Balancing Equations Worksheet for chemistry practice, featuring 37 problems to balance chemical equations.

Let's solve the Balancing Equations Worksheet step by step. I'll go through each chemical equation, balance it using the law of conservation of mass, and provide the balanced coefficients.

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🔁 How to Balance Chemical Equations:

1. Count atoms of each element on both sides.
2. Use coefficients (numbers in front of formulas) to balance atoms.
3. Start with elements that appear in only one compound on each side.
4. Save hydrogen and oxygen for last (especially in combustion reactions).
5. Check all atoms again after balancing.

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We'll now balance each equation:

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1) H₃PO₄ + KOH → K₃PO₄ + H₂O

- P: 1 = 1 ✔
- K: 1 on left → need 3 on right → put 3 in front of KOH
- H: 3 + 3 = 6 on left; H₂O has 2H → need 3 H₂O
- O: balance later

→ H₃PO₄ + 3KOH → K₃PO₄ + 3H₂O

✔ Balanced.

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2) K + B₂O₃ → K₂O + B

- B: 2 on left → need 2B on right
- O: 3 on left → need 3 K₂O → gives 6K on right → need 6K on left
- So: 6K + B₂O₃ → 3K₂O + 2B

✔ Balanced.

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3) HCl + NaOH → NaCl + H₂O

- This is an acid-base neutralization.
- Already balanced: HCl + NaOH → NaCl + H₂O

✔ Balanced.

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4) Na + NaNO₃ → Na₂O + N₂

This looks like a redox reaction. Let's analyze:

Left: Na, Na, N, O
Right: Na₂O (2Na, 1O), N₂ (2N)

But we have Na on both sides — likely decomposition or disproportionation.

Assume:
- Na reduces NO₃⁻ to N₂
- Na oxidizes to Na⁺

Try balancing:

Let’s suppose:
a Na + b NaNO₃ → c Na₂O + d N₂

Atoms:
- Na: a + b = 2c
- N: b = 2d
- O: 3b = c

Try values:
From O: c = 3b
Then Na: a + b = 2(3b) = 6b → a = 5b
From N: b = 2d → d = b/2

Choose b = 2 → then:
- b = 2
- d = 1
- c = 6
- a = 10

So: 10Na + 2NaNO₃ → 6Na₂O + N₂

Check:
- Na: 10 + 2 = 12; right: 6×2 = 12 ✔
- N: 2 → 2 ✔
- O: 6 → 6 ✔

✔ Balanced.

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5) C + S₈ → CS₂

S₈ has 8 S atoms. Each CS₂ has 1 C and 2 S.

So to use all S: need 4 CS₂ per S₈ → uses 4 C

→ 4C + S₈ → 4CS₂

✔ Balanced.

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6) Na + O₂ → Na₂O

- O₂ has 2 O → Na₂O has 1 O → need 2 Na₂O → 4 Na
- So: 4Na + O₂ → 2Na₂O

✔ Balanced.

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7) N₂ + O₂ → N₂O₅

- N₂ → N₂O₅: 2N, 5O
- O₂ provides 2O → need 5/2 O₂ → multiply whole eq by 2:

→ 2N₂ + 5O₂ → 2N₂O₅

✔ Balanced.

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8) H₃PO₄ + Mg(OH)₂ → Mg₃(PO₄)₂ + H₂O

Mg₃(PO₄)₂ has 3 Mg, 2 PO₄

So need:
- 2 H₃PO₄ (for 2 PO₄)
- 3 Mg(OH)₂ (for 3 Mg)

Now check H and O:
- Left: 2×3 = 6H from H₃PO₄, 3×2 = 6H from OH → total 12H
- Right: H₂O → each has 2H → need 6 H₂O

Also O: balance later

→ 2H₃PO₄ + 3Mg(OH)₂ → Mg₃(PO₄)₂ + 6H₂O

Check:
- P: 2 = 2 ✔
- Mg: 3 = 3 ✔
- H: 6 + 6 = 12 → 6×2 = 12 ✔
- O: left: 8 (from 2H₃PO₄) + 6 (from 3Mg(OH)₂) = 14; right: 8 (in PO₄) + 6 = 14 ✔

✔ Balanced.

