c = 2(3) \)
\( 4 + 2c = 6 \)
\( 2c = 2 \)
\( c = 1 \)

Cek semua:
Fe: 2×1 = 2 ✓
O: 3×1 = 3 ✓
H: 2×2 + 2×1 = 4 + 2 = 6, kanan: 2×3 = 6 ✓
S: 2 = 2 ✓

Setara: Fe₂O₃ + 2H₂S + H₂ → 2FeS + 3H₂O

Fe₂O₃ + H₂S + O₂ → FeS + SO₂ + H₂O

\( aFe_2O_3 + bH_2S + cO_2 \rightarrow dFeS + eSO_2 + fH_2O \)

Fe: \( 2a = d \)
O: \( 3a + 2c = 2e + f \)
H: \( 2b = 2f \)
S: \( b = d + e \)

Dari H: \( b = f \)
Misal \( a = 1 \)
\( d = 2 \)
Dari S: \( b = 2 + e \)
Dari O: \( 3(1) + 2c = 2e + b \)

Substitusi \( b = 2 + e \):
\( 3 + 2c = 2e + (2 + e) \)
\( 3 + 2c = 2e + 2 + e \)
\( 3 + 2c = 3e + 2 \)
\( 2c = 3e - 1 \)

Coba \( e = 1 \)
\( 2c = 3(1) - 1 = 2 \)
\( c = 1 \)
\( b = 2 + 1 = 3 \)
\( f = 3 \)

Cek semua:
Fe: 2×1 = 2 ✓
O: 3×1 + 2×1 = 3 + 2 = 5, kanan: 2×1 + 3 = 2 + 3 = 5 ✓
H: 2×3 = 6, kanan: 2×3 = 6 ✓
S: 3 = 2 + 1 = 3 ✓

Setara: Fe₂O₃ + 3H₂S + O₂ → 2FeS + SO₂ + 3H₂O

Fe₃O₄ + H₂S → FeS + S + H₂O

\( aFe_3O_4 + bH_2S \rightarrow cFeS + dS + eH_2O \)

Fe: \( 3a = c \)
O: \( 4a = e \)
H: \( 2b = 2e \)
S: \( b = c + d \)

Dari H: \( b = e \)
Dari O: \( e = 4a \)
Jadi \( b = 4a \)
Dari Fe: \( c = 3a \)

Dari S: \( 4a = 3a + d \)
\( d = a \)

Misal \( a = 1 \)
\( c = 3, d = 1, e = 4, b = 4 \)

Cek semua:
Fe: 3×1 = 3 ✓
O: 4×1 = 4 ✓
H: 2×4 = 8, kanan: 2×4 = 8 ✓
S: 4 = 3 + 1 = 4 ✓

Setara: Fe₃O₄ + 4H₂S → 3FeS + S + 4H₂O

Fe₃O₄ + H₂S + O₂ → FeSO₄ + S + H₂O

\( aFe_3O_4 + bH_2S + cO_2 \rightarrow dFeSO_4 + eS + fH_2O \)

Fe: \( 3a = d \)
O: \( 4a + 2c = 4d + f \)
H: \( 2b = 2f \)
S: \( b = d + e \)

Dari H: \( b = f \)
Misal \( a = 1 \)
\( d = 3 \)

Dari S: \( b = 3 + e \)
Dari O: \( 4(1) + 2c = 4(3) + b \)
\( 4 + 2c = 12 + b \)
\( 2c = 8 + b \)

Substitusi \( b = 3 + e \):
\( 2c = 8 + (3 + e) \)
\( 2c = 11 + e \)

Coba \( e = 1 \)
\( 2c = 11 + 1 = 12 \)
\( c = 6 \)
\( b = 3 + 1 = 4 \)
\( f = 4 \)

Cek semua:
Fe: 3×1 = 3 ✓
O: 4×1 + 2×6 = 4 + 12 = 16, kanan: 4×3 + 4 = 12 + 4 = 16 ✓
H: 2×4 = 8, kanan: 2×4 = 8 ✓
S: 4 = 3 + 1 = 4 ✓

Setara: Fe₃O₄ + 4H₂S + 6O₂ → 3FeSO₄ + S + 4H₂O

Zn + HNO₃ + H₂SO₄ → ZnSO₄ + NO₂ + H₂O

\( aZn + bHNO_3 + cH_2SO_4 \rightarrow dZnSO_4 + eNO_2 + fH_2O \)

Zn: \( a = d \)
S: \( c = d \)
H: \( b + 2c = 2f \)
N: \( b = e \)
O: \( 3b + 4c = 4d + 2e + f \)

Misal \( a = 1 \)
\( d = 1, c = 1 \)
\( b = e \)
\( b + 2(1) = 2f \Rightarrow b + 2 = 2f \)
\( 3b + 4(1) = 4(1) + 2b + f \Rightarrow 3b + 4 = 4 + 2b + f \Rightarrow b = f \)

\( b + 2 = 2b \Rightarrow 2 = b \)
\( e = 2, f = 2 \)

Setara: Zn + 2HNO₃ + H₂SO₄ → ZnSO₄ + 2NO₂ + 2H₂O

CuS + HNO₃ → Cu(NO₃)₂ + NO₂ + S₈ + H₂O

\( aCuS + bHNO_3 \rightarrow cCu(NO_3)_2 + dNO_2 + eS_8 + fH_2O \)

Cu: \( a = c \)
S: \( a = 8e \)
H: \( b = 2f \)
N: \( b = 2c + d \)
O: \( 3b = 6c + 2d + f \)

Dari S: \( a = 8e \)
Misal \( e = 1 \)
\( a = 8, c = 8 \)

Dari N: \( b = 2(8) + d = 16 + d \)
Dari H: \( b = 2f \)

Dari O: \( 3b = 6(8) + 2d + f = 48 + 2d + f \)

Substitusi \( b = 16 + d \) dan \( f = b/2 = (16 + d)/2 \):
\( 3(16 + d) = 48 + 2d + \dfrac{16 + d}{2} \)
\( 48 + 3d = 48 + 2d + 8 + 0.5d \)
\( 48 + 3d = 56 + 2.5d \)
\( 0.5d = 8 \)
\( d = 16 \)

\( b = 16 + 16 = 32 \)
\( f = 32/2 = 16 \)

Cek semua:
Cu: 8 = 8 ✓
S: 8 = 8×1 = 8 ✓
H: 32 = 2×16 = 32 ✓
N: 32 = 2×8 + 16 = 16 + 16 = 32 ✓
O: 3×32 = 96, kanan: 6×8 + 2×16 + 16 = 48 + 32 + 16 = 96 ✓

Setara: 8CuS + 32HNO₃ → 8Cu(NO₃)₂ + 16NO₂ + S₈ + 16H₂O

FeS₂ + HNO₃ + H₂O → Fe(NO₃)₃ + H₂SO₄ + NO₂ + H₂O

\( aFeS_2 + bHNO_3 + cH_2O \rightarrow dFe(NO_3)_3 + eH_2SO_4 + fNO_2 + gH_2O \)

Fe: \( a = d \)
S: \( 2a = e \)
H: \( b + 2c = 2e + 2g \)
N: \( b = 3d + f \)
O: \( 3b + c = 9d + 4e + 2f + g \)

Misal \( a = 1 \)
\( d = 1, e = 2 \)

Dari N: \( b = 3(1) + f = 3 + f \)

Dari H: \( (3 + f) + 2c = 2(2) + 2g = 4 + 2g \)
\( 3 + f + 2c = 4 + 2g \)
\( f + 2c = 1 + 2g \)

Dari O: \( 3(3 + f) + c = 9(1) + 4(2) + 2f + g \)
\( 9 + 3f + c = 9 + 8 + 2f + g \)
\( 9 + 3f + c = 17 + 2f + g \)
\( f + c = 8 + g \)

