Balancing Equations Worksheet featuring 22 chemical reactions for practice.

Let's go through each of the 22 chemical equations and balance them step by step. The goal is to ensure that the number of atoms of each element is the same on both sides of the equation (conservation of mass).

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1. Ca + O₂ → CaO

- Left: 1 Ca, 2 O
- Right: 1 Ca, 1 O
- Balance O: Put 2 CaO on right → now 2 O and 2 Ca
- Then put 2 Ca on left

✔ Balanced:
2Ca + O₂ → 2CaO

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2. H₂ + Cl₂ → HCl

- Left: 2 H, 2 Cl
- Right: 1 H, 1 Cl
- Need 2 HCl to balance

✔ Balanced:
H₂ + Cl₂ → 2HCl

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3. P₄ + S₈ → P₂S₅

- P₄ has 4 P; S₈ has 8 S
- P₂S₅ has 2 P and 5 S
- We need to make P and S match.

Find LCM:
- For P: 4 and 2 → LCM = 4 → use 2 P₂S₅ (gives 4 P, 10 S)
- But S₈ gives 8 S → not divisible by 10
- Try multiple of P₂S₅: 4 P₂S₅ → 8 P, 20 S
- P₄ → 4 P → need 2 P₄ (8 P)
- S₈ → 8 S → need 2.5 S₈ → not integer

Try 5 P₂S₅ → 10 P, 25 S → need 2.5 P₄ (10 P), 3.125 S₈ → no

Wait — let’s find smallest integers:

We want:
- P: 4a = 2b → a = b/2 → so b even
- S: 8c = 5b → 8c = 5b → b must be multiple of 8, c multiple of 5

Try b = 8 → then P₂S₅: 8 molecules → 16 P, 40 S
- So P₄: 16 P → need 4 P₄
- S₈: 40 S → need 5 S₈

✔ Balanced:
4P₄ + 5S₈ → 8P₂S₅

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4. CO + O₂ → CO₂

- Left: 1 C, 1 O from CO, 2 O from O₂ → total 1 C, 3 O
- Right: 1 C, 2 O
- Balance C and O

Try 2CO + O₂ → 2CO₂
- Left: 2 C, 2 O + 2 O = 4 O
- Right: 2 C, 4 O → balanced

✔ Balanced:
2CO + O₂ → 2CO₂

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5. C₂H₆ + O₂ → CO₂ + H₂O

- Combustion reaction.
- Left: 2 C, 6 H, 2 O
- Right: 1 C, 2 H, 3 O (from CO₂ and H₂O)

Balance C: 2CO₂
Balance H: 6 H → 3 H₂O
Now right: 2 C, 6 H, 4 O (from 2CO₂) + 3 O (from 3H₂O) = 7 O
Left: O₂ → need 7/2 = 3.5 O₂ → multiply all by 2

Multiply entire equation by 2:
- 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

✔ Balanced:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

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6. C₂H₅OH + O₂ → CO₂ + H₂O

- Ethanol combustion
- C₂H₅OH: 2 C, 6 H, 1 O
- Left: 2 C, 6 H, 1 O + 2 O per O₂
- Right: CO₂ and H₂O

Balance C: 2CO₂
Balance H: 6 H → 3 H₂O
Now right: 2 CO₂ → 4 O, 3 H₂O → 3 O → total 7 O
Left: 1 O from ethanol, rest from O₂ → need 6 O from O₂ → 3 O₂

Check:
- Left: C₂H₅OH + 3O₂ → 2 C, 6 H, 1 O + 6 O = 7 O
- Right: 2CO₂ + 3H₂O → 2 C, 6 H, 4 O + 3 O = 7 O

✔ Balanced:
C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O

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7. FeS₂ + O₂ → Fe₂O₃ + SO₂

- Left: Fe, 2 S, 2 O
- Right: 2 Fe, 3 O, 1 S, 2 O → total 1 S, 5 O? No: SO₂ has 1 S, 2 O

Need:
- Fe: 2 Fe → 2 FeS₂
- S: 2 FeS₂ → 4 S → need 4 SO₂
- O: 4 SO₂ → 8 O, Fe₂O₃ → 3 O → total 11 O
- Left: O₂ → need 11/2 = 5.5 → multiply by 2

