49 Balancing Chemical Equations Worksheets [with Answers] - Free Printable

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 equation with coefficients.

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🔁 How to Balance Chemical Equations:

1. Count atoms of each element on both sides.
2. Use coefficients (whole numbers) to balance atoms one at a time.
3. Start with elements that appear in only one compound on each side.
4. Save hydrogen and oxygen for last (common in combustion reactions).
5. Check your work!

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We'll now balance each reaction:

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1) H₃PO₄ + KOH → K₃PO₄ + H₂O

- P: 1 = 1 ✔
- K: 1 → 3 ⇒ need 3 KOH
- H: 3 + 3 = 6 on left; 2 on right ⇒ need 3 H₂O
- O: balance automatically

✔ Answer:
1 H₃PO₄ + 3 KOH → 1 K₃PO₄ + 3 H₂O

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2) K + B₂O₃ → K₂O + B

- B: 2 → 1 ⇒ need 2 B on right
- K: 1 → 2 ⇒ need 2 K on left
- O: 3 on left → 1 on right ⇒ need 3 K₂O
- So K: 6 needed on left

Try:
- 6K + B₂O₃ → 3K₂O + 2B

Check:
- K: 6 = 6 ✔
- B: 2 = 2 ✔
- O: 3 = 3 ✔

✔ Answer:
6 K + 1 B₂O₃ → 3 K₂O + 2 B

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3) HCl + NaOH → NaCl + H₂O

Simple acid-base neutralization.

- H: 1 + 1 = 2 → 2 in H₂O ✔
- Cl: 1 = 1 ✔
- Na: 1 = 1 ✔
- O: 1 = 1 ✔

✔ Answer:
1 HCl + 1 NaOH → 1 NaCl + 1 H₂O

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4) Na + NaNO₃ → Na₂O + N₂

This is a redox reaction. Let’s analyze:

Left: Na and NaNO₃ → two sources of Na
Right: Na₂O and N₂

- N: 1 → 2 ⇒ need 2 NO₃⁻ → so 2 NaNO₃
- O: 6 from 2 NO₃ ⇒ 6 O → need 3 Na₂O (each has 1 O)
- Na: from 2 NaNO₃ → 2 Na, plus extra Na metal → total Na needed: 6 (for 3 Na₂O)

So:
- 6 Na + 2 NaNO₃ → 3 Na₂O + N₂

Check:
- Na: 6 + 2 = 8; right: 3×2 = 6 → ✘

Wait — mistake.

Better approach:

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

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

From O: c = 3b
Then Na: a + b = 2(3b) = 6b → a = 5b

Set b = 2 → then:
- b = 2 → d = 1 (from N), c = 6, a = 10

So:
10 Na + 2 NaNO₃ → 6 Na₂O + 1 N₂

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

✔ Answer:
10 Na + 2 NaNO₃ → 6 Na₂O + 1 N₂

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5) C + S₈ → CS₂

S₈ has 8 S atoms → need 4 CS₂ (to use 8 S)

So: C + S₈ → 4 CS₂ → but C: 1 vs 4 → need 4 C

✔ Answer:
4 C + 1 S₈ → 4 CS₂

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6) Na + O₂ → Na₂O

- O₂ → 2 O → need 2 Na₂O (so 4 Na)
- So: 4 Na + O₂ → 2 Na₂O

✔ Answer:
4 Na + 1 O₂ → 2 Na₂O

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7) N₂ + O₂ → N₂O₅

- N₂: 2 N → N₂O₅ has 2 N → good
- O: 2 → 5 → need 5/2 O₂ → multiply by 2

→ 2 N₂ + 5 O₂ → 2 N₂O₅

✔ Answer:
2 N₂ + 5 O₂ → 2 N₂O₅

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8) H₃PO₄ + Mg(OH)₂ → Mg₃(PO₄)₂ + H₂O

- PO₄³⁻: 1 → 2 ⇒ need 2 H₃PO₄
- Mg²⁺: 1 → 3 ⇒ need 3 Mg(OH)₂
- H: 2×3 + 3×2 = 6+6=12 → H₂O: 12 H → 6 H₂O
- O: check later

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

Check:
- P: 2 = 2 ✔
- Mg: 3 = 3 ✔
- H: 6 + 6 = 12 → 6 H₂O → 12 H ✔
- O: 8 + 6 = 14 → right: 8 (in PO₄) + 6 = 14 ✔