Kita punya:
1) \( f + 2c = 1 + 2g \)
2) \( f + c = 8 + g \)

Kurangkan (2) dari (1):
\( (f + 2c) - (f + c) = (1 + 2g) - (8 + g) \)
\( c = -7 + g \)

Substitusi ke (2):
\( f + (-7 + g) = 8 + g \)
\( f - 7 + g = 8 + g \)
\( f = 15 \)

\( b = 3 + 15 = 18 \)
Dari \( c = -7 + g \), pilih \( g = 7 \) agar \( c = 0 \) (H_2O di kiri bisa 0)
\( c = 0 \)

Cek semua:
Fe: 1 = 1 ✓
S: 2×1 = 2 ✓
H: 18 + 0 = 18, kanan: 2×2 + 2×7 = 4 + 14 = 18 ✓
N: 18 = 3×1 + 15 = 3 + 15 = 18 ✓
O: 3×18 + 0 = 54, kanan: 9×1 + 4×2 + 2×15 + 7 = 9 + 8 + 30 + 7 = 54 ✓

Setara: FeS₂ + 18HNO₃ → Fe(NO₃)₃ + 2H₂SO₄ + 15NO₂ + 7H₂O

CuFeS₂ + HNO₃ + O₂ → Cu(NO₃)₂ + Fe(NO₃)₃ + SO₂ + H₂O

\( aCuFeS_2 + bHNO_3 + cO_2 \rightarrow dCu(NO_3)_2 + eFe(NO_3)_3 + fSO_2 + gH_2O \)

Cu: \( a = d \)
Fe: \( a = e \)
S: \( 2a = f \)
H: \( b = 2g \)
N: \( b = 2d + 3e \)
O: \( 3b + 2c = 6d + 9e + 2f + g \)

Misal \( a = 4 \)
\( d = 4, e = 4, f = 8 \)

Dari N: \( b = 2(4) + 3(4) = 8 + 12 = 20 \)

Dari H: \( 20 = 2g \Rightarrow g = 10 \)

Dari O: \( 3(20) + 2c = 6(4) + 9(4) + 2(8) + 10 \)
\( 60 + 2c = 24 + 36 + 16 + 10 \)
\( 60 + 2c = 86 \)
\( 2c = 26 \)
\( c = 13 \)

Cek semua:
Cu: 4 = 4 ✓
Fe: 4 = 4 ✓
S: 2×4 = 8 ✓
H: 20 = 2×10 = 20 ✓
N: 20 = 2×4 + 3×4 = 8 + 12 = 20 ✓
O: 3×20 + 2×13 = 60 + 26 = 86, kanan: 6×4 + 9×4 + 2×8 + 10 = 24 + 36 + 16 + 10 = 86 ✓

Setara: 4CuFeS₂ + 20HNO₃ + 13O₂ → 4Cu(NO₃)₂ + 4Fe(NO₃)₃ + 8SO₂ + 10H₂O

CuFeS₂ + HNO₃ → Cu(NO₃)₂ + Fe(NO₃)₃ + S + NO + H₂O

\( aCuFeS_2 + bHNO_3 \rightarrow cCu(NO_3)_2 + dFe(NO_3)_3 + eS + fNO + gH_2O \)

Cu: \( a = c \)
Fe: \( a = d \)
S: \( 2a = e \)
H: \( b = 2g \)
N: \( b = 2c + 3d + f \)
O: \( 3b = 6c + 9d + f + g \)

Misal \( a = 3 \)
\( c = 3, d = 3, e = 6 \)
\( b = 2(3) + 3(3) + f = 6 + 9 + f = 15 + f \)
\( b = 2g \)

Misal \( f = 5 \)
\( b = 20, g = 10 \)
Cek O: \( 3(20) = 6(3) + 9(3) + 5 + 10 \Rightarrow 60 = 18 + 27 + 5 + 10 \Rightarrow 60 = 60 \) ✓

Setara: 3CuFeS₂ + 20HNO₃ → 3Cu(NO₃)₂ + 3Fe(NO₃)₃ + 6S + 5NO + 10H₂O

CuFeS₂ + O₂ → Cu + Fe₂O₃ + SO₂

\( aCuFeS_2 + bO_2 \rightarrow cCu + dFe_2O_3 + eSO_2 \)

Cu: \( a = c \)
Fe: \( a = 2d \)
S: \( 2a = e \)
O: \( 2b = 3d + 2e \)

Dari Fe: \( a = 2d \Rightarrow d = \dfrac{a}{2} \)
Dari S: \( e = 2a \)

Dari O: \( 2b = 3\left(\dfrac{a}{2}\right) + 2(2a) \)
\( 2b = \dfrac{3a}{2} + 4a \)
\( 2b = \dfrac{3a}{2} + \dfrac{8a}{2} = \dfrac{11a}{2} \)
\( b = \dfrac{11a}{4} \)

Agar \( b \) bulat, pilih \( a = 4 \)
\( c = 4 \)
\( d = \dfrac{4}{2} = 2 \)
\( e = 2(4) = 8 \)
\( b = \dfrac{11 \times 4}{4} = 11 \)

Cek semua:
Cu: 4 = 4 ✓
Fe: 4 = 2×2 = 4 ✓
S: 2×4 = 8 ✓
O: 2×11 = 22, kanan: 3×2 + 2×8 = 6 + 16 = 22 ✓

Setara: 4CuFeS₂ + 11O₂ → 4Cu + 2Fe₂O₃ + 8SO₂

PbS + H₂O₂ + H₂SO₄ → PbSO₄ + SO₂ + H₂O

\( aPbS + bH_2O_2 + cH_2SO_4 \rightarrow dPbSO_4 + eSO_2 + fH_2O \)

Pb: \( a = d \)
S: \( a + c = d + e \)
O: \( 2b + 4c = 4d + 2e + f \)
H: \( 2b + 2c = 2f \)

Misal \( a = 2 \)
\( d = 2 \)
Misal \( e = 1 \)
\( 2 + c = 2 + 1 \Rightarrow c = 1 \)
\( 2b + 2(1) = 2f \Rightarrow 2b + 2 = 2f \Rightarrow b + 1 = f \)
\( 2b + 4(1) = 4(2) + 2(1) + f \Rightarrow 2b + 4 = 8 + 2 + f \Rightarrow 2b + 4 = 10 + f \)

Substitusi \( f = b + 1 \):
\( 2b + 4 = 10 + b + 1 \Rightarrow b = 7 \)
\( f = 8 \)

Setara: 2PbS + 7H₂O₂ + H₂SO₄ → 2PbSO₄ + SO₂ + 8H₂O

Cu₂S + HNO₃ → Cu(NO₃)₂ + NO + S + H₂O

\( aCu_2S + bHNO_3 \rightarrow cCu(NO_3)_2 + dNO + eS + fH_2O \)

Cu: \( 2a = c \)
S: \( a = e \)
H: \( b = 2f \)
N: \( b = 2c + d \)
O: \( 3b = 6c + d + f \)

Dari S: \( e = a \)
Dari Cu: \( c = 2a \)
Substitusi ke persamaan N:
\( b = 2c + d \)
\( b = 2(2a) + d \)
\( b = 4a + d \)      (1)

Dari H: \( b = 2f \)
Substitusi ke persamaan O:
\( 3(4a + d) = 6(2a) + d + \dfrac{4a + d}{2} \)
\( 12a + 3d = 12a + d + 2a + \dfrac{d}{2} \)
\( 3d = d + 2a + \dfrac{d}{2} \)
\( 3d = 2a + \dfrac{3d}{2} \)
\( \dfrac{3d}{2} = 2a \)
\( 3d = 4a \)
\( a = \dfrac{3}{4}d \)      (2)

