Comprehensive chemistry worksheet featuring 37 unbalanced chemical equations for students to practice balancing.

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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🔧 Rules for Balancing Chemical Equations:

1. Count atoms of each element on both sides.
2. Use coefficients (numbers in front of compounds) to balance atoms.
3. Never change subscripts (e.g., H₂O → H₂O₂ is invalid).
4. Start with elements that appear in only one compound on each side.
5. Balance polyatomic ions as a group if they remain unchanged.
6. Check all atoms at the end.

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Let’s begin:

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1)

H₃PO₄ + KOH → K₃PO₄ + H₂O

- P: 1 on both sides ✔
- K: 1 on left, 3 on right → put 3 KOH
- H: 3 from H₃PO₄ + 3 from KOH = 6 H; right: 2 H₂O → 2 H per molecule → need 3 H₂O
- O: check later

Try:
> H₃PO₄ + 3KOH → K₃PO₄ + 3H₂O

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

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2)

K + B₂O₃ → K₂O + B

- B: 2 on left → 2 on right → 2B
- K: 2 on right → 2K on left
- O: 3 on left → 3 K₂O needed? But K₂O has 1 O → need 3 K₂O → 6 K

So:
- 6 K → 3 K₂O
- B₂O₃ → 2 B

Now:
> 6 K + 1 B₂O₃ → 3 K₂O + 2 B

✔ Balanced

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3)

HCl + NaOH → NaCl + H₂O

This is an acid-base neutralization.

> HCl + NaOH → NaCl + H₂O

All atoms already balanced: 1 H, 1 Cl, 1 Na, 1 O, 1 H from OH → total H: 2 → H₂O

✔ Already balanced:
1 HCl + 1 NaOH → 1 NaCl + 1 H₂O

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4)

Na + NaNO₃ → Na₂O + N₂

This looks like a redox reaction.

Left: Na, NaNO₃ → two Na sources

Right: Na₂O (2 Na), N₂ (2 N)

Let’s count:
- N: 1 in NaNO₃ → need 2 NaNO₃ → 2 N → N₂
- So use 2 NaNO₃
- Then we have 2 Na from NaNO₃ + some Na metal → total Na atoms?

Products: Na₂O has 2 Na → so total Na atoms = 2

But 2 NaNO₃ contributes 2 Na → so no extra Na needed? But Na metal is reactant.

Wait — this seems odd.

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

N: b = 2d → let d=1 → b=2

Na: a + b = 2c → a + 2 = 2c

O: 3b = c → 3×2 = 6 → c=6 → 2c=12 → a+2=12 → a=10

So:
> 10 Na + 2 NaNO₃ → 6 Na₂O + 1 N₂

Check:
- Na: 10 + 2 = 12 → 6×2 = 12 ✔
- N: 2 → 2 ✔
- O: 2×3 = 6 → 6×1 = 6 ✔

✔ 10 Na + 2 NaNO₃ → 6 Na₂O + N₂

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5)

C + S₈ → CS₂

S₈ is a molecule of 8 sulfur atoms.

CS₂ has 1 C and 2 S.

So to get even number of S, need multiple CS₂.

Each S₈ gives 8 S → can make 4 CS₂ (since 4×2=8 S)

So:
> C + S₈ → 4 CS₂

But C: left=1, right=4 → need 4 C

> 4 C + S₈ → 4 CS₂

✔ Balanced

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6)

Na + O₂ → Na₂O

Na₂O has 2 Na and 1 O

O₂ has 2 O → need 2 Na₂O → 4 Na

So:
> 4 Na + O₂ → 2 Na₂O

✔ Balanced

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7)

N₂ + O₂ → N₂O₅

N₂O₅ has 2 N and 5 O

N₂ → 1 N₂ → OK

O₂ → 2 O → need 5/2 → multiply whole equation by 2

> 2 N₂ + 5 O₂ → 2 N₂O₅

Check:
- N: 4 → 4 ✔
- O: 10 → 10 ✔

✔ 2 N₂ + 5 O₂ → 2 N₂O₅

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8)

H₃PO₄ + Mg(OH)₂ → Mg₃(PO₄)₂ + H₂O

Mg₃(PO₄)₂ has 3 Mg and 2 PO₄

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

Now:
- H: from 2 H₃PO₄ → 6 H; 3 Mg(OH)₂ → 6 H → total 12 H
- O: many, but H₂O: 12 H → 6 H₂O

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

Check:
- P: 2 → 2 ✔
- Mg: 3 → 3 ✔
- O: Left: 2×4 + 3×2 = 8+6=14; Right: 8 (from PO₄) + 6 = 14 ✔
- H: 2×3 + 3×2 = 6+6=12 → 6 H₂O → 12 H ✔

✔ Balanced

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9)

NaOH + H₂CO₃ → Na₂CO₃ + H₂O

H₂CO₃ is carbonic acid.

