Balancing Equations Worksheet featuring 20 chemical reactions to practice balancing.

Let's solve each of these chemical equations by balancing them. Balancing a chemical equation means ensuring the number of atoms of each element is equal on both sides (reactants and products), following the Law of Conservation of Mass.

We’ll go through each one step-by-step, showing how to balance it.

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1) Na₃PO₄ + KOH → NaOH + K₃PO₄

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

We need 3 Na on right → put 3 in front of NaOH
We need 3 K on left → put 3 in front of KOH

✔ Balanced:
1 Na₃PO₄ + 3 KOH → 3 NaOH + 1 K₃PO₄

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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 on right → 2 LiF
Now F: 2 on both sides
Li: 2 on both sides

✔ Balanced:
1 MgF₂ + 1 Li₂CO₃ → 1 MgCO₃ + 2 LiF

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3) P₄ + O₂ → P₂O₃

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

We need 2 P₂O₃ → gives 4P, 6O
So O₂ must be 3 → 3×2 = 6 O

P₄ already has 4P → matches

✔ Balanced:
1 P₄ + 3 O₂ → 2 P₂O₃

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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

Be(NO₃)₂ needs 2 NO₃ → so need 2 RbNO₃
Then Rb: 2 on left → need 2 RbF on right → F: 2 on both sides

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

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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

Cu(NO₃)₂ has 2 NO₃ → need 2 AgNO₃
Then Ag: 2 on left → need 2 Ag on right

✔ Balanced:
2 AgNO₃ + 1 Cu → 1 Cu(NO₃)₂ + 2 Ag

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6) CF₄ + Br₂ → CBr₄ + F₂

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

CBr₄ needs 4 Br → need 2 Br₂ → 2×2 = 4 Br
F: 4 on left → but only 2 on right → need 2 F₂ → 4 F

✔ Balanced:
1 CF₄ + 2 Br₂ → 1 CBr₄ + 2 F₂

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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

Cu(CN)₂ has 2 CN → so need 2 HCN
Then H: 2 on left → matches H₂SO₄

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

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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 → 3 F and 3 Cs
So need 3 Cs on left

✔ Balanced:
1 GaF₃ + 3 Cs → 3 CsF + 1 Ga

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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

All atoms match as written.

✔ Balanced:
1 BaS + 1 PtF₂ → 1 BaF₂ + 1 PtS

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10) N₂ + H₂ → NH₃

- Left: N=2, H=2
- Right: N=1, H=3

Need 2 NH₃ → 2N, 6H
So H₂ must be 3 → 6H
N₂ provides 2N → perfect

✔ Balanced:
1 N₂ + 3 H₂ → 2 NH₃

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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 → 2F and 2Na
Then 2 NaBr → 2Na, 2Br
But Br₂ has 2Br → okay

F₂ needs 2F → from 2 NaF → yes

✔ Balanced:
2 NaF + 1 Br₂ → 2 NaBr + 1 F₂

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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

PbCl₂ needs 2 Cl → need 2 HCl
Then H: 2 from HCl + 2 from OH → total 4H on left
On right: H₂O has 2H → need 2 H₂O → 4H

O: Left: 2 from OH + 2 from HCl? No — HCl has no O

Wait:

Pb(OH)₂ has: Pb, 2O, 2H
HCl has: H, Cl

So total left: Pb=1, O=2, H=2+1=3, Cl=1

Right: H₂O → H=2, O=1; PbCl₂ → Pb=1, Cl=2

So need 2 HCl → Cl=2, H=2 → total H=2 (from HCl) + 2 (from OH) = 4H
But H₂O has only 2H per molecule → need 2 H₂O → 4H, 2O

Left: O=2 from Pb(OH)₂ → matches

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

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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

Al₂(SO₄)₃ has 2Al, 3SO₄ → so need 2 AlBr₃ and 3 K₂SO₄

Then:
- Al: 2 on both sides
- Br: 2×3=6 → need 6 KBr
- K: 3×2=6 → matches 6 KBr
- S: 3 on both sides
- O: 3×4=12 → matches

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

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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

Need 2 H₂O → 4H, 2O
Then O₂ needed: CO₂ has 2O, H₂O has 2O → total 4O → need 2 O₂

✔ Balanced:
1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O

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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

Ca₃(PO₄)₂ needs 3 Ca and 2 PO₄ → so need 2 Na₃PO₄ and 3 CaCl₂

Then:
- Na: 2×3=6 → need 6 NaCl
- Cl: 3×2=6 → matches
- Ca: 3 → matches
- P: 2 → matches
- O: 2×4=8 → matches

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

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16) K + Cl₂ → KCl

- Left: K=1, Cl=2
- Right: K=1, Cl=1

Need 2 KCl → 2K, 2Cl
So need 2 K on left

✔ Balanced:
2 K + 1 Cl₂ → 2 KCl

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17) Al + HCl → H₂ + AlCl₃

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

AlCl₃ needs 3 Cl → need 3 HCl
Then H: 3H → need 3/2 H₂ → but we want whole numbers

Multiply all by 2:

→ 2 Al + 6 HCl → 3 H₂ + 2 AlCl₃

✔ Balanced:
2 Al + 6 HCl → 3 H₂ + 2 AlCl₃

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18) N₂ + F₂ → NF₃

- Left: N=2, F=2
- Right: N=1, F=3

Need 2 NF₃ → 2N, 6F
So need 3 F₂ → 6F

N₂ gives 2N → perfect

✔ Balanced:
1 N₂ + 3 F₂ → 2 NF₃

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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 SO₂ → 2S, 4O
Need 2 Li₂Se → 2Li₂, 2Se → but Li₂Se has 2Li, 1Se → 2×2=4Li → too many

Wait:

We need SSe₂ → 1S, 2Se → so need 2 Li₂Se → 2Se, 4Li
But Li₂Se has 2Li → 2×2=4Li → okay

But then Li₂O → needs 2Li → so need 2 Li₂O → 4Li, 2O

But SO₂ has 2O → 2 SO₂ → 4O → too much

Let’s try:

- Want 1 SSe₂ → needs 1S, 2Se
- So need 1 SO₂ and 2 Li₂Se → gives: S=1, O=2, Li=4, Se=2
- Products: SSe₂ and Li₂O → need 2 Li₂O → 4Li, 2O

Perfect!

✔ Balanced:
1 SO₂ + 2 Li₂Se → 1 SSe₂ + 2 Li₂O

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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

(NH₄)₂SO₄ has 2N, 8H → so need 2 NH₃ → 2N, 6H
Then H₂SO₄ has 2H → total H: 6+2=8 → matches

✔ Balanced:
2 NH₃ + 1 H₂SO₄ → 1 (NH₄)₂SO₄

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## ✔ Final Answer: All 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) 1 SO₂ + 2 Li₂Se → 1 SSe₂ + 2 Li₂O
20) 2 NH₃ + 1 H₂SO₄ → 1 (NH₄)₂SO₄

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✔ All equations are now balanced! Let me know if you'd like explanations for any specific ones.