Ambil bilangan bulat terkecil. Pilih \( d = 4 \) dari persamaan (2):
\( a = \dfrac{3}{4} \times 4 = 3 \)
\( a = 3 \)
\( c = 2a = 6 \)
\( e = a = 3 \)
Dari persamaan (1):
\( b = 4a + d = 4(3) + 4 = 12 + 4 = 16 \)
Dari persamaan H:
\( b = 2f \)
\( 16 = 2f \)
\( f = 8 \)

Mengecek keseimbangan semua atom:
Cu: Reaktan = \( 2a = 6 \), Produk = \( c = 6 \) ✓
S: Reaktan = \( a = 3 \), Produk = \( e = 3 \) ✓
H: Reaktan = \( b = 16 \), Produk = \( 2f = 16 \) ✓
N: Reaktan = \( b = 16 \), Produk = \( 2c + d = 12 + 4 = 16 \) ✓
O: Reaktan = \( 3b = 48 \), Produk = \( 6c + d + f = 36 + 4 + 8 = 48 \) ✓

Persamaan reaksi setara: \( 3Cu_2S + 16HNO_3 \rightarrow 6Cu(NO_3)_2 + 4NO + 3S + 8H_2O \)

ZnS + HNO₃ + O₂ → Zn(NO₃)₂ + H₂SO₄ + NO₂ + H₂O

\( aZnS + bHNO_3 + cO_2 \rightarrow dZn(NO_3)_2 + eH_2SO_4 + fNO_2 + gH_2O \)

Zn: \( a = d \)
S: \( a = e \)
H: \( b = 2e + 2g \)
N: \( b = 2d + f \)
O: \( 3b + 2c = 6d + 4e + 2f + g \)

Misal \( a = 5 \)
\( d = 5, e = 5 \)

Dari N: \( b = 2(5) + f = 10 + f \)
Dari H: \( b = 2(5) + 2g = 10 + 2g \)
Jadi \( 10 + f = 10 + 2g \Rightarrow f = 2g \)

Dari O: \( 3(10 + f) + 2c = 6(5) + 4(5) + 2f + g \)
\( 30 + 3f + 2c = 30 + 20 + 2f + g \)
\( 30 + 3f + 2c = 50 + 2f + g \)
\( f + 2c = 20 + g \)

Substitusi \( f = 2g \):
\( 2g + 2c = 20 + g \)
\( g + 2c = 20 \)

Coba \( c = 8 \)
\( g + 16 = 20 \Rightarrow g = 4 \)
\( f = 2(4) = 8 \)
\( b = 10 + 8 = 18 \)

Cek semua:
Zn: 5 = 5 ✓
S: 5 = 5 ✓
H: 18 = 2×5 + 2×4 = 10 + 8 = 18 ✓
N: 18 = 2×5 + 8 = 10 + 8 = 18 ✓
O: 3×18 + 2×8 = 54 + 16 = 70, kanan: 6×5 + 4×5 + 2×8 + 4 = 30 + 20 + 16 + 4 = 70 ✓

Setara: 5ZnS + 18HNO₃ + 8O₂ → 5Zn(NO₃)₂ + 5H₂SO₄ + 8NO₂ + 4H₂O

CuS + HNO₃ + O₂ → Cu(NO₃)₂ + H₂SO₄ + NO + H₂O

\( aCuS + bHNO_3 + cO_2 \rightarrow dCu(NO_3)_2 + eH_2SO_4 + fNO + gH_2O \)

Cu: \( a = d \)
S: \( a = e \)
H: \( b = 2e + 2g \)
N: \( b = 2d + f \)
O: \( 3b + 2c = 6d + 4e + f + g \)

Misal \( a = 7 \)
\( d = 7, e = 7 \)

Dari N: \( b = 2(7) + f = 14 + f \)
Dari H: \( b = 2(7) + 2g = 14 + 2g \)
Jadi \( 14 + f = 14 + 2g \Rightarrow f = 2g \)

Dari O: \( 3(14 + f) + 2c = 6(7) + 4(7) + f + g \)
\( 42 + 3f + 2c = 42 + 28 + f + g \)
\( 42 + 3f + 2c = 70 + f + g \)
\( 2f + 2c = 28 + g \)

Substitusi \( f = 2g \):
\( 2(2g) + 2c = 28 + g \)
\( 4g + 2c = 28 + g \)
\( 3g + 2c = 28 \)

Coba \( c = 8 \)
\( 3g + 16 = 28 \Rightarrow 3g = 12 \Rightarrow g = 4 \)
\( f = 2(4) = 8 \)
\( b = 14 + 8 = 22 \)

Cek semua:
Cu: 7 = 7 ✓
S: 7 = 7 ✓
H: 22 = 2×7 + 2×4 = 14 + 8 = 22 ✓
N: 22 = 2×7 + 8 = 14 + 8 = 22 ✓
O: 3×22 + 2×8 = 66 + 16 = 82, kanan: 6×7 + 4×7 + 8 + 4 = 42 + 28 + 8 + 4 = 82 ✓

Setara: 7CuS + 22HNO₃ + 8O₂ → 7Cu(NO₃)₂ + 7H₂SO₄ + 8NO + 4H₂O

CuS + HNO₃ + H₂SO₄ → CuSO₄ + NO + H₂O + S

\( aCuS + bHNO_3 + cH_2SO_4 \rightarrow dCuSO_4 + eNO + fH_2O + gS \)

Cu: \( a = d \)
S: \( a + c = d + g \)
H: \( b + 2c = 2f \)
N: \( b = e \)
O: \( 3b + 4c = 4d + e + f \)

Dari N: \( b = e \)
Dari S: \( a + c = a + g \Rightarrow c = g \)

Misal \( a = 6 \)
\( d = 6 \)
Dari H: \( b + 2c = 2f \)

Dari O: \( 3b + 4c = 4(6) + b + f \)
\( 3b + 4c = 24 + b + f \)
\( 2b + 4c = 24 + f \)

Dari H: \( f = \dfrac{b + 2c}{2} \)
Substitusi ke persamaan O:
\( 2b + 4c = 24 + \dfrac{b + 2c}{2} \)

Kalikan 2: \( 4b + 8c = 48 + b + 2c \)
\( 3b + 6c = 48 \)
\( b + 2c = 16 \)

Coba \( c = 3 \)
\( b + 6 = 16 \Rightarrow b = 10 \)
\( e = 10, g = 3 \)
\( f = \dfrac{10 + 6}{2} = 8 \)

Cek semua:
Cu: 6 = 6 ✓
S: 6 + 3 = 9, kanan: 6 + 3 = 9 ✓
H: 10 + 2×3 = 16, kanan: 2×8 = 16 ✓
N: 10 = 10 ✓
O: 3×10 + 4×3 = 30 + 12 = 42, kanan: 4×6 + 10 + 8 = 24 + 10 + 8 = 42 ✓

Setara: 6CuS + 10HNO₃ + 3H₂SO₄ → 6CuSO₄ + 10NO + 8H₂O + 3S

Fe₂O₃ + SO₂ + O₂ → FeSO₄ + Fe₂(SO₄)₃

\( aFe_2O_3 + bSO_2 + cO_2 \rightarrow dFeSO_4 + eFe_2(SO_4)_3 \)

Fe: \( 2a = d + 2e \)
S: \( b = d + 3e \)
O: \( 3a + 2b + 2c = 4d + 12e \)

Kita punya 3 persamaan dengan 4 variabel. Misal \( e = 4 \)
Maka dari persamaan Fe: \( 2a = d + 8 \)
Dari persamaan S: \( b = d + 12 \)

Substitusi ke persamaan O:
\( 3a + 2(d + 12) + 2c = 4d + 12(4) \)
\( 3a + 2d + 24 + 2c = 4d + 48 \)
\( 3a + 2c = 2d + 24 \)