Na₂CO₃ has 2 Na → need 2 NaOH

Then:
> 2 NaOH + H₂CO₃ → Na₂CO₃ + H₂O

H: 2 + 2 = 4 → H₂O has 2 H → need 2 H₂O

But now O: left: 2+3=5; right: 3+2=5 → ok

H: 2 from NaOH + 2 from H₂CO₃ = 4 H → 2 H₂O → 4 H ✔

So:
> 2 NaOH + H₂CO₃ → Na₂CO₃ + 2 H₂O

✔ Balanced

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10)

KOH + HBr → KBr + H₂O

Acid-base: H⁺ + OH⁻ → H₂O

So:
> KOH + HBr → KBr + H₂O

All atoms balanced: 1 each → ✔

✔ 1 KOH + 1 HBr → 1 KBr + 1 H₂O

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11)

Na + O₂ → Na₂O

Same as #6:
> 4 Na + O₂ → 2 Na₂O

✔

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12)

Al(OH)₃ + H₂CO₃ → Al₂(CO₃)₃ + H₂O

Al₂(CO₃)₃ → needs 2 Al and 3 CO₃

So:
- 2 Al(OH)₃
- 3 H₂CO₃

Now H: 2×3 + 3×2 = 6+6=12 → 6 H₂O

O: check later

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

Check:
- Al: 2 → 2 ✔
- C: 3 → 3 ✔
- O: left: 2×3 + 3×3 = 6+9=15; right: 9 (in CO₃) + 6 = 15 ✔
- H: 6+6=12 → 6 H₂O → 12 H ✔

✔ Balanced

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13)

Al + S₈ → Al₂S₃

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

So:
- 24 S → 8 S₈ → 8 S₈
- 24 S → 8 Al₂S₃ (since each has 3 S)
- Al: 8 Al₂S₃ → 16 Al

So:
> 16 Al + S₈ → 8 Al₂S₃

Wait — S₈ has 8 S → 8 S₈ → 64 S → too much

Better:

We want 3 S per Al₂S₃ → total S must be divisible by 3 and 8 → LCM(3,8)=24

So:
- 24 S → 3 Al₂S₃ (3×8=24 S) → no, 3 Al₂S₃ has 9 S → wrong

Al₂S₃ has 3 S → to get 24 S → 8 Al₂S₃

→ 8 × 3 = 24 S → 3 S per unit

S₈ provides 8 S → so 3 S₈ → 24 S

Al: 8 Al₂S₃ → 16 Al

So:
> 16 Al + 3 S₈ → 8 Al₂S₃

Check:
- Al: 16 → 16 ✔
- S: 3×8=24 → 8×3=24 ✔

✔ 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

✔ Balanced

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15)

Mg + Cl₂ → MgCl₂

MgCl₂ has 1 Mg, 2 Cl → Cl₂ has 2 Cl → so:
> Mg + Cl₂ → MgCl₂

✔ Already balanced

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16)

Rb + RbNO₃ → Rb₂O + N₂

Redox: Rb reduces NO₃⁻ to N₂, oxidizes to Rb⁺

Let’s balance:

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

N: b = 2d → d = b/2

O: b = c → c = b

Rb: a + b = 2c = 2b → a = b

So a = b, c = b, d = b/2

To eliminate fraction, set b=2 → then a=2, c=2, d=1

So:
> 2 Rb + 2 RbNO₃ → 2 Rb₂O + N₂

Check:
- Rb: 2 + 2 = 4 → 2×2 = 4 ✔
- N: 2 → 2 ✔
- O: 2×3 = 6 → 2×1 = 2 → not equal ✘

Wait! RbNO₃ has 3 O → 2 RbNO₃ → 6 O → Rb₂O has 1 O → 2 Rb₂O → 2 O → not enough

So O: left: 3b → right: c → so 3b = c

Earlier: c = b → contradiction

So fix:

From earlier:
- N: b = 2d → d = b/2
- O: 3b = c
- Rb: a + b = 2c

Substitute:
a + b = 2(3b) = 6b → a = 5b

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

So:
> 10 Rb + 2 RbNO₃ → 6 Rb₂O + N₂

Check:
- Rb: 10 + 2 = 12 → 6×2 = 12 ✔
- N: 2 → 2 ✔
- O: 2×3 = 6 → 6×1 = 6 ✔

✔ 10 Rb + 2 RbNO₃ → 6 Rb₂O + N₂

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17)

C₆H₆ + O₂ → CO₂ + H₂O

Combustion of benzene.