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9) NaOH + H₂CO₃ → Na₂CO₃ + H₂O

- CO₃²⁻ from H₂CO₃ → Na₂CO₃ needs 2 Na
- So need 2 NaOH
- H₂CO₃ has 2H → forms 1 H₂O

→ 2NaOH + H₂CO₃ → Na₂CO₃ + 2H₂O

Wait: H₂CO₃ → 2H⁺ + CO₃²⁻, so 2NaOH → 2H₂O?

Actually:
- 2NaOH + H₂CO₃ → Na₂CO₃ + 2H₂O

Check:
- Na: 2 = 2 ✔
- O: 2+3=5 vs 3+2=5 ✔
- H: 2+2=4 → 4 ✔

✔ Balanced.

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10) KOH + HBr → KBr + H₂O

Simple acid-base:
→ KOH + HBr → KBr + H₂O

✔ Balanced.

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11) Na + O₂ → Na₂O

Same as #6: 4Na + O₂ → 2Na₂O

✔ Balanced.

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12) Al(OH)₃ + H₂CO₃ → Al₂(CO₃)₃ + H₂O

Al₂(CO₃)₃ has 2 Al, 3 CO₃

So need:
- 2 Al(OH)₃
- 3 H₂CO₃

Now H: left: 2×3 = 6H from OH, 3×2 = 6H from H₂CO₃ → 12H
Right: H₂O → need 6 H₂O

→ 2Al(OH)₃ + 3H₂CO₃ → Al₂(CO₃)₃ + 6H₂O

Check:
- Al: 2 = 2 ✔
- C: 3 = 3 ✔
- O: left: 6 + 9 = 15? Wait:
- Al(OH)₃: 2×3 = 6O, 2×3H = 6H
- H₂CO₃: 3×3 = 9O, 3×2 = 6H → total O: 6+9=15, H: 6+6=12
- Right: Al₂(CO₃)₃: 9O, 6H₂O: 6O → total 15O ✔
- H: 12 → 12 ✔

✔ Balanced.

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13) Al + S₈ → Al₂S₃

S₈ has 8 S → Al₂S₃ has 3 S → LCM of 8 and 3 is 24

So: 3 S₈ = 24 S → 8 Al₂S₃ → 16 Al

→ 16Al + 3S₈ → 8Al₂S₃

Check:
- Al: 16 = 16 ✔
- S: 24 = 24 ✔

✔ Balanced.

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14) Cs + N₂ → Cs₃N

N₂ → 2N → need 2 Cs₃N → 6 Cs

→ 6Cs + N₂ → 2Cs₃N

✔ Balanced.

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15) Mg + Cl₂ → MgCl₂

→ Mg + Cl₂ → MgCl₂

✔ Balanced.

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16) Rb + RbNO₃ → Rb₂O + N₂

Disproportionation.

Let’s suppose:
a Rb + b RbNO₃ → c Rb₂O + d N₂

Atoms:
- Rb: a + b = 2c
- N: b = 2d
- O: 3b = c

From O: c = 3b
Then Rb: a + b = 2(3b) = 6b → a = 5b
From N: b = 2d → d = b/2

Let b = 2 → d = 1, c = 6, a = 10

→ 10Rb + 2RbNO₃ → 6Rb₂O + N₂

Check:
- Rb: 10 + 2 = 12; right: 6×2 = 12 ✔
- N: 2 → 2 ✔
- O: 6 → 6 ✔

✔ Balanced.

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17) C₆H₆ + O₂ → CO₂ + H₂O

Combustion of benzene.