So:
- 4 FeS₂ → 4 Fe, 8 S
- 2 Fe₂O₃ → 4 Fe, 6 O
- 8 SO₂ → 8 S, 16 O
- Total O on right: 6 + 16 = 22 O → need 11 O₂

Left: 4 FeS₂ + 11 O₂ → 2 Fe₂O₃ + 8 SO₂

✔ Balanced:
4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂

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8. P₄ + O₃ → P₂O₅

- P₄ → 4 P
- P₂O₅ → 2 P → need 2 P₂O₅ → 4 P, 10 O
- O₃ → 3 O per molecule → need 10/3 → not integer

Try 3 P₂O₅ → 6 P → need 1.5 P₄ → not good

Try 2 P₂O₅ → 4 P → 1 P₄ → good
- O needed: 10 O → O₃ provides 3 O → need 10/3 → not integer

Try 3 P₄ → 12 P → need 6 P₂O₅ → 12 P, 30 O
- O₃ → 3 O → need 10 O₃

✔ Balanced:
3P₄ + 10O₃ → 6P₂O₅

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9. NH₃ + O₂ → N₂O₃ + H₂O

- Left: N, 3 H, 2 O
- Right: 2 N, 3 O, 2 H, 1 O → total 2 N, 4 O, 2 H

Balance N: 2 NH₃ → 2 N
H: 6 H → 3 H₂O
O: Right: N₂O₃ → 3 O, 3 H₂O → 3 O → total 6 O
Left: O₂ → need 3 O₂

Check:
- Left: 2NH₃ + 3O₂ → 2 N, 6 H, 6 O
- Right: N₂O₃ + 3H₂O → 2 N, 3 O + 3 O = 6 O, 6 H

✔ Balanced:
2NH₃ + 3O₂ → N₂O₃ + 3H₂O

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10. Si₅H₁₁ + Br₂ → Si + HBr

- Left: 5 Si, 11 H, 2 Br
- Right: 1 Si, 1 H, 1 Br

Balance Si: 5 Si → 5 Si
H: 11 H → 11 HBr
Br: 11 Br → need 11/2 Br₂ → 5.5 Br₂

Multiply by 2:
- 2Si₅H₁₁ + 11Br₂ → 10Si + 22HBr

✔ Balanced:
2Si₅H₁₁ + 11Br₂ → 10Si + 22HBr

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11. FeCl₃ + NaOH → Fe(OH)₃ + NaCl

- Left: Fe, 3 Cl, Na, O, H
- Right: Fe, 3 O, 3 H, Na, Cl

Balance Fe: 1 each
Cl: 3 on left → 3 NaCl on right
Na: 3 Na → 3 NaOH on left
OH: 3 OH → 3 NaOH → matches

✔ Balanced:
FeCl₃ + 3NaOH → Fe(OH)₃ + 3NaCl

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12. Fe + H₂O → Fe₂O₃ + H₂

- Left: Fe, 2 H, 1 O
- Right: 2 Fe, 3 O, 2 H

Balance Fe: 2 Fe on left
O: 3 O → need 3 H₂O
H: 6 H → 3 H₂

Check:
- Left: 2Fe + 3H₂O → 2 Fe, 6 H, 3 O
- Right: Fe₂O₃ + 3H₂ → 2 Fe, 3 O, 6 H

✔ Balanced:
2Fe + 3H₂O → Fe₂O₃ + 3H₂

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13. C₁₀H₂₂ + Cl₂ → C + HCl

- Left: 10 C, 22 H, 2 Cl
- Right: 1 C, 1 H, 1 Cl

Balance C: 10 C → 10 C
H: 22 H → 22 HCl
Cl: 22 Cl → 11 Cl₂

✔ Balanced:
C₁₀H₂₂ + 11Cl₂ → 10C + 22HCl

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14. FeCl₃ + Na₂SO₄ → Fe₂(SO₄)₃ + NaCl

- Left: Fe, 3 Cl, 2 Na, S, 4 O
- Right: 2 Fe, 3 S, 12 O, Na, Cl

Balance Fe: 2 Fe → 2 FeCl₃
SO₄: 3 SO₄ → 3 Na₂SO₄
Na: 6 Na → 6 NaCl
Cl: 6 Cl → 6 NaCl