C₆H₆ → 6 CO₂ + 3 H₂O → but H: 6 H → 3 H₂O → 6 H → good

O: right: 6×2 + 3×1 = 12+3=15 → O₂ → 15/2 → so multiply by 2

> 2 C₆H₆ + 15 O₂ → 12 CO₂ + 6 H₂O

Check:
- C: 12 → 12 ✔
- H: 12 → 6×2=12 ✔
- O: 30 → 12×2 + 6×1 = 24+6=30 ✔

✔ 2 C₆H₆ + 15 O₂ → 12 CO₂ + 6 H₂O

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18)

N₂ + H₂ → NH₃

Classic Haber process.

> N₂ + 3 H₂ → 2 NH₃

✔ Balanced

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19)

C₁₀H₂₂ + O₂ → CO₂ + H₂O

Combustion of decane.

C₁₀H₂₂ → 10 CO₂ + 11 H₂O

H: 22 → 11 H₂O → 22 H ✔

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

> 2 C₁₀H₂₂ + 31 O₂ → 20 CO₂ + 22 H₂O

Check:
- C: 20 → 20 ✔
- H: 44 → 22×2=44 ✔
- O: 62 → 40+22=62 ✔

✔ 2 C₁₀H₂₂ + 31 O₂ → 20 CO₂ + 22 H₂O

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20)

Al(OH)₃ + HBr → AlBr₃ + H₂O

Al(OH)₃ has 3 OH → AlBr₃ has 3 Br → so need 3 HBr

> Al(OH)₃ + 3 HBr → AlBr₃ + 3 H₂O

H: 3 + 3 = 6 → 3 H₂O → 6 H ✔

O: 3 + 3 = 6 → 3 H₂O → 3 O → missing? Wait

Al(OH)₃ has 3 O, 3 H → HBr has no O → total O: 3 → H₂O: 3 → 3 O ✔

H: 3 (from OH) + 3 (from HBr) = 6 → 3 H₂O → 6 H ✔

Br: 3 → 3 ✔

✔ 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

H: 10 → 5 H₂O → 10 H ✔

O: 8 + 5 = 13 → O₂ → 13/2 → ×2

> 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

✔

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22)

C₃H₈ + O₂ → CO₂ + H₂O

Propane: C₃H₈ → 3 CO₂ + 4 H₂O

H: 8 → 4 H₂O → 8 H ✔

O: 6 + 4 = 10 → O₂ → 5 O₂

> C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

✔

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23)

Li + AlCl₃ → LiCl + Al

Single displacement.

AlCl₃ → 3 Cl → need 3 LiCl → 3 Li

Al: 1 → 1

> 3 Li + AlCl₃ → 3 LiCl + Al

✔

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24)

C₂H₆ + O₂ → CO₂ + H₂O

Ethane: C₂H₆ → 2 CO₂ + 3 H₂O

H: 6 → 3 H₂O → 6 H ✔

O: 4 + 3 = 7 → O₂ → 7/2 → ×2

> 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

✔

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25)

NH₄OH + H₃PO₄ → (NH₄)₃PO₄ + H₂O

(NH₄)₃PO₄ has 3 NH₄⁺ → need 3 NH₄OH

H₃PO₄ → 1

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

Check:
- N: 3 → 3 ✔
- H: 3×5 + 3 = 15+3=18 → 3 H₂O → 6 H → wait

NH₄OH: 5 H (NH₄ has 4, OH has 1) → 3×5 = 15 H

H₃PO₄: 3 H → total 18 H

Right: (NH₄)₃PO₄: 3×4 = 12 H; 3 H₂O: 6 H → total 18 ✔

O: left: 3×1 + 4 = 7 → right: 4 + 3 = 7 ✔

P: 1 → 1 ✔

✔ 3 NH₄OH + H₃PO₄ → (NH₄)₃PO₄ + 3 H₂O

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26)

Rb + P → Rb₃P

Rb₃P → 3 Rb → 3 Rb

> 3 Rb + P → Rb₃P

✔

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27)

CH₄ + O₂ → CO₂ + H₂O

Methane combustion.