✔ Answer:
2 H₃PO₄ + 3 Mg(OH)₂ → 1 Mg₃(PO₄)₂ + 6 H₂O

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9) NaOH + H₂CO₃ → Na₂CO₃ + H₂O

- CO₃²⁻: 1 → 1 ✔
- Na: 1 → 2 ⇒ need 2 NaOH
- H: 2 + 2 = 4 → H₂O: 2 H₂O → 4 H ✔

So:
2 NaOH + H₂CO₃ → Na₂CO₃ + 2 H₂O

✔ Answer:
2 NaOH + 1 H₂CO₃ → 1 Na₂CO₃ + 2 H₂O

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10) KOH + HBr → KBr + H₂O

Acid-base: 1:1 ratio

✔ Answer:
1 KOH + 1 HBr → 1 KBr + 1 H₂O

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11) Na + O₂ → Na₂O

Same as #6 → 4 Na + 1 O₂ → 2 Na₂O

✔ Answer:
4 Na + 1 O₂ → 2 Na₂O

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12) Al(OH)₃ + H₂CO₃ → Al₂(CO₃)₃ + H₂O

- Al: 1 → 2 ⇒ need 2 Al(OH)₃
- CO₃²⁻: 1 → 3 ⇒ need 3 H₂CO₃
- H: 2×3 + 3×2 = 6+6=12 → H₂O: 6 H₂O
- O: check later

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

Check:
- Al: 2 = 2 ✔
- C: 3 = 3 ✔
- O: 6 + 9 = 15 → right: 9 + 6 = 15 ✔
- H: 6 + 6 = 12 → 6 H₂O → 12 H ✔

✔ Answer:
2 Al(OH)₃ + 3 H₂CO₃ → 1 Al₂(CO₃)₃ + 6 H₂O

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13) Al + S₈ → Al₂S₃

- S₈ → 8 S → need 8/3 Al₂S₃? Not integer

Better: Al₂S₃ has 2 Al, 3 S

So find LCM of 8 and 3 → 24 S

So: 3 S₈ = 24 S → 8 Al₂S₃ (since 8×3=24 S)

Al: 8×2 = 16 → need 16 Al

✔ Answer:
16 Al + 3 S₈ → 8 Al₂S₃

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14) Cs + N₂ → Cs₃N

- N₂ → 2 N → need 2 Cs₃N → 6 Cs
- So: 6 Cs + N₂ → 2 Cs₃N

✔ Answer:
6 Cs + 1 N₂ → 2 Cs₃N

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15) Mg + Cl₂ → MgCl₂

- Mg: 1 → 1 ✔
- Cl: 2 → 2 ✔

✔ Answer:
1 Mg + 1 Cl₂ → 1 MgCl₂

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16) Rb + RbNO₃ → Rb₂O + N₂

Similar to #4.

Let’s balance:

RbNO₃ provides Rb and NO₃⁻

N: 1 → 2 in N₂ ⇒ need 2 RbNO₃

O: 6 → need 3 Rb₂O (3×1 O)

Rb: from 2 RbNO₃ → 2 Rb, plus extra Rb → total Rb = 6 (for 3 Rb₂O)

So: 4 Rb + 2 RbNO₃ → 3 Rb₂O + N₂

Check:
- Rb: 4 + 2 = 6; right: 6 ✔
- N: 2 = 2 ✔
- O: 6 = 3 → wait, 3 Rb₂O has 3 O → but 2 RbNO₃ has 6 O → ✘

Wait: RbNO₃ has 3 O per molecule → 2 × 3 = 6 O → need 6 O → 6 Rb₂O?

But Rb₂O has 1 O → so need 6 Rb₂O → 12 Rb

From 2 RbNO₃ → 2 Rb

So need 10 Rb extra → total Rb = 12

N: 2 → 1 N₂ ✔

So:
10 Rb + 2 RbNO₃ → 6 Rb₂O + 1 N₂

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

✔ Answer:
10 Rb + 2 RbNO₃ → 6 Rb₂O + 1 N₂

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17) C₆H₆ + O₂ → CO₂ + H₂O

Combustion of benzene.

