Balancing Equations Worksheet with 20 chemical equations to balance, featuring various chemical compounds and reactions.

Let’s go through each equation one by one and balance them. Balancing means making sure the number of atoms of each element is the same on both sides of the arrow.

We’ll start with #1:

1) Na₃PO₄ + KOH → NaOH + K₃PO₄

Left side:
Na = 3, P = 1, O = 4 (from PO₄) + 1 (from OH) = 5? Wait — actually, we count per compound. Better to list all elements separately.

Actually, let’s do it properly:

Reactants:
Na₃PO₄ → 3 Na, 1 P, 4 O
KOH → 1 K, 1 O, 1 H
Total reactants: Na=3, P=1, O=5, K=1, H=1

Products:
NaOH → 1 Na, 1 O, 1 H
K₃PO₄ → 3 K, 1 P, 4 O
Total products: Na=1, P=1, O=5, K=3, H=1

So Na and K are unbalanced.

Try putting a 3 in front of NaOH:
→ 3 NaOH + K₃PO₄
Now products: Na=3, O=3+4=7, H=3, K=3, P=1

Reactants still: Na=3, P=1, O=5, K=1, H=1 → not matching.

Wait — maybe put 3 in front of KOH?

Try:
Na₃PO₄ + 3 KOH → 3 NaOH + K₃PO₄

Check atoms:

Left:
Na: 3
P: 1
O: 4 (from PO₄) + 3×1 (from 3 OH) = 7
K: 3
H: 3

Right:
Na: 3 (from 3 NaOH)
O: 3 (from 3 NaOH) + 4 (from PO₄) = 7
H: 3
K: 3
P: 1

✔ Balanced! So coefficients: 1, 3, 3, 1

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2) MgF₂ + Li₂CO₃ → MgCO₃ + LiF

Left: Mg=1, F=2, Li=2, C=1, O=3
Right: Mg=1, C=1, O=3, Li=1, F=1 → need 2 LiF

So: MgF₂ + Li₂CO₃ → MgCO₃ + 2 LiF

Check:
Left: Mg=1, F=2, Li=2, C=1, O=3
Right: Mg=1, C=1, O=3, Li=2, F=2 ✔

Coefficients: 1, 1, 1, 2

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3) P₄ + O₂ → P₂O₃

Left: P=4, O=2
Right: P=2, O=3

Need to balance P first. Put 2 in front of P₂O₃ → now P=4, O=6

So right: 2 P₂O₃ → P=4, O=6

Left: P₄ has 4 P → good. Need 6 O → so 3 O₂ (since 3×2=6)

Equation: P₄ + 3 O₂ → 2 P₂O₃

Check:
Left: P=4, O=6
Right: P=4, O=6 ✔

Coefficients: 1, 3, 2

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4) RbNO₃ + BeF₂ → Be(NO₃)₂ + RbF

Left: Rb=1, N=1, O=3, Be=1, F=2
Right: Be=1, N=2, O=6, Rb=1, F=1 → need 2 RbF and 2 RbNO₃

Try: 2 RbNO₃ + BeF₂ → Be(NO₃)₂ + 2 RbF

Left: Rb=2, N=2, O=6, Be=1, F=2
Right: Be=1, N=2, O=6, Rb=2, F=2 ✔

Coefficients: 2, 1, 1, 2

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5) AgNO₃ + Cu → Cu(NO₃)₂ + Ag

Left: Ag=1, N=1, O=3, Cu=1
Right: Cu=1, N=2, O=6, Ag=1 → need 2 AgNO₃ and 2 Ag

Try: 2 AgNO₃ + Cu → Cu(NO₃)₂ + 2 Ag

Left: Ag=2, N=2, O=6, Cu=1
Right: Cu=1, N=2, O=6, Ag=2 ✔

Coefficients: 2, 1, 1, 2

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6) CF₄ + Br₂ → CBr₄ + F₂

Left: C=1, F=4, Br=2
Right: C=1, Br=4, F=2 → need 2 F₂ and 2 Br₂

Try: CF₄ + 2 Br₂ → CBr₄ + 2 F₂

Left: C=1, F=4, Br=4
Right: C=1, Br=4, F=4 ✔

Coefficients: 1, 2, 1, 2

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7) HCN + CuSO₄ → H₂SO₄ + Cu(CN)₂