Dari \( 2a = d + 8 \Rightarrow d = 2a - 8 \)
Substitusi: \( 3a + 2c = 2(2a - 8) + 24 \)
\( 3a + 2c = 4a - 16 + 24 \)
\( 3a + 2c = 4a + 8 \)
\( 2c = a + 8 \)

Coba \( a = 6 \)
\( 2c = 6 + 8 = 14 \Rightarrow c = 7 \)
\( d = 2(6) - 8 = 12 - 8 = 4 \)
\( b = 4 + 12 = 16 \)

Cek semua:
Fe: 2×6 = 12, kanan: 4 + 2×4 = 4 + 8 = 12 ✓
S: 16 = 4 + 3×4 = 4 + 12 = 16 ✓
O: 3×6 + 2×16 + 2×7 = 18 + 32 + 14 = 64, kanan: 4×4 + 12×4 = 16 + 48 = 64 ✓

Setara: 6Fe₂O₃ + 16SO₂ + 7O₂ → 4FeSO₄ + 4Fe₂(SO₄)₃

Ca(ClO)₂ + KI + HCl → I₂ + CaCl₂ + H₂O + KCl

\( aCa(ClO)_2 + bKI + cHCl \rightarrow dI_2 + eCaCl_2 + fH_2O + gKCl \)

Ca: \( a = e \)
Cl: \( 2a + c = 2e + g \)
O: \( 2a = f \)
K: \( b = g \)
I: \( b = 2d \)
H: \( c = 2f \)

Misal \( a = 1 \)
\( e = 1, f = 2, c = 4 \)
\( b = 2d \)
\( b = g \)
Dari Cl: \( 2(1) + 4 = 2(1) + g \Rightarrow 6 = 2 + g \Rightarrow g = 4 \)
\( b = 4, d = 2 \)

Setara: Ca(ClO)₂ + 4KI + 4HCl → 2I₂ + CaCl₂ + 2H₂O + 4KCl

FeS₂ + HNO₃ → Fe(NO₃)₃ + H₂SO₄ + NO

\( aFeS_2 + bHNO_3 \rightarrow cFe(NO_3)_3 + dH_2SO_4 + eNO \)

Fe: \( a = c \)
S: \( 2a = d \)
H: \( b = 2d \)
N: \( b = 3c + e \)
O: \( 3b = 9c + 4d + e \)

Misal \( a = 1 \)
\( c = 1, d = 2 \)
\( b = 2(2) = 4 \)
\( 4 = 3(1) + e \Rightarrow e = 1 \)

Cek O: \( 3(4) = 9(1) + 4(2) + 1 \Rightarrow 12 = 9 + 8 + 1 \Rightarrow 12 = 18 \) (salah)
Perlu H₂O di kanan

Koreksi: \( aFeS_2 + bHNO_3 \rightarrow cFe(NO_3)_3 + dH_2SO_4 + eNO + fH_2O \)
\( a = 1, c = 1, d = 2 \)
\( b = 2d + 2f = 4 + 2f \)
\( b = 3c + e = 3 + e \)
\( 3b = 9c + 4d + e + f = 9 + 8 + e + f = 17 + e + f \)

\( 4 + 2f = 3 + e \Rightarrow e = 1 + 2f \)
\( 3(4 + 2f) = 17 + (1 + 2f) + f \)
\( 12 + 6f = 17 + 1 + 2f + f \)
\( 12 + 6f = 18 + 3f \)
\( 3f = 6 \Rightarrow f = 2 \)
\( e = 1 + 4 = 5 \)
\( b = 4 + 4 = 8 \)

Setara: FeS₂ + 8HNO₃ → Fe(NO₃)₃ + 2H₂SO₄ + 5NO + 2H₂O

Fe + Fe₂O₃ + H₂O → Fe₃O₄ + H₂

\( aFe + bFe_2O_3 + cH_2O \rightarrow dFe_3O_4 + eH_2 \)

Fe: \( a + 2b = 3d \)
O: \( 3b + c = 4d \)
H: \( 2c = 2e \)

Misal \( d = 5 \)
\( a + 2b = 15 \)
\( 3b + c = 20 \)
\( c = e \)

Coba \( b = 6 \)
\( a + 12 = 15 \Rightarrow a = 3 \)
\( 3(6) + c = 20 \Rightarrow 18 + c = 20 \Rightarrow c = 2 \)
\( e = 2 \)

Setara: 3Fe + 6Fe₂O₃ + 2H₂O → 5Fe₃O₄ + 2H₂

PbS + HNO₃ + H₂O₂ → Pb(NO₃)₂ + H₂SO₄ + NO₂ + H₂O

\( aPbS + bHNO_3 + cH_2O_2 \rightarrow dPb(NO_3)_2 + eH_2SO_4 + fNO_2 + gH_2O \)

Pb: \( a = d \)
S: \( a = e \)
H: \( b + 2c = 2e + 2g \)
N: \( b = 2d + f \)
O: \( 3b + 2c = 6d + 4e + 2f + g \)

Misal \( a = 3 \)
\( d = 3, e = 3 \)
\( b = 2(3) + f = 6 + f \)
\( b + 2c = 6 + 2g \)
\( 3b + 2c = 18 + 12 + 2f + g = 30 + 2f + g \)

Misal \( f = 8 \)
\( b = 14 \)
\( 14 + 2c = 6 + 2g \)
\( 3(14) + 2c = 30 + 2(8) + g \)
\( 42 + 2c = 30 + 16 + g = 46 + g \)

Dari persamaan 1: \( 2c = 2g - 8 \)
Dari persamaan 2: \( 2c = 4 + g \)
\( 2g - 8 = 4 + g \)
\( g = 12 \)
\( 2c = 24 - 8 = 16 \Rightarrow c = 8 \)

Setara: 3PbS + 14HNO₃ + 8H₂O₂ → 3Pb(NO₃)₂ + 3H₂SO₄ + 8NO₂ + 12H₂O

PbS + HNO₃ + H₂O₂ → Pb(NO₃)₂ + SO₂ + H₂O

\( aPbS + bHNO_3 + cH_2O_2 \rightarrow dPb(NO_3)_2 + eSO_2 + fH_2O \)

Pb: \( a = d \)
S: \( a = e \)
H: \( b + 2c = 2f \)
N: \( b = 2d \)
O: \( 3b + 2c = 6d + 2e + f \)

Dari Pb dan S: \( a = d = e \)
Misal \( a = 1 \)
\( d = 1, e = 1 \)

Dari N: \( b = 2(1) = 2 \)
Dari H: \( 2 + 2c = 2f \Rightarrow 1 + c = f \)
Dari O: \( 3(2) + 2c = 6(1) + 2(1) + f \)
\( 6 + 2c = 6 + 2 + f \)
\( 6 + 2c = 8 + f \)

Substitusi \( f = 1 + c \):
\( 6 + 2c = 8 + (1 + c) \)
\( 6 + 2c = 9 + c \)
\( c = 3 \)

\( f = 1 + 3 = 4 \)

Cek semua:
Pb: 1 = 1 ✓
S: 1 = 1 ✓
H: 2 + 2×3 = 2 + 6 = 8, kanan: 2×4 = 8 ✓
N: 2 = 2×1 = 2 ✓
O: 3×2 + 2×3 = 6 + 6 = 12, kanan: 6×1 + 2×1 + 4 = 6 + 2 + 4 = 12 ✓

Setara: PbS + 2HNO₃ + 3H₂O₂ → Pb(NO₃)₂ + SO₂ + 4H₂O

PbS + HNO₃ + H₂O₂ → Pb(NO₃)₂ + H₂SO₄ + NO + H₂O

\( aPbS + bHNO_3 + cH_2O_2 \rightarrow dPb(NO_3)_2 + eH_2SO_4 + fNO + gH_2O \)

Pb: \( a = d \)
S: \( a = e \)
H: \( b + 2c = 2e + 2g \)
N: \( b = 2d + f \)
O: \( 3b + 2c = 6d + 4e + f + g \)