C₆H₆ → 6CO₂ + 3H₂O

But H: 6H → 3H₂O → 6H ✔

O: right: 6×2 + 3 = 12 + 3 = 15 O → need 15/2 O₂ → multiply by 2:

→ 2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O

✔ Balanced.

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18) N₂ + H₂ → NH₃

Classic: N₂ + 3H₂ → 2NH₃

✔ Balanced.

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19) C₁₀H₂₂ + O₂ → CO₂ + H₂O

C₁₀H₂₂ → 10CO₂ + 11H₂O

O: right: 10×2 + 11 = 20 + 11 = 31 → need 31/2 O₂

Multiply by 2: 2C₁₀H₂₂ + 31O₂ → 20CO₂ + 22H₂O

✔ Balanced.

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20) Al(OH)₃ + HBr → AlBr₃ + H₂O

Al(OH)₃ has 3 OH → need 3 HBr

→ Al(OH)₃ + 3HBr → AlBr₃ + 3H₂O

✔ Balanced.

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21) CH₃CH₂CH₂CH₃ + O₂ → CO₂ + H₂O

Butane: C₄H₁₀

→ C₄H₁₀ + O₂ → 4CO₂ + 5H₂O

O: right: 8 + 5 = 13 → need 13/2 O₂ → multiply by 2:

→ 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

✔ Balanced.

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22) C₃H₈ + O₂ → CO₂ + H₂O

Propane: C₃H₈ → 3CO₂ + 4H₂O

O: 6 + 4 = 10 → need 5 O₂

→ C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

✔ Balanced.

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23) Li + AlCl₃ → LiCl + Al

Reduction of Al³⁺ by Li

AlCl₃ → Al + 3Cl⁻ → need 3 LiCl

So: 3Li + AlCl₃ → 3LiCl + Al

✔ Balanced.

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24) C₂H₆ + O₂ → CO₂ + H₂O

Ethane: C₂H₆ → 2CO₂ + 3H₂O

O: 4 + 3 = 7 → need 7/2 O₂ → ×2:

→ 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

✔ Balanced.

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25) NH₄OH + H₃PO₄ → (NH₄)₃PO₄ + H₂O

(NH₄)₃PO₄ has 3 NH₄⁺ → need 3 NH₄OH

→ 3NH₄OH + H₃PO₄ → (NH₄)₃PO₄ + 3H₂O

✔ Balanced.

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26) Rb + P → Rb₃P

Need 3 Rb per P → 3Rb + P → Rb₃P

✔ Balanced.

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27) CH₄ + O₂ → CO₂ + H₂O

Methane: CH₄ → CO₂ + 2H₂O

O: 2 + 1 = 3 → need 3/2 O₂ → ×2:

→ 2CH₄ + 3O₂ → 2CO₂ + 4H₂O

✔ Balanced.

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28) Al(OH)₃ + H₂SO₄ → Al₂(SO₄)₃ + H₂O

Al₂(SO₄)₃ has 2 Al, 3 SO₄

So need:
- 2 Al(OH)₃
- 3 H₂SO₄

Now H: left: 2×3 = 6H from OH, 3×2 = 6H from H₂SO₄ → 12H
→ 6 H₂O

→ 2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O

✔ Balanced.

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29) Na + Cl₂ → NaCl

→ 2Na + Cl₂ → 2NaCl

✔ Balanced.

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30) Rb + S₈ → Rb₂S

S₈ has 8 S → need 8 Rb₂S → 16 Rb

→ 16Rb + S₈ → 8Rb₂S

✔ Balanced.

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31) H₃PO₄ + Ca(OH)₂ → Ca₃(PO₄)₂ + H₂O

Ca₃(PO₄)₂ → 3 Ca, 2 PO₄

So:
- 2 H₃PO₄
- 3 Ca(OH)₂

H: left: 6H from acid, 6H from base → 12H → 6 H₂O

→ 2H₃PO₄ + 3Ca(OH)₂ → Ca₃(PO₄)₂ + 6H₂O

✔ Balanced.

