Balancing Equations Worksheet featuring 20 chemical reactions to balance, published by Cavalcade Publishing in 2004.

Let's solve each of these chemical equations by balancing them. The goal is to ensure that the number of atoms of each element is the same on both sides of the equation.

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1) Na₃PO₄ + KOH → NaOH + K₃PO₄

- Left: Na=3, P=1, O=4+1=5, K=1, H=1
- Right: Na=1, O=1+4=5, H=1, K=3, P=1

We need 3 Na on right → 3 NaOH
Need 3 K on left → 3 KOH

So:
1 Na₃PO₄ + 3 KOH → 3 NaOH + 1 K₃PO₄

✔ Balanced.

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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 → not balanced

Li and F need balancing. Need 2 LiF on right → 2 LiF

Now right has: Li=2, F=2 → good

So:
1 MgF₂ + 1 Li₂CO₃ → 1 MgCO₃ + 2 LiF

✔ Balanced.

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3) P₄ + O₂ → P₂O₃

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

We need even number of P on right. Use 2 P₂O₃ → P=4, O=6

Then need O₂ = 3 (since 3×2=6)

So:
1 P₄ + 3 O₂ → 2 P₂O₃

✔ Balanced.

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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 → unbalanced

Need 2 NO₃ on right → 2 RbNO₃
Then 2 Rb on left → 2 RbF on right

So:
2 RbNO₃ + 1 BeF₂ → 1 Be(NO₃)₂ + 2 RbF

✔ Balanced.

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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 more NO₃

Use 2 AgNO₃ → gives 2 Ag, 2 N, 6 O
Right needs 2 Ag → 2 Ag

So:
2 AgNO₃ + 1 Cu → 1 Cu(NO₃)₂ + 2 Ag

✔ Balanced.

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6) CF₄ + Br₂ → CBr₄ + F₂

- Left: C=1, F=4, Br=2
- Right: C=1, Br=4, F=2 → unbalanced

Need 4 Br on left → 2 Br₂
Need 4 F on right → 2 F₂

So:
1 CF₄ + 2 Br₂ → 1 CBr₄ + 2 F₂

✔ Balanced.

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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 → unbalanced

Need 2 HCN → 2 H, 2 C, 2 N
Then right: Cu(CN)₂ has 2 CN → good

But H₂SO₄ has 2 H → now H balanced

So:
2 HCN + 1 CuSO₄ → 1 H₂SO₄ + 1 Cu(CN)₂

✔ Balanced.

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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 F on right → 3 CsF

So need 3 Cs on left

1 GaF₃ + 3 Cs → 3 CsF + 1 Ga

✔ Balanced.

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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!

1 BaS + 1 PtF₂ → 1 BaF₂ + 1 PtS

✔ Balanced.

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10) N₂ + H₂ → NH₃

- Left: N=2, H=2
- Right: N=1, H=3 → need multiples

Use 2 NH₃ → N=2, H=6 → need 3 H₂

So:
1 N₂ + 3 H₂ → 2 NH₃

✔ Balanced.

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11) NaF + Br₂ → NaBr + F₂

- Left: Na=1, F=1, Br=2
- Right: Na=1, Br=1, F=2 → unbalanced

Need 2 NaF → 2 Na, 2 F
Need 2 NaBr → 2 Na, 2 Br
But Br₂ gives 2 Br → OK

F₂ on right needs 2 F → so 2 NaF gives 2 F → good

So:
2 NaF + 1 Br₂ → 2 NaBr + 1 F₂

✔ Balanced.

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12) Pb(OH)₂ + HCl → H₂O + PbCl₂

- Left: Pb=1, O=2, H=2+1=3, Cl=1
- Right: H=2, O=1, Pb=1, Cl=2 → unbalanced

Need 2 HCl → Cl=2, H=2
Then total H on left: 2 (from OH) + 2 (from HCl) = 4 H
Right: H₂O has 2 H → need 2 H₂O

So:
1 Pb(OH)₂ + 2 HCl → 2 H₂O + 1 PbCl₂

Check:
- Left: Pb=1, O=2+2=4? Wait — Pb(OH)₂ has 2 O and 2 H, plus 2 HCl → 2 H and 2 Cl
- Total: Pb=1, O=2, H=4, Cl=2
- Right: 2 H₂O → 4 H, 2 O; PbCl₂ → Pb, 2 Cl → all match

✔ Balanced.

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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 → unbalanced

Need 2 Al on left → 2 AlBr₃ → Al=2, Br=6
Need 3 SO₄ → 3 K₂SO₄ → K=6, S=3, O=12
Right: Al₂(SO₄)₃ → Al=2, S=3, O=12
KBr → need 6 KBr → K=6, Br=6

So:
2 AlBr₃ + 3 K₂SO₄ → 6 KBr + 1 Al₂(SO₄)₃

✔ Balanced.

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14) CH₄ + O₂ → CO₂ + H₂O

Classic combustion.

- Left: C=1, H=4, O=2
- Right: C=1, O=2+1=3, H=2 → unbalanced

Need 2 H₂O → H=4 → good
Now O: 2 (CO₂) + 1 (H₂O) = 3 → but H₂O has 2 O → 2 H₂O → 2 O from water

Total O on right: CO₂ has 2, 2 H₂O has 2 → total 4 O → need 2 O₂

So:
1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O

✔ Balanced.