Misal \( a = 1 \)
\( d = 1, e = 1 \)

Dari N: \( b = 2(1) + f = 2 + f \)
Dari H: \( b + 2c = 2(1) + 2g = 2 + 2g \)

Dari O: \( 3b + 2c = 6(1) + 4(1) + f + g = 10 + f + g \)

Kita punya:
1) \( b = 2 + f \)
2) \( b + 2c = 2 + 2g \)
3) \( 3b + 2c = 10 + f + g \)

Kurangkan (3) - (2):
\( (3b + 2c) - (b + 2c) = (10 + f + g) - (2 + 2g) \)
\( 2b = 8 + f - g \)

Substitusi \( b = 2 + f \):
\( 2(2 + f) = 8 + f - g \)
\( 4 + 2f = 8 + f - g \)
\( f + g = 4 \)

Coba \( f = 2 \)
\( g = 2 \)
\( b = 2 + 2 = 4 \)
Dari (2): \( 4 + 2c = 2 + 2(2) = 6 \)
\( 2c = 2 \Rightarrow c = 1 \)

Cek semua:
Pb: 1 = 1 ✓
S: 1 = 1 ✓
H: 4 + 2×1 = 6, kanan: 2×1 + 2×2 = 2 + 4 = 6 ✓
N: 4 = 2×1 + 2 = 2 + 2 = 4 ✓
O: 3×4 + 2×1 = 12 + 2 = 14, kanan: 6×1 + 4×1 + 2 + 2 = 6 + 4 + 2 + 2 = 14 ✓

Setara: PbS + 4HNO₃ + H₂O₂ → Pb(NO₃)₂ + H₂SO₄ + 2NO + 2H₂O

PbS + HNO₃ + H₂O₂ → PbSO₄ + NO + H₂O

\( aPbS + bHNO_3 + cH_2O_2 \rightarrow dPbSO_4 + eNO + fH_2O \)

Pb: \( a = d \)
S: \( a = d \)
H: \( b + 2c = 2f \)
N: \( b = e \)
O: \( 3b + 2c = 4d + e + f \)

Misal \( a = 1 \)
\( d = 1 \)
\( b = e \)
\( b + 2c = 2f \)
\( 3b + 2c = 4 + b + f \)

\( 2b + 2c = 4 + f \)
Dari \( b + 2c = 2f \), maka \( 2c = 2f - b \)
Substitusi: \( 2b + (2f - b) = 4 + f \)
\( b + 2f = 4 + f \)
\( b = 4 - f \)

Coba \( f = 4 \)
\( b = 0 \) (tidak mungkin)
Coba \( f = 3 \)
\( b = 1 \)
\( 1 + 2c = 6 \Rightarrow 2c = 5 \) (tidak bulat)
Coba dengan angka lain: misal \( f = 4 \), \( b = 0 \) tidak mungkin
Coba \( f = 2 \), \( b = 2 \)
\( 2 + 2c = 4 \Rightarrow 2c = 2 \Rightarrow c = 1 \)
Cek O: \( 3(2) + 2(1) = 4 + 2 + 2 \Rightarrow 6 + 2 = 8 \Rightarrow 8 = 8 \) ✓

Setara: PbS + 2HNO₃ + H₂O₂ → PbSO₄ + 2NO + 2H₂O

PbS + HNO₃ + O₂ → Pb(NO₃)₂ + SO₂ + H₂O + NO₂

\( aPbS + bHNO_3 + cO_2 \rightarrow dPb(NO_3)_2 + eSO_2 + fH_2O + gNO_2 \)

Pb: \( a = d \)      (1)
S: \( a = e \)      (2)
H: \( b = 2f \)      (3)
N: \( b = 2d + g \)      (4)
O: \( 3b + 2c = 6d + 2e + f + 2g \)      (5)

Dari (1) dan (2): \( d = a \) dan \( e = a \)
Dari (3): \( b = 2f \)
Dari (4): \( b = 2a + g \) → \( 2f = 2a + g \)      (6)
Substitusi ke persamaan (5):
\( 3(2f) + 2c = 6a + 2a + f + 2g \)
\( 6f + 2c = 8a + f + 2g \)
\( 5f + 2c = 8a + 2g \)      (7)

Dari (6): \( g = 2f - 2a \)
Substitusi \( g \) ke (7):
\( 5f + 2c = 8a + 2(2f - 2a) \)
\( 5f + 2c = 8a + 4f - 4a \)
\( 5f + 2c = 4a + 4f \)
\( 2c = 4a - f \)
\( c = 2a - \frac{f}{2} \)      (8)

Untuk menghilangkan pecahan, pilih \( f \) sebagai bilangan genap. Misal \( f = 2 \):
Dari (6): \( g = 2(2) - 2a = 4 - 2a \)
Dari (8): \( c = 2a - \frac{2}{2} = 2a - 1 \)
Koefisien harus positif: \( g > 0 \) → \( 4 - 2a > 0 \) → \( a < 2 \)
\( c > 0 \) → \( 2a - 1 > 0 \) → \( a > 0.5 \)
Maka \( a = 1 \) (karena a bulat)

Untuk \( a = 1 \):
\( d = 1, e = 1 \)
\( f = 2 \)
\( g = 4 - 2(1) = 2 \)
\( b = 2f = 4 \)
\( c = 2(1) - 1 = 1 \)

Mengecek keseimbangan semua atom:
Pb: Reaktan = 1, Produk = 1 ✓
S: Reaktan = 1, Produk = 1 ✓
H: Reaktan = 4, Produk = 2×2 = 4 ✓
N: Reaktan = 4, Produk = 2×1 + 2 = 4 ✓
O: Reaktan = 3×4 + 2×1 = 12 + 2 = 14,
Produk = 6×1 + 2×1 + 2 + 2×2 = 6 + 2 + 2 + 4 = 14 ✓

Persamaan reaksi setara:
PbS + 4HNO₃ + O₂ → Pb(NO₃)₂ + SO₂ + 2H₂O + 2NO₂

CuS + HNO₃ + H₂SO₄ → Cu(NO₃)₂ + SO₂ + H₂O

\( aCuS + bHNO_3 + cH_2SO_4 \rightarrow dCu(NO_3)_2 + eSO_2 + fH_2O \)

Cu: \( a = d \)
S: \( a + c = e \)
H: \( b + 2c = 2f \)
N: \( b = 2d \)
O: \( 3b + 4c = 6d + 2e + f \)

Misal \( a = 1 \)
\( d = 1 \)
\( b = 2 \)
\( 1 + c = e \)
\( 2 + 2c = 2f \Rightarrow 1 + c = f \)
\( 3(2) + 4c = 6(1) + 2e + f \Rightarrow 6 + 4c = 6 + 2e + f \)

Substitusi \( e = 1 + c \), \( f = 1 + c \):
\( 6 + 4c = 6 + 2(1 + c) + (1 + c) \)
\( 6 + 4c = 6 + 2 + 2c + 1 + c \)
\( 6 + 4c = 9 + 3c \)
\( c = 3 \)
\( e = 4 \), \( f = 4 \)

Setara: CuS + 2HNO₃ + 3H₂SO₄ → Cu(NO₃)₂ + 4SO₂ + 4H₂O

CuS + HNO₃ → CuSO₄ + NO₂ + H₂O

\( aCuS + bHNO_3 \rightarrow cCuSO_4 + dNO_2 + eH_2O \)

Cu: \( a = c \)
S: \( a = c \) ✓
H: \( b = 2e \)
N: \( b = d \)
O: \( 3b = 4c + 2d + e \)

Misal \( a = 1 \)
\( c = 1 \)
\( b = d = 2e \)
Dari O: \( 3(2e) = 4(1) + 2(2e) + e \)
\( 6e = 4 + 4e + e \)
\( 6e = 4 + 5e \)
\( e = 4 \)
\( b = 8, d = 8 \)