> CH₄ + 2 O₂ → CO₂ + 2 H₂O

✔

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28)

Al(OH)₃ + H₂SO₄ → Al₂(SO₄)₃ + H₂O

Al₂(SO₄)₃ → 2 Al, 3 SO₄

So:
- 2 Al(OH)₃
- 3 H₂SO₄

H: 2×3 + 3×2 = 6+6=12 → 6 H₂O

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

Check:
- Al: 2 → 2 ✔
- S: 3 → 3 ✔
- O: left: 2×3 + 3×4 = 6+12=18; right: 12 (SO₄) + 6 = 18 ✔
- H: 6+6=12 → 6 H₂O → 12 H ✔

✔

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29)

Na + Cl₂ → NaCl

Need 2 Na → 2 NaCl

> 2 Na + Cl₂ → 2 NaCl

✔

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30)

Rb + S₈ → Rb₂S

S₈ → 8 S → Rb₂S → 2 Rb per S → so 8 S → 16 Rb

> 16 Rb + S₈ → 8 Rb₂S

✔

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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: 2×3 + 3×2 = 6+6=12 → 6 H₂O

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

Check:
- P: 2 → 2 ✔
- Ca: 3 → 3 ✔
- O: left: 2×4 + 3×2 = 8+6=14; right: 8 + 6 = 14 ✔
- H: 6+6=12 → 6 H₂O → 12 H ✔

✔

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32)

NH₃ + HCl → NH₄Cl

Simple:
> NH₃ + HCl → NH₄Cl

✔

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33)

Li + H₂O → LiOH + H₂

Li + H₂O → LiOH + H₂

Balance:
- Li: 1 → 1
- H: 2 → 1 in LiOH + 2 in H₂ → total 3 → not balanced

Try:
> 2 Li + 2 H₂O → 2 LiOH + H₂

H: 4 → 2 (in LiOH) + 2 (in H₂) = 4 ✔
O: 2 → 2 ✔
Li: 2 → 2 ✔

✔ 2 Li + 2 H₂O → 2 LiOH + H₂

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34)

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

Complex redox.

Assume:
- Ca₃(PO₄)₂ → 3 Ca, 2 P, 8 O
- SiO₂ → Si, 2 O
- C → C
- Products: CaSiO₃, CO, P

Goal: reduce P⁵⁺ to P⁰, oxidize C to CO

Each P⁵⁺ → P⁰ → gains 5 e⁻ → 2 P → gain 10 e⁻

Each C → CO → loses 2 e⁻ → need 5 C

Also:
- Ca: 3 → 3 CaSiO₃ → need 3 SiO₂
- Si: 3 → 3 SiO₂ → 3 SiO₂
- O: left: 8 (from PO₄) + 3×2 = 8+6=14; right: 3×3 (CaSiO₃) + 5×1 (CO) = 9+5=14 ✔
- C: 5 → 5 CO ✔
- P: 2 → 2 P ✔

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

✔

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35)

NH₃ + O₂ → N₂ + H₂O

Ammonia oxidation.

N: 2 → N₂ → 2 NH₃

H: 6 → 3 H₂O

O: 3 → 3/2 O₂ → ×2

> 4 NH₃ + 3 O₂ → 2 N₂ + 6 H₂O

Check:
- N: 4 → 4 ✔
- H: 12 → 6×2=12 ✔
- O: 6 → 6 ✔

✔

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36)

FeS₂ + O₂ → Fe₂O₃ + SO₂

Pyrite roasting.

Fe: 2 → 2 FeS₂ → 2 FeS₂ → 2 Fe → Fe₂O₃ → 1 Fe₂O₃

S: 2×2=4 → 4 SO₂

O: right: 3 (Fe₂O₃) + 4×2 = 3+8=11 → O₂ → 11/2 → ×2

So:
> 4 FeS₂ + 11 O₂ → 2 Fe₂O₃ + 8 SO₂

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

✔

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37)

C + SO₂ → CS₂ + CO

C + SO₂ → CS₂ + CO

S: 1 → 1 → OK

C: 1 → 1 → CS₂ has 1 C, CO has 1 → total 2 C → need 2 C

So:
> 2 C + SO₂ → CS₂ + CO

But S: 1 → 1 ✔

C: 2 → 1+1=2 ✔

O: 2 → 1 → not balanced

CO has 1 O → need 2 CO → 2 O

So:
> 2 C + SO₂ → CS₂ + 2 CO

But C: 2 → 1+2=3 → not balanced

Try:
> 3 C + SO₂ → CS₂ + 2 CO

C: 3 → 1+2=3 ✔
S: 1 → 1 ✔
O: 2 → 2 ✔

✔ 3 C + SO₂ → CS₂ + 2 CO

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## ✔ Final 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 | CH₄ + 2 O₂ → CO₂ + 2 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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