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15) Na₃PO₄ + CaCl₂ → NaCl + Ca₃(PO₄)₂

- Left: Na=3, P=1, O=4, Ca=1, Cl=2
- Right: Na=1, Cl=1, Ca=3, P=2, O=8 → unbalanced

Need 2 PO₄ → 2 Na₃PO₄ → Na=6, P=2, O=8
Need 3 Ca → 3 CaCl₂ → Ca=3, Cl=6
Right: Ca₃(PO₄)₂ → Ca=3, P=2, O=8
NaCl → need 6 NaCl → Na=6, Cl=6

So:
2 Na₃PO₄ + 3 CaCl₂ → 6 NaCl + 1 Ca₃(PO₄)₂

✔ Balanced.

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16) K + Cl₂ → KCl

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

So need 2 K on left

2 K + 1 Cl₂ → 2 KCl

✔ Balanced.

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17) Al + HCl → H₂ + AlCl₃

- Left: Al=1, H=1, Cl=1
- Right: H=2, Al=1, Cl=3 → unbalanced

Need 3 Cl → 3 HCl → H=3, Cl=3
But H₂ → 2 H → need 3/2 H₂ → better use whole numbers

Multiply by 2:

2 Al → 2 Al
6 HCl → 6 H, 6 Cl
→ 3 H₂ (6 H), 2 AlCl₃ (2 Al, 6 Cl)

So:
2 Al + 6 HCl → 3 H₂ + 2 AlCl₃

✔ Balanced.

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18) N₂ + F₂ → NF₃

- Left: N=2, F=2
- Right: N=1, F=3 → need multiple

Use 2 NF₃ → N=2, F=6 → need 3 F₂ → F=6

So:
1 N₂ + 3 F₂ → 2 NF₃

✔ Balanced.

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19) SO₂ + Li₂Se → SSE₂ + Li₂O

- Left: S=1, O=2, Li=2, Se=1
- Right: S=2, Se=2, Li=2, O=1 → unbalanced

Need 2 S on right → 2 SSE₂ → S=2, Se=4 → need 4 Se on left → 2 Li₂Se → Se=2 → no

Wait: SSE₂ has one S and two Se → so 2 SSE₂ → S=2, Se=4

Left: SO₂ → S=1 → need 2 SO₂ → S=2, O=4
Li₂Se → need 2 Li₂Se → Li=4, Se=2 → but we need Se=4 → need 4 Li₂Se → Li=8, Se=4

Right: 2 SSE₂ → S=2, Se=4
Li₂O → need 4 Li₂O → Li=8, O=4

Left: 2 SO₂ → O=4, S=2
4 Li₂Se → Li=8, Se=4

So:
2 SO₂ + 4 Li₂Se → 2 SSE₂ + 4 Li₂O

✔ Balanced.

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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 N, 8 H

So need 2 NH₃ → N=2, H=6
Plus H₂SO₄ → H=2 → total H=8 → good

So:
2 NH₃ + 1 H₂SO₄ → 1 (NH₄)₂SO₄

✔ Balanced.

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## ✔ Final Balanced Equations:

1) 1 Na₃PO₄ + 3 KOH → 3 NaOH + 1 K₃PO₄
2) 1 MgF₂ + 1 Li₂CO₃ → 1 MgCO₃ + 2 LiF
3) 1 P₄ + 3 O₂ → 2 P₂O₃
4) 2 RbNO₃ + 1 BeF₂ → 1 Be(NO₃)₂ + 2 RbF
5) 2 AgNO₃ + 1 Cu → 1 Cu(NO₃)₂ + 2 Ag
6) 1 CF₄ + 2 Br₂ → 1 CBr₄ + 2 F₂
7) 2 HCN + 1 CuSO₄ → 1 H₂SO₄ + 1 Cu(CN)₂
8) 1 GaF₃ + 3 Cs → 3 CsF + 1 Ga
9) 1 BaS + 1 PtF₂ → 1 BaF₂ + 1 PtS
10) 1 N₂ + 3 H₂ → 2 NH₃
11) 2 NaF + 1 Br₂ → 2 NaBr + 1 F₂
12) 1 Pb(OH)₂ + 2 HCl → 2 H₂O + 1 PbCl₂
13) 2 AlBr₃ + 3 K₂SO₄ → 6 KBr + 1 Al₂(SO₄)₃
14) 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O
15) 2 Na₃PO₄ + 3 CaCl₂ → 6 NaCl + 1 Ca₃(PO₄)₂
16) 2 K + 1 Cl₂ → 2 KCl
17) 2 Al + 6 HCl → 3 H₂ + 2 AlCl₃
18) 1 N₂ + 3 F₂ → 2 NF₃
19) 2 SO₂ + 4 Li₂Se → 2 SSE₂ + 4 Li₂O
20) 2 NH₃ + 1 H₂SO₄ → 1 (NH₄)₂SO₄

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📌 How to Balance Chemical Equations:

1. Count atoms of each element on both sides.
2. Adjust coefficients (numbers in front) to balance atoms.
3. Start with elements that appear in only one compound on each side.
4. Save elements like H and O for last if they appear in multiple compounds.
5. Check all atoms again after balancing.

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