Balancing chemical equations worksheet for practice.

Sure! Let’s go through each chemical equation one by one and balance them using the law of conservation of mass — meaning the number of atoms of each element must be equal on both sides of the equation.

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1. CH₄ + O₂ → CO₂ + H₂O

This is the combustion of methane.

- Left: C=1, H=4, O=2
- Right: C=1, H=2, O=3

Balance H first: put 2 in front of H₂O → now H=4 on right.
Now O: right has 2 (from CO₂) + 2 (from 2H₂O) = 4 O atoms → need 2 O₂ on left.

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

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2. Na⁺ + Cl⁻ → NaCl

This is an ionic combination forming sodium chloride.

- Already balanced as written: 1 Na⁺ and 1 Cl⁻ make 1 NaCl.

✔ Balanced:
1 Na⁺ + 1 Cl⁻ → 1 NaCl

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3. Al + O₂ → Al₂O₃

Aluminum reacts with oxygen to form aluminum oxide.

- Left: Al=1, O=2
- Right: Al=2, O=3

Find LCM for O: 6 O atoms → need 3 O₂ (6 O) on left, and 2 Al₂O₃ (6 O) on right → but that gives 4 Al on right → so need 4 Al on left.

✔ Balanced:
4 Al + 3 O₂ → 2 Al₂O₃

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4. N₂ + H₂ → NH₃

Synthesis of ammonia (Haber process).

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

Balance N: put 2 in front of NH₃ → now N=2, H=6 on right → need 3 H₂ on left.

✔ Balanced:
1 N₂ + 3 H₂ → 2 NH₃

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5. CO(g) + H₂(g) → C₈H₁₈(l) + H₂O

This is a Fischer-Tropsch type reaction producing octane and water.

- Left: C=1, O=1, H=2
- Right: C=8, H=18+2=20 (from C₈H₁₈ and H₂O), O=1

We need 8 C on left → 8 CO
That gives 8 O on left → need 8 H₂O on right → which adds 16 H from water → plus 18 H from octane = 34 H total on right.

Each H₂ gives 2 H → need 17 H₂ on left.

✔ Balanced:
8 CO + 17 H₂ → 1 C₈H₁₈ + 8 H₂O

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6. Fe₂O₃(s) + CO(g) → Fe(l) + CO₂(g)

Reduction of iron(III) oxide with carbon monoxide.

- Left: Fe=2, O=3+1=4, C=1
- Right: Fe=1, O=2, C=1

Balance Fe: put 2 in front of Fe.
Now O: right has 2 CO₂ → 4 O atoms → matches left if we have 3 CO → produces 3 CO₂? Wait:

Actually, standard balanced equation:

Fe₂O₃ + 3CO → 2Fe + 3CO₂

Check:
Left: Fe=2, O=3+3=6, C=3
Right: Fe=2, O=6, C=3 → ✔

✔ Balanced:
1 Fe₂O₃ + 3 CO → 2 Fe + 3 CO₂

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7. H₂SO₄ + Pb(OH)₄ → Pb(SO₄)₂ + H₂O

Sulfuric acid reacts with lead(IV) hydroxide.

Note: Pb(SO₄)₂ implies Pb⁴⁺ and SO₄²⁻ → correct formula.

- Left: H=2+4=6, S=1, O=4+4=8, Pb=1
- Right: Pb=1, S=2, O=8+1=9, H=2

Need 2 H₂SO₄ → gives 2 S, 8 O, 4 H → then Pb(OH)₄ gives 4 H and 4 O → total H=8, O=12, S=2, Pb=1

Right: Pb(SO₄)₂ → Pb=1, S=2, O=8 → need 4 H₂O → H=8, O=4 → total O=12 ✔

✔ Balanced:
2 H₂SO₄ + 1 Pb(OH)₄ → 1 Pb(SO₄)₂ + 4 H₂O

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8. Al + HCl → AlCl₃ + H₂

Aluminum reacts with hydrochloric acid.

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

Balance Cl: need 3 HCl → gives 3 H → need 3/2 H₂ → multiply entire equation by 2.

So: 2 Al + 6 HCl → 2 AlCl₃ + 3 H₂

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

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9. Ca₃(PO₄)₂ + H₂SO₄ → CaSO₄ + Ca(H₂PO₄)₂

This is a double displacement or acid-base reaction.

Let’s count atoms:

Left: Ca=3, P=2, O=8+4=12, H=2, S=1
Right: Ca=1 (from CaSO₄) + 1 (from Ca(H₂PO₄)₂) = 2 → not enough!

Actually, this reaction should produce calcium sulfate and calcium dihydrogen phosphate.

Balanced version:

Ca₃(PO₄)₂ + 2 H₂SO₄ → 2 CaSO₄ + Ca(H₂PO₄)₂

Check:
Left: Ca=3, P=2, O=8 (from PO₄) + 8 (from 2 SO₄) = 16? Wait:

Ca₃(PO₄)₂: Ca=3, P=2, O=8
2 H₂SO₄: H=4, S=2, O=8
Total left: Ca=3, P=2, H=4, S=2, O=16

Right: 2 CaSO₄ → Ca=2, S=2, O=8
Ca(H₂PO₄)₂ → Ca=1, H=4, P=2, O=8
Total right: Ca=3, P=2, H=4, S=2, O=16 → ✔

✔ Balanced:
1 Ca₃(PO₄)₂ + 2 H₂SO₄ → 2 CaSO₄ + 1 Ca(H₂PO₄)₂

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10. H₃PO₄ + HCl → PCl₅ + H₂O

Phosphoric acid reacting with HCl to form phosphorus pentachloride? This is unusual — typically PCl₅ is made from P₄ and Cl₂, but let’s balance as given.

Left: H=3+1=4, P=1, O=4, Cl=1
Right: P=1, Cl=5, H=2, O=1

We need 5 Cl on left → 5 HCl → gives 5 H → total H=3+5=8 → need 4 H₂O on right → gives 8 H and 4 O → matches left O=4.

✔ Balanced:
1 H₃PO₄ + 5 HCl → 1 PCl₅ + 4 H₂O

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## ✔ Final Answer Key:

1. 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O
2. 1 Na⁺ + 1 Cl⁻ → 1 NaCl
3. 4 Al + 3 O₂ → 2 Al₂O₃
4. 1 N₂ + 3 H₂ → 2 NH₃
5. 8 CO + 17 H₂ → 1 C₈H₁₈ + 8 H₂O
6. 1 Fe₂O₃ + 3 CO → 2 Fe + 3 CO₂
7. 2 H₂SO₄ + 1 Pb(OH)₄ → 1 Pb(SO₄)₂ + 4 H₂O
8. 2 Al + 6 HCl → 2 AlCl₃ + 3 H₂
9. 1 Ca₃(PO₄)₂ + 2 H₂SO₄ → 2 CaSO₄ + 1 Ca(H₂PO₄)₂
10. 1 H₃PO₄ + 5 HCl → 1 PCl₅ + 4 H₂O

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Let me know if you want explanations for any specific step or if you’d like to see how to balance using algebraic methods!