C₆H₆ → 6 CO₂ + 3 H₂O (but H: 6 → 6 H in 3 H₂O → yes)

But O: 6×2 + 3×1 = 12 + 3 = 15 → need 15/2 O₂

Multiply all by 2:

2 C₆H₆ + 15 O₂ → 12 CO₂ + 6 H₂O

✔ Answer:
2 C₆H₆ + 15 O₂ → 12 CO₂ + 6 H₂O

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18) N₂ + H₂ → NH₃

Classic Haber process.

N₂ + 3 H₂ → 2 NH₃

✔ Answer:
1 N₂ + 3 H₂ → 2 NH₃

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19) C₁₀H₂₂ + O₂ → CO₂ + H₂O

C₁₀H₂₂ → 10 CO₂ + 11 H₂O (H: 22 → 11 H₂O)

O: 10×2 + 11×1 = 20 + 11 = 31 → need 31/2 O₂ → ×2

→ 2 C₁₀H₂₂ + 31 O₂ → 20 CO₂ + 22 H₂O

✔ Answer:
2 C₁₀H₂₂ + 31 O₂ → 20 CO₂ + 22 H₂O

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20) Al(OH)₃ + HBr → AlBr₃ + H₂O

- Al: 1 = 1 ✔
- Br: 1 → 3 ⇒ need 3 HBr
- H: 3 + 3 = 6 → 3 H₂O
- O: 3 → 3 ✔

So:
Al(OH)₃ + 3 HBr → AlBr₃ + 3 H₂O

✔ Answer:
1 Al(OH)₃ + 3 HBr → 1 AlBr₃ + 3 H₂O

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21) CH₃CH₂CH₂CH₃ + O₂ → CO₂ + H₂O

Butane: C₄H₁₀

C₄H₁₀ → 4 CO₂ + 5 H₂O

O: 8 + 5 = 13 → 13/2 O₂ → ×2

→ 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

✔ Answer:
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

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22) C₃H₈ + O₂ → CO₂ + H₂O

Propane combustion:

C₃H₈ → 3 CO₂ + 4 H₂O

O: 6 + 4 = 10 → 5 O₂

✔ Answer:
1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

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23) Li + AlCl₃ → LiCl + Al

Redox: Li reduces Al³⁺

AlCl₃ → Al + 3 Cl⁻ → needs 3 Li⁺ → 3 LiCl

So: 3 Li + AlCl₃ → 3 LiCl + Al

✔ Answer:
3 Li + 1 AlCl₃ → 3 LiCl + 1 Al

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24) C₂H₆ + O₂ → CO₂ + H₂O

Ethane: C₂H₆ → 2 CO₂ + 3 H₂O

O: 4 + 3 = 7 → 7/2 O₂ → ×2

→ 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

✔ Answer:
2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

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25) NH₄OH + H₃PO₄ → (NH₄)₃PO₄ + H₂O

- PO₄³⁻: 1 → 1 ✔
- NH₄⁺: 1 → 3 ⇒ need 3 NH₄OH
- H: 3×1 + 3 = 6 → H₂O: 3 H₂O
- O: check

So:
3 NH₄OH + H₃PO₄ → (NH₄)₃PO₄ + 3 H₂O

✔ Answer:
3 NH₄OH + 1 H₃PO₄ → 1 (NH₄)₃PO₄ + 3 H₂O

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26) Rb + P → Rb₃P

- P: 1 → 1 ✔
- Rb: 3 → need 3 Rb

✔ Answer:
3 Rb + 1 P → 1 Rb₃P

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27) CH₄ + O₂ → CO₂ + H₂O

Methane:

CH₄ → CO₂ + 2 H₂O

O: 2 + 1 = 3 → 3/2 O₂ → ×2

→ 2 CH₄ + 3 O₂ → 2 CO₂ + 4 H₂O

✔ Answer:
2 CH₄ + 3 O₂ → 2 CO₂ + 4 H₂O

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28) Al(OH)₃ + H₂SO₄ → Al₂(SO₄)₃ + H₂O

- Al: 1 → 2 ⇒ need 2 Al(OH)₃
- SO₄²⁻: 1 → 3 ⇒ need 3 H₂SO₄
- H: 6 + 6 = 12 → 6 H₂O