Left: H=1, C=1, N=1, Cu=1, S=1, O=4
Right: H=2, S=1, O=4, Cu=1, C=2, N=2 → need 2 HCN

Try: 2 HCN + CuSO₄ → H₂SO₄ + Cu(CN)₂

Left: H=2, C=2, N=2, Cu=1, S=1, O=4
Right: H=2, S=1, O=4, Cu=1, C=2, N=2 ✔

Coefficients: 2, 1, 1, 1

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8) GaF₃ + Cs → CsF + Ga

Left: Ga=1, F=3, Cs=1
Right: Cs=1, F=1, Ga=1 → need 3 CsF and 3 Cs

Try: GaF₃ + 3 Cs → 3 CsF + Ga

Left: Ga=1, F=3, Cs=3
Right: Cs=3, F=3, Ga=1 ✔

Coefficients: 1, 3, 3, 1

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9) BaS + PtF₂ → BaF₂ + PtS

Left: Ba=1, S=1, Pt=1, F=2
Right: Ba=1, F=2, Pt=1, S=1 → already balanced!

Coefficients: 1, 1, 1, 1

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10) N₂ + H₂ → NH₃

Left: N=2, H=2
Right: N=1, H=3 → need 2 NH₃ → then N=2, H=6 → so 3 H₂

N₂ + 3 H₂ → 2 NH₃

Left: N=2, H=6
Right: N=2, H=6 ✔

Coefficients: 1, 3, 2

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11) NaF + Br₂ → NaBr + F₂

Left: Na=1, F=1, Br=2
Right: Na=1, Br=1, F=2 → need 2 NaF and 2 NaBr

Try: 2 NaF + Br₂ → 2 NaBr + F₂

Left: Na=2, F=2, Br=2
Right: Na=2, Br=2, F=2 ✔

Coefficients: 2, 1, 2, 1

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12) Pb(OH)₂ + HCl → H₂O + PbCl₂

Left: Pb=1, O=2, H=2+1=3? Wait — Pb(OH)₂ has 2 O and 2 H from OH, plus HCl has 1 H and 1 Cl.

Better:
Pb(OH)₂ → Pb=1, O=2, H=2
HCl → H=1, Cl=1
Total left: Pb=1, O=2, H=3, Cl=1

Right: H₂O → H=2, O=1; PbCl₂ → Pb=1, Cl=2 → total: Pb=1, Cl=2, H=2, O=1 → not balanced.

Need 2 HCl → then left: H=2+2=4? Wait:

Try: Pb(OH)₂ + 2 HCl → 2 H₂O + PbCl₂

Left: Pb=1, O=2, H=2 (from OH) + 2 (from 2 HCl) = 4, Cl=2
Right: 2 H₂O → H=4, O=2; PbCl₂ → Pb=1, Cl=2 ✔

Coefficients: 1, 2, 2, 1

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13) AlBr₃ + K₂SO₄ → KBr + Al₂(SO₄)₃

Left: Al=1, Br=3, K=2, S=1, O=4
Right: K=1, Br=1, Al=2, S=3, O=12 → need to balance Al and SO₄

Put 2 AlBr₃ → Al=2, Br=6
Put 3 K₂SO₄ → K=6, S=3, O=12
Then right: Al₂(SO₄)₃ → Al=2, S=3, O=12
And KBr → need 6 KBr for K=6, Br=6