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32) NH₃ + HCl → NH₄Cl

→ NH₃ + HCl → NH₄Cl

✔ Balanced.

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33) Li + H₂O → LiOH + H₂

Li + H₂O → LiOH + H₂

Balance:
- H: 2 on left → 1 in LiOH, 2 in H₂ → total 3H → not good

Try:
2Li + 2H₂O → 2LiOH + H₂

H: 4 → 2 + 2 = 4 ✔
O: 2 = 2 ✔
Li: 2 = 2 ✔

→ 2Li + 2H₂O → 2LiOH + H₂

✔ Balanced.

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34) Ca₃(PO₄)₂ + SiO₂ + C → CaSiO₃ + CO + P

This is a reduction reaction (phosphate ore smelting).

Target: reduce P from PO₄³⁻ to P⁰

Each P goes from +5 to 0 → gains 5e⁻

C → CO → loses 2e⁻

So need 5C per P, but there are 2P in Ca₃(PO₄)₂

So need 10C for 2P → produce 10CO

Also, Ca₃(PO₄)₂ → 3Ca²⁺ → need 3 SiO₂ → 3CaSiO₃

So:
- Ca₃(PO₄)₂
- 3SiO₂
- 10C → 10CO + 2P

→ Ca₃(PO₄)₂ + 3SiO₂ + 10C → 3CaSiO₃ + 10CO + 2P

Check:
- Ca: 3 = 3 ✔
- P: 2 = 2 ✔
- Si: 3 = 3 ✔
- O: left: 8 (PO₄) + 6 (SiO₂) = 14; right: 9 (CaSiO₃) + 10 (CO) = 19 → wait!

Wait: PO₄ has 4O × 2 = 8O; SiO₂: 3×2 = 6O; C: 10C → no O → total O: 14

Right: CaSiO₃: 3×3 = 9O; CO: 10O → total 19 → too many!

Mistake: CO comes from C + O → but where does O come from?

Actually, O from phosphate and silica goes into CaSiO₃ and CO.

But let's recheck:

Left:
- Ca₃(PO₄)₂: 3Ca, 2P, 8O
- 3SiO₂: 3Si, 6O
- 10C: 10C
→ Total: 3Ca, 2P, 3Si, 10C, 14O

Right:
- 3CaSiO₃: 3Ca, 3Si, 9O
- 10CO: 10C, 10O
- 2P: 2P
→ Total: 3Ca, 2P, 3Si, 10C, 19O → too much O

Problem: CO can't have more O than available.

Wait: maybe the reaction consumes O from PO₄ and SiO₂.

But we have only 14 O atoms.

But 3CaSiO₃ uses 9O, 10CO uses 10O → 19 needed → impossible.

Ah! We must not have 10CO.

Re-think: each P reduced requires 5e⁻, C → CO loses 2e⁻ → so need 5/2 C per P → 5C per 2P → 5C → 5CO

But earlier I said 10C → too much.

Let’s balance electrons.

Ca₃(PO₄)₂ → 2P + ...
Each P⁵⁺ → P⁰ → gain 5e⁻ → total 10e⁻ gained

C → CO → lose 2e⁻ → so need 5C

So 5C → 5CO

Now:
- Ca₃(PO₄)₂ → 3Ca²⁺ → need 3SiO₂ → 3CaSiO₃
- 3SiO₂ provides 6O
- Ca₃(PO₄)₂ has 8O → total 14O
- 3CaSiO₃ uses 9O → 5O left → used in 5CO → yes!

So:
- Ca₃(PO₄)₂ + 3SiO₂ + 5C → 3CaSiO₃ + 5CO + 2P

Check:
- Ca: 3 = 3 ✔
- P: 2 = 2 ✔
- Si: 3 = 3 ✔
- C: 5 = 5 ✔
- O: left: 8 + 6 = 14; right: 9 + 5 = 14 ✔
- H: none ✔

✔ Balanced.