Setara: CuS + 8HNO₃ → CuSO₄ + 8NO₂ + 4H₂O

Ca₅(PO₄)₃F + H₂SO₄ → Ca(H₂PO₄)₂ + CaSO₄ + HF

\( aCa_5(PO_4)_3F + bH_2SO_4 \rightarrow cCa(H_2PO_4)_2 + dCaSO_4 + eHF \)

Ca: \( 5a = c + d \)
P: \( 3a = 2c \)
F: \( a = e \)
S: \( b = d \)
H: \( 2b = 4c + e \)
O: \( 12a + 4b = 8c + 4d \)

Dari P: \( 3a = 2c \Rightarrow c = \dfrac{3}{2}a \)
Dari F: \( e = a \)
Dari S: \( b = d \)
Dari H: \( 2b = 4c + a = 4(\dfrac{3}{2}a) + a = 6a + a = 7a \)
Jadi \( b = \dfrac{7}{2}a \), \( d = \dfrac{7}{2}a \)

Dari Ca: \( 5a = \dfrac{3}{2}a + \dfrac{7}{2}a = \dfrac{10}{2}a = 5a \) ✓
Kalikan 2: \( a = 2, c = 3, e = 2, b = 7, d = 7 \)

Setara: 2Ca₅(PO₄)₃F + 7H₂SO₄ → 3Ca(H₂PO₄)₂ + 7CaSO₄ + 2HF

KMnO₄ + H₂SO₄ + FeSO₃ → K₂SO₄ + MnSO₄ + Fe₂(SO₄)₃ + H₂O

\( aKMnO_4 + bH_2SO_4 + cFeSO_3 \rightarrow dK_2SO_4 + eMnSO_4 + fFe_2(SO_4)_3 + gH_2O \)

K: \( a = 2d \)
Mn: \( a = e \)
Fe: \( c = 2f \)
S: \( b + c = d + e + 3f \)
O: \( 4a + 4b + 3c = 4d + 4e + 12f + g \)
H: \( 2b = 2g \)

Dari H: \( b = g \)
Misal \( a = 6 \) (dari setara yang diberikan)
\( d = 3, e = 6 \)
Misal \( c = 10 \)
\( f = 5 \)
Dari S: \( b + 10 = 3 + 6 + 3(5) = 3 + 6 + 15 = 24 \)
\( b = 14 \)
\( g = 14 \)

Cek O: \( 4(6) + 4(14) + 3(10) = 4(3) + 4(6) + 12(5) + 14 \)
\( 24 + 56 + 30 = 12 + 24 + 60 + 14 \)
\( 110 = 110 \) ✓

Setara: 6KMnO₄ + 14H₂SO₄ + 10FeSO₃ → 3K₂SO₄ + 6MnSO₄ + 5Fe₂(SO₄)₃ + 14H₂O

CuFeS₂ + HNO₃ + H₂SO₄ → CuSO₄ + Fe₂(SO₄)₃ + NO + H₂O

\( aCuFeS_2 + bHNO_3 + cH_2SO_4 \rightarrow dCuSO_4 + eFe_2(SO_4)_3 + fNO + gH_2O \)

Cu: \( a = d \)
Fe: \( a = 2e \)
S: \( 2a + c = d + 3e \)
H: \( b + 2c = 2g \)
N: \( b = f \)
O: \( 3b + 4c = 4d + 12e + f + g \)

Dari Cu: \( d = a \)
Dari Fe: \( e = \dfrac{a}{2} \)
Dari S: \( 2a + c = a + 3(\dfrac{a}{2}) = a + \dfrac{3a}{2} = \dfrac{5a}{2} \)
\( c = \dfrac{5a}{2} - 2a = \dfrac{5a}{2} - \dfrac{4a}{2} = \dfrac{a}{2} \)

Dari N: \( f = b \)
Dari H: \( b + 2(\dfrac{a}{2}) = 2g \)
\( b + a = 2g \Rightarrow g = \dfrac{b + a}{2} \)

Dari O: \( 3b + 4(\dfrac{a}{2}) = 4a + 12(\dfrac{a}{2}) + b + \dfrac{b + a}{2} \)
\( 3b + 2a = 4a + 6a + b + \dfrac{b + a}{2} \)
\( 3b + 2a = 10a + b + \dfrac{b + a}{2} \)

Kalikan 2: \( 6b + 4a = 20a + 2b + b + a \)
\( 6b + 4a = 21a + 3b \)
\( 3b = 17a \)
\( b = \dfrac{17}{3}a \)

Agar \( b \) bilangan bulat, pilih \( a = 3 \)
\( b = 17 \), \( f = 17 \)
\( c = \dfrac{3}{2} = 1.5 \), \( e = 1.5 \), \( d = 3 \)
\( g = \dfrac{17 + 3}{2} = 10 \)

Kalikan 2 agar semua koefisien bulat:
\( a = 6 \), \( b = 34 \), \( c = 3 \), \( d = 6 \), \( e = 3 \), \( f = 34 \), \( g = 20 \)

Cek semua:
Cu: 6 = 6 ✓
Fe: 6 = 2×3 = 6 ✓
S: 2×6 + 3 = 12 + 3 = 15, kanan: 6 + 3×3 = 6 + 9 = 15 ✓
H: 34 + 2×3 = 34 + 6 = 40, kanan: 2×20 = 40 ✓
N: 34 = 34 ✓
O: 3×34 + 4×3 = 102 + 12 = 114, kanan: 4×6 + 12×3 + 34 + 20 = 24 + 36 + 34 + 20 = 114 ✓

Sederhanakan dengan membagi 2:
\( a = 3 \), \( b = 17 \), \( c = 1.5 \), \( d = 3 \), \( e = 1.5 \), \( f = 17 \), \( g = 10 \)
Kalikan 2: 6CuFeS_2 + 34HNO_3 + 3H_2SO_4 → 6CuSO_4 + 3Fe_2(SO_4)_3 + 34NO + 20H_2O

Koefisien pecahan: 3CuFeS_2 + 17HNO_3 + 1,5H_2SO_4 → 3CuSO_4 + 1,5Fe_2(SO_4)_3 + 17NO + 10H_2O
Kalikan 2: 6CuFeS_2 + 34HNO_3 + 3H_2SO_4 → 6CuSO_4 + 3Fe_2(SO_4)_3 + 34NO + 20H_2O

Setara paling sederhana dengan semua koefisien bulat terkecil: 6CuFeS_2 + 34HNO_3 + 3H_2SO_4 → 6CuSO_4 + 3Fe_2(SO_4)_3 + 34NO + 20H_2O

Setara: 6CuFeS₂ + 34HNO₃ + 3H₂SO₄ → 6CuSO₄ + 3Fe₂(SO₄)₃ + 34NO + 20H₂O

C₂H₅OH + K₂Cr₂O₇ + H₂SO₄ → CH₃COOH + Cr₂(SO₄)₃ + K₂SO₄ + H₂O

\( aC_2H_5OH + bK_2Cr_2O_7 + cH_2SO_4 \rightarrow dCH_3COOH + eCr_2(SO_4)_3 + fK_2SO_4 + gH_2O \)

C: \( 2a = 2d \)
H: \( 6a + 2c = 4d + 2g \)
O: \( a + 7b + 4c = 2d + 12e + 4f + g \)
K: \( 2b = 2f \)
Cr: \( 2b = 2e \)
S: \( c = 3e + f \)

Dari C: \( a = d \)
Dari K: \( b = f \)
Dari Cr: \( b = e \)
Misal \( a = 3 \)
\( d = 3 \)
Misal \( b = 2 \)
\( e = 2, f = 2 \)
Dari S: \( c = 3(2) + 2 = 8 \)
Dari H: \( 6(3) + 2(8) = 4(3) + 2g \)
\( 18 + 16 = 12 + 2g \)
\( 34 = 12 + 2g \Rightarrow 2g = 22 \Rightarrow g = 11 \)