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

✔ Answer:
2 Al(OH)₃ + 3 H₂SO₄ → 1 Al₂(SO₄)₃ + 6 H₂O

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29) Na + Cl₂ → NaCl

- Cl₂ → 2 Cl ⇒ need 2 NaCl → 2 Na

So: 2 Na + Cl₂ → 2 NaCl

✔ Answer:
2 Na + 1 Cl₂ → 2 NaCl

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30) Rb + S₈ → Rb₂S

- S₈ → 8 S → need 8 Rb₂S → 16 Rb

So: 16 Rb + S₈ → 8 Rb₂S

✔ Answer:
16 Rb + 1 S₈ → 8 Rb₂S

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31) H₃PO₄ + Ca(OH)₂ → Ca₃(PO₄)₂ + H₂O

- PO₄³⁻: 1 → 2 ⇒ need 2 H₃PO₄
- Ca²⁺: 1 → 3 ⇒ need 3 Ca(OH)₂
- H: 6 + 6 = 12 → 6 H₂O

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

✔ Answer:
2 H₃PO₄ + 3 Ca(OH)₂ → 1 Ca₃(PO₄)₂ + 6 H₂O

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32) NH₃ + HCl → NH₄Cl

Simple: 1:1

✔ Answer:
1 NH₃ + 1 HCl → 1 NH₄Cl

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33) Li + H₂O → LiOH + H₂

Li + H₂O → LiOH + H₂

H: 2 → 1 in LiOH + 1 in H₂ → 2 H → ok

But H₂ is diatomic → need 2 H from water → 2 H₂O

So:
2 Li + 2 H₂O → 2 LiOH + H₂

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

✔ Answer:
2 Li + 2 H₂O → 2 LiOH + 1 H₂

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

This is a complex metallurgical reaction.

Let’s balance step by step.

Ca₃(PO₄)₂ → 3 Ca → 3 CaSiO₃ → need 3 SiO₂

P: 2 → 2 P

C: used to reduce → produces CO

Each P requires reduction: PO₄³⁻ → P → gains 5 e⁻

But let’s look at known reaction:

Typical reaction:
Ca₃(PO₄)₂ + 3 SiO₂ + 5 C → 3 CaSiO₃ + 5 CO + 2 P

Check:
- Ca: 3 = 3 ✔
- P: 2 = 2 ✔
- Si: 3 = 3 ✔
- O: 8 + 6 = 14 → 3×3 = 9 in CaSiO₃ + 5 in CO = 14 ✔
- C: 5 = 5 ✔

✔ Answer:
1 Ca₃(PO₄)₂ + 3 SiO₂ + 5 C → 3 CaSiO₃ + 5 CO + 2 P

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35) NH₃ + O₂ → N₂ + H₂O

Ammonia oxidation:

2 NH₃ → N₂ + 3 H₂O

H: 6 → 6 → 3 H₂O

O: 3 → need 3/2 O₂ → ×2

→ 4 NH₃ + 3 O₂ → 2 N₂ + 6 H₂O

✔ Answer:
4 NH₃ + 3 O₂ → 2 N₂ + 6 H₂O

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36) FeS₂ + O₂ → Fe₂O₃ + SO₂

Pyrite oxidation.

FeS₂ → Fe₂O₃ and SO₂

Fe: 2 → need 2 FeS₂

S: 4 → 4 SO₂

O: 3 in Fe₂O₃ + 8 in SO₂ = 11 → need 11/2 O₂ → ×2

So:
4 FeS₂ + 11 O₂ → 2 Fe₂O₃ + 8 SO₂

Check:
- Fe: 4 = 4 ✔
- S: 8 = 8 ✔
- O: 22 = 6 + 16 = 22 ✔

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

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37) C + SO₂ → CS₂ + CO

C + SO₂ → CS₂ + CO

S: 1 → 1 ✔

But C: 1 → 1 in CS₂ + 1 in CO → 2 C → need 2 C

So:
2 C + SO₂ → CS₂ + CO

But O: 2 → 1 in CO → need 2 CO → 2 C

Wait: 2 C + SO₂ → CS₂ + 2 CO

Now C: 2 = 1 + 2 = 3 → no

Try:
2 C + SO₂ → CS₂ + 2 CO → C: 2 vs 3 → ✘

Try:
3 C + SO₂ → CS₂ + 2 CO

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

✔ Answer:
3 C + 1 SO₂ → 1 CS₂ + 2 CO

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## ✔ Final Balanced Answers:

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

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