✔ Balanced:
2FeCl₃ + 3Na₂SO₄ → Fe₂(SO₄)₃ + 6NaCl

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15. Al + HNO₃ → Al(NO₃)₃ + H₂

- Left: Al, H, N, 3 O
- Right: Al, 3 N, 9 O, 2 H

Balance Al: 1 each
NO₃: 3 → need 3 HNO₃
H: 3 H → need 3/2 H₂ → not integer

So multiply by 2:
- 2Al + 6HNO₃ → 2Al(NO₃)₃ + 3H₂

✔ Balanced:
2Al + 6HNO₃ → 2Al(NO₃)₃ + 3H₂

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16. K + H₂O → KOH + H₂

- Left: K, 2 H, 1 O
- Right: K, O, H, 2 H

Balance K: 1 each
H: Left 2 H, Right: 1 in KOH, 2 in H₂ → total 3 H → mismatch

Try 2K + 2H₂O → 2KOH + H₂
- Left: 2 K, 4 H, 2 O
- Right: 2 K, 2 O, 2 H in KOH, 2 H in H₂ → total 4 H → yes

✔ Balanced:
2K + 2H₂O → 2KOH + H₂

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17. C₃H₈ + S₈ → CS₂ + H₂S

- Left: 3 C, 8 H, 8 S
- Right: 1 C, 2 S, 2 H, 1 S → total 1 C, 3 S, 2 H

Balance C: 3 C → 3 CS₂
S: 3 CS₂ → 6 S, plus H₂S → let say x H₂S → x S
Total S: 6 + x → must equal 8 → x = 2 → 2 H₂S
H: 8 H → 2 H₂S → 4 H → missing 4 H → but only 2 H per H₂S → wait

Wait: H₂S has 2 H → 2 H₂S → 4 H → but we have 8 H → need 4 H₂S → 4 S

But CS₂ uses 6 S → total S = 6 + 4 = 10 → but S₈ gives only 8 S → too much

Try:
- C₃H₈ → 3 C, 8 H
- Let a CS₂, b H₂S → 3 C → a=3
- H: 8 H → 2b = 8 → b=4
- S: 3×2 + 4 = 6+4=10 S → need 10/8 = 1.25 S₈ → not integer

Try 4 C₃H₈ → 12 C, 32 H
→ 12 CS₂ → 24 S
→ 16 H₂S → 32 H → good
→ total S = 24 + 16 = 40 → need 40/8 = 5 S₈

✔ Balanced:
4C₃H₈ + 5S₈ → 12CS₂ + 16H₂S

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18. MgCl₂ + Al(OH)₃ → AlCl₃ + Mg(OH)₂

- Double displacement
- MgCl₂ + Al(OH)₃ → AlCl₃ + Mg(OH)₂

Balance Cl: 2 vs 3 → LCM = 6
→ 3 MgCl₂ → 6 Cl
→ 2 AlCl₃ → 6 Cl
Mg: 3 → 3 Mg(OH)₂
Al: 2 → 2 Al(OH)₃
OH: 2 Al(OH)₃ → 6 OH → 3 Mg(OH)₂ → 6 OH → good

✔ Balanced:
3MgCl₂ + 2Al(OH)₃ → 2AlCl₃ + 3Mg(OH)₂

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19. ZnFeS₂ + O₂ → Zn + FeO + SO₂

- Left: Zn, Fe, 2 S, 2 O
- Right: Zn, Fe, O, S, 2 O

Balance S: 2 S → 2 SO₂
O: 2 SO₂ → 4 O, FeO → 1 O → total 5 O → need 5/2 O₂

Zn and Fe: 1 each → fine

So:
ZnFeS₂ + 5/2 O₂ → Zn + FeO + 2SO₂

Multiply by 2:
2ZnFeS₂ + 5O₂ → 2Zn + 2FeO + 4SO₂

✔ Balanced:
2ZnFeS₂ + 5O₂ → 2Zn + 2FeO + 4SO₂

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20. P₄O₁₀ + H₂O → H₃PO₄

- P₄O₁₀ → 4 P, 10 O
- H₃PO₄ → 1 P, 4 O, 3 H, 1 O → total 5 O

Need 4 H₃PO₄ → 4 P, 16 O, 12 H
Left: P₄O₁₀ → 4 P, 10 O → need 6 more O → H₂O → 1 O per molecule → 6 H₂O → 6 O, 12 H