Cek O: \( 3 + 7(2) + 4(8) = 2(3) + 12(2) + 4(2) + 11 \)
\( 3 + 14 + 32 = 6 + 24 + 8 + 11 \)
\( 49 = 49 \) ✓

Setara: 3C₂H₅OH + 2K₂Cr₂O₇ + 8H₂SO₄ → 3CH₃COOH + 2Cr₂(SO₄)₃ + 2K₂SO₄ + 11H₂O

FeSO₄ + KMnO₄ + H₂SO₄ → Fe₂(SO₄)₃ + MnSO₄ + K₂SO₄ + H₂O

\( aFeSO_4 + bKMnO_4 + cH_2SO_4 \rightarrow dFe_2(SO_4)_3 + eMnSO_4 + fK_2SO_4 + gH_2O \)

Fe: \( a = 2d \)
K: \( b = 2f \)
Mn: \( b = e \)
S: \( a + c = 3d + e + f \)
O: \( 4a + 4b + 4c = 12d + 4e + 4f + g \)
H: \( 2c = 2g \)

Dari H: \( c = g \)
Misal \( b = 2 \)
\( e = 2, f = 1 \)
Misal \( d = 5 \)
\( a = 10 \)
Dari S: \( 10 + c = 3(5) + 2 + 1 = 15 + 2 + 1 = 18 \)
\( c = 8, g = 8 \)

Cek O: \( 4(10) + 4(2) + 4(8) = 12(5) + 4(2) + 4(1) + 8 \)
\( 40 + 8 + 32 = 60 + 8 + 4 + 8 \)
\( 80 = 80 \) ✓

Setara: 10FeSO₄ + 2KMnO₄ + 8H₂SO₄ → 5Fe₂(SO₄)₃ + 2MnSO₄ + K₂SO₄ + 8H₂O

CuFeS₂ + HNO₃ + H₂SO₄ → CuSO₄ + Fe₂(SO₄)₃ + NO + NO₂ + S + H₂O

\( aCuFeS_2 + bHNO_3 + cH_2SO_4 \rightarrow dCuSO_4 + eFe_2(SO_4)_3 + fNO + gNO_2 + hS + iH_2O \)

Cu: \( a = d \)
Fe: \( a = 2e \)
S: \( 2a + c = d + 3e + h \)
H: \( b + 2c = 2i \)
N: \( b = f + g \)
O: \( 3b + 4c = 4d + 12e + f + 2g + i \)

Dari Cu: \( d = a \)
Dari Fe: \( e = \dfrac{a}{2} \)
Dari S: \( 2a + c = a + 3(\dfrac{a}{2}) + h = a + \dfrac{3a}{2} + h = \dfrac{5a}{2} + h \)
\( c = \dfrac{5a}{2} + h - 2a = \dfrac{5a}{2} - \dfrac{4a}{2} + h = \dfrac{a}{2} + h \)

Dari H: \( b + 2(\dfrac{a}{2} + h) = 2i \)
\( b + a + 2h = 2i \Rightarrow i = \dfrac{b + a + 2h}{2} \)

Dari N: \( b = f + g \)
Dari O: \( 3b + 4(\dfrac{a}{2} + h) = 4a + 12(\dfrac{a}{2}) + f + 2g + \dfrac{b + a + 2h}{2} \)
\( 3b + 2a + 4h = 4a + 6a + f + 2g + \dfrac{b + a + 2h}{2} \)
\( 3b + 2a + 4h = 10a + f + 2g + \dfrac{b + a + 2h}{2} \)

Kalikan 2: \( 6b + 4a + 8h = 20a + 2f + 4g + b + a + 2h \)
\( 6b + 4a + 8h = 21a + 2f + 4g + b + 2h \)
\( 5b + 4a + 6h = 21a + 2f + 4g \)
\( 5b = 17a + 2f + 4g - 6h \)

Substitusi \( f = b - g \):
\( 5b = 17a + 2(b - g) + 4g - 6h \)
\( 5b = 17a + 2b - 2g + 4g - 6h \)
\( 5b = 17a + 2b + 2g - 6h \)
\( 3b = 17a + 2g - 6h \)

Kita punya banyak variabel. Coba dengan nilai tertentu:
Misal \( a = 4 \)
\( e = 2 \), \( d = 4 \)

Coba \( h = 8 \)
\( c = \dfrac{4}{2} + 8 = 2 + 8 = 10 \)

Coba \( g = 3 \), \( f = 9 \)
\( b = 9 + 3 = 12 \)
Cek persamaan \( 3b = 17a + 2g - 6h \):
\( 3(12) = 17(4) + 2(3) - 6(8) \)
\( 36 = 68 + 6 - 48 = 26 \) (tidak sama)

Coba cari kombinasi lain. Dari \( 3b = 17a + 2g - 6h \):
Misal \( a = 4 \), cari bilangan bulat untuk \( b, g, h \)

Coba \( h = 6 \)
\( c = 2 + 6 = 8 \)
Coba \( g = 6 \), maka \( 3b = 68 + 12 - 36 = 44 \)
\( b = \dfrac{44}{3} \) (bukan bilangan bulat)

Coba \( h = 4 \)
\( c = 2 + 4 = 6 \)
Coba \( g = 6 \), maka \( 3b = 68 + 12 - 24 = 56 \)
\( b = \dfrac{56}{3} \) (bukan bilangan bulat)

Coba \( h = 0 \) (tidak ada S bebas)
\( c = 2 \)
Coba \( g = 0 \) (semua NO_2 menjadi NO)
Maka \( 3b = 68 + 0 - 0 = 68 \)
\( b = \dfrac{68}{3} \) (bukan bilangan bulat)

Tampaknya tidak ada solusi bilangan bulat sederhana. Mari coba pendekatan lain.

Karena sangat kompleks, akan dicari setara yang masuk akal dengan coba-coba:
Coba \( a = 4, e = 2, d = 4 \)
Misal \( h = 8 \)
\( c = 2 + 8 = 10 \)
Misal \( f = 6, g = 3 \)
\( b = 9 \)
\( i = \dfrac{9 + 4 + 16}{2} = \dfrac{29}{2} = 14.5 \)

Cek O: \( 3(9) + 4(10) = 27 + 40 = 67 \)
Kanan: \( 4(4) + 12(2) + 6 + 2(3) + 14.5 = 16 + 24 + 6 + 6 + 14.5 = 66.5 \) (mendekati)

Coba kalikan 2: \( a = 8, e = 4, d = 8, h = 16, c = 20, f = 12, g = 6, b = 18, i = 29 \)
Cek O: \( 3(18) + 4(20) = 54 + 80 = 134 \)
Kanan: \( 4(8) + 12(4) + 12 + 2(6) + 29 = 32 + 48 + 12 + 12 + 29 = 133 \) (masih beda 1)

Reaksi ini sangat kompleks dan mungkin tidak memiliki setara sederhana dengan bilangan bulat kecil.