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35) NH₃ + O₂ → N₂ + H₂O

Ammonia oxidation.

N: 2N on right → need 2NH₃

H: 6H → 3H₂O

O: 3O → need 3/2 O₂

→ 2NH₃ + 3/2O₂ → N₂ + 3H₂O

Multiply by 2: 4NH₃ + 3O₂ → 2N₂ + 6H₂O

✔ Balanced.

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36) FeS₂ + O₂ → Fe₂O₃ + SO₂

FeS₂ → Fe₂O₃ + SO₂

Need 2 Fe → 2 FeS₂

S: 4 S → 4 SO₂

O: right: 3 (Fe₂O₃) + 8 (SO₂) = 11 O → need 11/2 O₂

→ 2FeS₂ + 11/2O₂ → Fe₂O₃ + 4SO₂

Multiply by 2: 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂

✔ Balanced.

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37) C + SO₂ → CS₂ + CO

Carbon reduces SO₂ to CS₂

S: SO₂ → CS₂ → need 1 C per S

But CO also formed → carbon used in two ways.

Let’s suppose:
a C + b SO₂ → c CS₂ + d CO

S: b = c
O: 2b = d
C: a = c + d

Substitute:
a = c + 2b
But c = b → a = b + 2b = 3b

So: a = 3b, c = b, d = 2b

Let b = 1 → a = 3, c = 1, d = 2

→ 3C + SO₂ → CS₂ + 2CO

Check:
- C: 3 = 1 + 2 = 3 ✔
- S: 1 = 1 ✔
- O: 2 = 2 ✔

✔ Balanced.

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## ✔ Final Answers (Balanced Equations):

1) H₃PO₄ + 3KOH → K₃PO₄ + 3H₂O
2) 6K + B₂O₃ → 3K₂O + 2B
3) HCl + NaOH → NaCl + H₂O
4) 10Na + 2NaNO₃ → 6Na₂O + N₂
5) 4C + S₈ → 4CS₂
6) 4Na + O₂ → 2Na₂O
7) 2N₂ + 5O₂ → 2N₂O₅
8) 2H₃PO₄ + 3Mg(OH)₂ → Mg₃(PO₄)₂ + 6H₂O
9) 2NaOH + H₂CO₃ → Na₂CO₃ + 2H₂O
10) KOH + HBr → KBr + H₂O
11) 4Na + O₂ → 2Na₂O
12) 2Al(OH)₃ + 3H₂CO₃ → Al₂(CO₃)₃ + 6H₂O
13) 16Al + 3S₈ → 8Al₂S₃
14) 6Cs + N₂ → 2Cs₃N
15) Mg + Cl₂ → MgCl₂
16) 10Rb + 2RbNO₃ → 6Rb₂O + N₂
17) 2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O
18) N₂ + 3H₂ → 2NH₃
19) 2C₁₀H₂₂ + 31O₂ → 20CO₂ + 22H₂O
20) Al(OH)₃ + 3HBr → AlBr₃ + 3H₂O
21) 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
22) C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
23) 3Li + AlCl₃ → 3LiCl + Al
24) 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
25) 3NH₄OH + H₃PO₄ → (NH₄)₃PO₄ + 3H₂O
26) 3Rb + P → Rb₃P
27) 2CH₄ + 3O₂ → 2CO₂ + 4H₂O
28) 2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O
29) 2Na + Cl₂ → 2NaCl
30) 16Rb + S₈ → 8Rb₂S
31) 2H₃PO₄ + 3Ca(OH)₂ → Ca₃(PO₄)₂ + 6H₂O
32) NH₃ + HCl → NH₄Cl
33) 2Li + 2H₂O → 2LiOH + H₂
34) Ca₃(PO₄)₂ + 3SiO₂ + 5C → 3CaSiO₃ + 5CO + 2P
35) 4NH₃ + 3O₂ → 2N₂ + 6H₂O
36) 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
37) 3C + SO₂ → CS₂ + 2CO

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