Right: 4 H₃PO₄ → 4 P, 16 O, 12 H
Left: P₄O₁₀ + 6H₂O → 4 P, 10 O + 6 O = 16 O, 12 H

✔ Balanced:
P₄O₁₀ + 6H₂O → 4H₃PO₄

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21. KClO₃ → KCl + KClO₄

- Decomposition
- Left: K, Cl, 3 O
- Right: K, Cl, K, Cl, 4 O → 2 K, 2 Cl, 4 O

Not balanced.

Try:
KClO₃ → KCl + KClO₄
But atoms don't match.

This is a disproportionation reaction.

Let’s assume:
a KClO₃ → b KCl + c KClO₄

K: a = b + c
Cl: a = b + c → same
O: 3a = 4c

From O: 3a = 4c → a = 4k, c = 3k → then b = a - c = 4k - 3k = k

So k=1: a=4, b=1, c=3

Check:
4 KClO₃ → 1 KCl + 3 KClO₄
- K: 4 → 1 + 3 = 4
- Cl: 4 → 1 + 3 = 4
- O: 12 → 3×4 = 12

✔ Balanced:
4KClO₃ → KCl + 3KClO₄

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22. Mg₃N₂ + H₂O → Mg(OH)₂ + NH₃

- Left: 3 Mg, 2 N, 2 H, 1 O
- Right: Mg, 2 O, 2 H, N, 3 H → total 1 Mg, 2 O, 5 H, 1 N

Balance Mg: 3 → 3 Mg(OH)₂
N: 2 → 2 NH₃
H: 3 Mg(OH)₂ → 6 H, 2 NH₃ → 6 H → total 12 H
O: 3 Mg(OH)₂ → 6 O → need 6 H₂O

Check:
- Left: Mg₃N₂ + 6H₂O → 3 Mg, 2 N, 12 H, 6 O
- Right: 3Mg(OH)₂ + 2NH₃ → 3 Mg, 6 O, 6 H, 2 N, 6 H → total 12 H

✔ Balanced:
Mg₃N₂ + 6H₂O → 3Mg(OH)₂ + 2NH₃

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✔ Final Balanced Equations:

1. 2Ca + O₂ → 2CaO
2. H₂ + Cl₂ → 2HCl
3. 4P₄ + 5S₈ → 8P₂S₅
4. 2CO + O₂ → 2CO₂
5. 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
6. C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
7. 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
8. 3P₄ + 10O₃ → 6P₂O₅
9. 2NH₃ + 3O₂ → N₂O₃ + 3H₂O
10. 2Si₅H₁₁ + 11Br₂ → 10Si + 22HBr
11. FeCl₃ + 3NaOH → Fe(OH)₃ + 3NaCl
12. 2Fe + 3H₂O → Fe₂O₃ + 3H₂
13. C₁₀H₂₂ + 11Cl₂ → 10C + 22HCl
14. 2FeCl₃ + 3Na₂SO₄ → Fe₂(SO₄)₃ + 6NaCl
15. 2Al + 6HNO₃ → 2Al(NO₃)₃ + 3H₂
16. 2K + 2H₂O → 2KOH + H₂
17. 4C₃H₈ + 5S₈ → 12CS₂ + 16H₂S
18. 3MgCl₂ + 2Al(OH)₃ → 2AlCl₃ + 3Mg(OH)₂
19. 2ZnFeS₂ + 5O₂ → 2Zn + 2FeO + 4SO₂
20. P₄O₁₀ + 6H₂O → 4H₃PO₄
21. 4KClO₃ → KCl + 3KClO₄
22. Mg₃N₂ + 6H₂O → 3Mg(OH)₂ + 2NH₃

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📝 Summary:

All equations are now balanced using the inspection method, ensuring atom conservation. Some required finding least common multiples or handling fractional coefficients (then multiplying through). These are standard techniques in balancing chemical equations.

Let me know if you'd like explanations for any specific one!