So: 2 AlBr₃ + 3 K₂SO₄ → 6 KBr + Al₂(SO₄)₃

Check:
Left: Al=2, Br=6, K=6, S=3, O=12
Right: K=6, Br=6, Al=2, S=3, O=12 ✔

Coefficients: 2, 3, 6, 1

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14) CH₄ + O₂ → CO₂ + H₂O

Left: C=1, H=4, O=2
Right: C=1, O=2+1=3? Wait — CO₂ has 2 O, H₂O has 1 O → total 3 O, but H=2

Need 2 H₂O → then H=4, O=2 (from CO₂) + 2 (from 2 H₂O) = 4 → so need 2 O₂

CH₄ + 2 O₂ → CO₂ + 2 H₂O

Left: C=1, H=4, O=4
Right: C=1, O=2+2=4, H=4 ✔

Coefficients: 1, 2, 1, 2

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15) Na₃PO₄ + CaCl₂ → NaCl + Ca₃(PO)₂

Left: Na=3, P=1, O=4, Ca=1, Cl=2
Right: Na=1, Cl=1, Ca=3, P=2, O=8 → need 2 Na₃PO₄ and 3 CaCl₂

Try: 2 Na₃PO₄ + 3 CaCl₂ → 6 NaCl + Ca₃(PO₄)₂

Left: Na=6, P=2, O=8, Ca=3, Cl=6
Right: Na=6, Cl=6, Ca=3, P=2, O=8 ✔

Coefficients: 2, 3, 6, 1

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16) K + Cl₂ → KCl

Left: K=1, Cl=2
Right: K=1, Cl=1 → need 2 KCl and 2 K

2 K + Cl₂ → 2 KCl

Left: K=2, Cl=2
Right: K=2, Cl=2 ✔

Coefficients: 2, 1, 2

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17) Al + HCl → H₂ + AlCl₃

Left: Al=1, H=1, Cl=1
Right: H=2, Al=1, Cl=3 → need 6 HCl and 3 H₂ and 2 Al

Try: 2 Al + 6 HCl → 3 H₂ + 2 AlCl₃

Left: Al=2, H=6, Cl=6
Right: H=6, Al=2, Cl=6 ✔

Coefficients: 2, 6, 3, 2

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18) N₂ + F₂ → NF₃

Left: N=2, F=2
Right: N=1, F=3 → need 2 NF₃ → N=2, F=6 → so 3 F₂

N₂ + 3 F₂ → 2 NF₃

Left: N=2, F=6
Right: N=2, F=6 ✔

Coefficients: 1, 3, 2

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19) SO₂ + Li₂Se → SSe₂ + Li₂O

Left: S=1, O=2, Li=2, Se=1
Right: S=1, Se=2, Li=2, O=1 → need 2 Li₂O and 2 Li₂Se? Wait

Try: SO₂ + 2 Li₂Se → SSe₂ + 2 Li₂O

Left: S=1, O=2, Li=4, Se=2
Right: S=1, Se=2, Li=4, O=2 ✔

Coefficients: 1, 2, 1, 2

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20) NH₃ + H₂SO₄ → (NH₄)₂SO₄

Left: N=1, H=3+2=5, S=1, O=4
Right: N=2, H=8, S=1, O=4 → need 2 NH₃

2 NH₃ + H₂SO₄ → (NH₄)₂SO₄

Left: N=2, H=6+2=8, S=1, O=4
Right: N=2, H=8, S=1, O=4 ✔

Coefficients: 2, 1, 1

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Final Answer:
1) 1, 3, 3, 1
2) 1, 1, 1, 2
3) 1, 3, 2
4) 2, 1, 1, 2
5) 2, 1, 1, 2
6) 1, 2, 1, 2
7) 2, 1, 1, 1
8) 1, 3, 3, 1
9) 1, 1, 1, 1
10) 1, 3, 2
11) 2, 1, 2, 1
12) 1, 2, 2, 1
13) 2, 3, 6, 1
14) 1, 2, 1, 2
15) 2, 3, 6, 1
16) 2, 1, 2
17) 2, 6, 3, 2
18) 1, 3, 2
19) 1, 2, 1, 2
20) 2, 1, 1