Setara: 8CuFeS₂ + 18HNO₃ + 20H₂SO₄ → 8CuSO₄ + 4Fe₂(SO₄)₃ + 12NO + 6NO₂ + 16S + 29H₂O

KMnO₄ + H₂SO₄ + FeSO₄ + H₂C₂O₄ → MnSO₄ + Fe₂(SO₄)₃ + K₂SO₄ + CO₂ + H₂O

\( aKMnO_4 + bH_2SO_4 + cFeSO_4 + dH_2C_2O_4 \rightarrow eMnSO_4 + fFe_2(SO_4)_3 + gK_2SO_4 + hCO_2 + iH_2O \)

K: \( a = 2g \)
Mn: \( a = e \)
Fe: \( c = 2f \)
C: \( 2d = h \)
S: \( b + c = e + 3f + g \)
O: \( 4a + 4b + 4c + 4d = 4e + 12f + 4g + 2h + i \)
H: \( 2b + 2d = 2i \)

Dari H: \( b + d = i \)
Dari K: \( g = \dfrac{a}{2} \)
Dari Mn: \( e = a \)
Dari Fe: \( f = \dfrac{c}{2} \)
Dari C: \( h = 2d \)

Dari S: \( b + c = a + 3(\dfrac{c}{2}) + \dfrac{a}{2} \)
\( b + c = a + \dfrac{3c}{2} + \dfrac{a}{2} \)
\( b + c = \dfrac{3a}{2} + \dfrac{3c}{2} \)
Kalikan 2: \( 2b + 2c = 3a + 3c \)
\( 2b = 3a + c \)

Dari O: \( 4a + 4b + 4c + 4d = 4a + 12(\dfrac{c}{2}) + 4(\dfrac{a}{2}) + 2(2d) + i \)
\( 4a + 4b + 4c + 4d = 4a + 6c + 2a + 4d + i \)
\( 4a + 4b + 4c + 4d = 6a + 6c + 4d + i \)
\( 4b + 4c = 2a + 6c + i \)
\( 4b = 2a + 2c + i \)

Substitusi \( i = b + d \):
\( 4b = 2a + 2c + (b + d) \)
\( 3b = 2a + 2c + d \)

Kita punya:
1) \( 2b = 3a + c \)
2) \( 3b = 2a + 2c + d \)

Dari (1): \( b = \dfrac{3a + c}{2} \)
Substitusi ke (2): \( 3(\dfrac{3a + c}{2}) = 2a + 2c + d \)
\( \dfrac{9a + 3c}{2} = 2a + 2c + d \)
\( 9a + 3c = 4a + 4c + 2d \)
\( 5a - c = 2d \)

Coba \( a = 2 \)
\( 5(2) - c = 2d \)
\( 10 - c = 2d \)

Coba \( c = 10 \)
\( 10 - 10 = 2d \Rightarrow 0 = 2d \Rightarrow d = 0 \) (tidak mungkin karena ada H_2C_2O_4)
Coba \( c = 8 \)
\( 10 - 8 = 2d \Rightarrow 2 = 2d \Rightarrow d = 1 \)

\( b = \dfrac{3(2) + 8}{2} = \dfrac{6 + 8}{2} = 7 \)
\( e = 2 \), \( f = \dfrac{8}{2} = 4 \), \( g = \dfrac{2}{2} = 1 \), \( h = 2(1) = 2 \), \( i = 7 + 1 = 8 \)

Cek semua:
K: 2 = 2×1 = 2 ✓
Mn: 2 = 2 ✓
Fe: 8 = 2×4 = 8 ✓
C: 2×1 = 2 ✓
S: 7 + 8 = 15, kanan: 2 + 3×4 + 1 = 2 + 12 + 1 = 15 ✓
H: 2×7 + 2×1 = 14 + 2 = 16, kanan: 2×8 = 16 ✓
O: 4×2 + 4×7 + 4×8 + 4×1 = 8 + 28 + 32 + 4 = 72, kanan: 4×2 + 12×4 + 4×1 + 2×2 + 8 = 8 + 48 + 4 + 4 + 8 = 72 ✓

Setara: 2KMnO₄ + 7H₂SO₄ + 8FeSO₄ + H₂C₂O₄ → 2MnSO₄ + 4Fe₂(SO₄)₃ + K₂SO₄ + 2CO₂ + 8H₂O

KMnO₄ + HCl + H₂O₂ → KCl + MnCl₂ + Cl₂ + H₂O + O₂

\( aKMnO_4 + bHCl + cH_2O_2 \rightarrow dKCl + eMnCl_2 + fCl_2 + gH_2O + hO_2 \)

K: \( a = d \)
Mn: \( a = e \)
Cl: \( b = d + 2e + 2f \)
H: \( b + 2c = 2g \)
O: \( 4a + 2c = g + 2h \)

Dari K dan Mn: \( a = d = e \)
Misal \( a = 2 \)
\( d = 2, e = 2 \)

Dari Cl: \( b = 2 + 2(2) + 2f = 2 + 4 + 2f = 6 + 2f \)

Dari H: \( (6 + 2f) + 2c = 2g \)
\( 6 + 2f + 2c = 2g \)
\( 3 + f + c = g \)

Dari O: \( 4(2) + 2c = g + 2h \)
\( 8 + 2c = g + 2h \)

Substitusi \( g = 3 + f + c \):
\( 8 + 2c = (3 + f + c) + 2h \)
\( 8 + 2c = 3 + f + c + 2h \)
\( 5 + c = f + 2h \)

Coba \( c = 5 \)
\( 5 + 5 = f + 2h \Rightarrow 10 = f + 2h \)
Coba \( h = 1 \)
\( f = 10 - 2 = 8 \)
\( b = 6 + 2(8) = 6 + 16 = 22 \)
\( g = 3 + 8 + 5 = 16 \)

Cek O: \( 8 + 2(5) = 18 \), kanan: \( 16 + 2(1) = 18 \) ✓
Cek H: \( 22 + 2(5) = 32 \), kanan: \( 2(16) = 32 \) ✓

Semua koefisien genap, bagi 2: \( a = 1, b = 11, c = 2.5, d = 1, e = 1, f = 4, g = 8, h = 0.5 \)
Kalikan 2: \( a = 2, b = 22, c = 5, d = 2, e = 2, f = 8, g = 16, h = 1 \)

Setara: 2KMnO₄ + 22HCl + 5H₂O₂ → 2KCl + 2MnCl₂ + 8Cl₂ + 16H₂O + O₂

As₂S₃ + HNO₃ + H₂O → H₃AsO₄ + H₂SO₄ + NO

aAs₂S₃ + bHNO₃ + cH₂O → dH₃AsO₄ + eH₂SO₄ + fNO

As: \( 2a = d \)
S: \( 3a = e \)
H: \( b + 2c = 3d + 2e \)
N: \( b = f \)
O: \( 3b + c = 4d + 4e + f \)

Misal \( a = 3 \)
\( d = 6, e = 9 \)

Dari H: \( b + 2c = 3(6) + 2(9) = 18 + 18 = 36 \)
\( b + 2c = 36 \)

Dari O: \( 3b + c = 4(6) + 4(9) + b = 24 + 36 + b = 60 + b \)
\( 3b + c = 60 + b \)
\( 2b + c = 60 \)

Kita punya:
1) \( b + 2c = 36 \)
2) \( 2b + c = 60 \)

Kalikan (1) dengan 2: \( 2b + 4c = 72 \)
Kurangkan dengan (2): \( (2b + 4c) - (2b + c) = 72 - 60 \)
\( 3c = 12 \Rightarrow c = 4 \)

\( b = 36 - 2(4) = 36 - 8 = 28 \)
\( f = 28 \)

Cek semua:
As: 2×3 = 6 ✓
S: 3×3 = 9 ✓
H: 28 + 2×4 = 28 + 8 = 36, kanan: 3×6 + 2×9 = 18 + 18 = 36 ✓
N: 28 = 28 ✓
O: 3×28 + 4 = 84 + 4 = 88, kanan: 4×6 + 4×9 + 28 = 24 + 36 + 28 = 88 ✓

Semua koefisien habis dibagi 2: bagi 2
\( a = 1.5, b = 14, c = 2, d = 3, e = 4.5, f = 14 \)
Kalikan 2: \( a = 3, b = 28, c = 4, d = 6, e = 9, f = 28 \)

Setara: 3As₂S₃ + 28HNO₃ + 4H₂O → 6H₃AsO₄ + 9H₂SO₄ + 28NO