Balancing Equations worksheet for chemistry students to practice balancing chemical reactions.

Let's solve each of these chemical equations by balancing them. Balancing a chemical equation means ensuring the number of atoms of each element is equal on both sides (reactants and products), following the Law of Conservation of Mass.

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1. CH₄ + O₂ → CO₂ + H₂O

Unbalanced:
CH₄ + O₂ → CO₂ + H₂O

- Carbon: 1 on both sides ✔
- Hydrogen: 4 on left, 2 on right → need 2 H₂O
- Oxygen: 2 on left, 2 (CO₂) + 1 (H₂O) = 3 → not balanced

Try:
- 2 H₂O → gives 4 H and 2 O from water
- Now O: 2 (from CO₂) + 2 (from 2 H₂O) = 4 O → so need 2 O₂

Now:
CH₄ + 2O₂ → CO₂ + 2H₂O

Check:
- C: 1 = 1 ✔
- H: 4 = 4 ✔
- O: 4 = 2 (CO₂) + 2 (2H₂O) = 4 ✔

✔ Balanced: CH₄ + 2O₂ → CO₂ + 2H₂O

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2. Na⁺ + Cl⁻ → NaCl

This is an ionic combination.

Na⁺ + Cl⁻ → NaCl

- One Na⁺ and one Cl⁻ make one NaCl

✔ Balanced: Na⁺ + Cl⁻ → NaCl

(Note: This is already balanced as written.)

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3. Al + O₂ → Al₂O₃

Aluminum reacts with oxygen to form aluminum oxide.

Unbalanced:
Al + O₂ → Al₂O₃

- Al: 1 on left, 2 on right → need 2 Al
- O: 2 on left, 3 on right → LCM of 2 and 3 is 6

So:
- 3 O₂ → 6 O
- 2 Al₂O₃ → 4 Al and 6 O

So:
4Al + 3O₂ → 2Al₂O₃

Check:
- Al: 4 = 4 ✔
- O: 6 = 6 ✔

✔ Balanced: 4Al + 3O₂ → 2Al₂O₃

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4. N₂ + H₂ → NH₃

Nitrogen and hydrogen form ammonia.

N₂ + H₂ → NH₃

- N: 2 on left, 1 on right → need 2 NH₃
- H: 2 on left, 6 on right → need 3 H₂

So:
N₂ + 3H₂ → 2NH₃

Check:
- N: 2 = 2 ✔
- H: 6 = 6 ✔

✔ Balanced: N₂ + 3H₂ → 2NH₃

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5. CO(g) + H₂(g) → C₈H₁₈(l) + H₂O

This is a synthesis of octane from CO and H₂ — unusual, but let’s balance it.

C₈H₁₈ has:
- 8 C
- 18 H

Each CO has 1 C, each H₂ has 2 H

We need:
- 8 CO for 8 C
- 18 H → 9 H₂ molecules

But also produces H₂O. So we must account for oxygen.

From 8 CO → 8 O atoms → must go into H₂O → so 8 H₂O → 16 H

But C₈H₁₈ has 18 H, so total H needed: 18 (in C₈H₁₈) + 16 (in H₂O) = 34 H → 17 H₂ molecules

So:

8CO + 17H₂ → C₈H₁₈ + 8H₂O

Check:
- C: 8 = 8 ✔
- O: 8 = 8 ✔
- H: 34 = 18 (C₈H₁₈) + 16 (8H₂O) = 34 ✔

✔ Balanced: 8CO + 17H₂ → C₈H₁₈ + 8H₂O

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6. Fe₂O₃(s) + CO(g) → Fe(l) + CO₂(g)

Iron(III) oxide reduced by carbon monoxide.

Fe₂O₃ + CO → Fe + CO₂

- Fe: 2 on left → need 2 Fe on right
- O: 3 in Fe₂O₃ + 1 in CO → total O on left depends on CO
- Each CO turns into CO₂ → uses one O atom

Fe₂O₃ has 3 O atoms → needs 3 CO → produce 3 CO₂

So:
Fe₂O₃ + 3CO → 2Fe + 3CO₂

Check:
- Fe: 2 = 2 ✔
- O: 3 + 3 = 6 → 3 CO₂ has 6 O ✔
- C: 3 = 3 ✔

✔ Balanced: Fe₂O₃ + 3CO → 2Fe + 3CO₂

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7. H₂SO₄ + Pb(OH)₄ → Pb(SO₄)₂ + H₂O

Acid-base reaction.

Pb(OH)₄ is lead(IV) hydroxide, Pb(SO₄)₂ is lead(IV) sulfate.

Note: Pb(SO₄)₂ has 2 SO₄ groups → needs 2 H₂SO₄

So:
2H₂SO₄ + Pb(OH)₄ → Pb(SO₄)₂ + H₂O

Now check H and O:

Left:
- H: 2×2 = 4 (from H₂SO₄) + 4 (from Pb(OH)₄) = 8 H
- O: 8 (from 2 H₂SO₄) + 4 (from Pb(OH)₄) = 12 O

Right:
- Pb(SO₄)₂: 8 O (from SO₄)
- H₂O: each has 1 O and 2 H

We have 8 H → need 4 H₂O

So:
2H₂SO₄ + Pb(OH)₄ → Pb(SO₄)₂ + 4H₂O

Check:
- H: 4 (H₂SO₄) + 4 (Pb(OH)₄) = 8 → 4 H₂O has 8 H ✔
- O: 8 (H₂SO₄) + 4 (Pb(OH)₄) = 12 → 8 (SO₄) + 4 (H₂O) = 12 ✔
- S: 2 = 2 ✔
- Pb: 1 = 1 ✔

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

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8. Al + HCl → AlCl₃ + H₂

Aluminum reacts with hydrochloric acid.

Al + HCl → AlCl₃ + H₂

- Al: 1 = 1 ✔
- Cl: 1 on left, 3 on right → need 3 HCl
- H: 3 on left → need 3/2 H₂ → use fractions or multiply

Try:
Al + 3HCl → AlCl₃ + 3/2 H₂

Multiply whole equation by 2:
2Al + 6HCl → 2AlCl₃ + 3H₂

Check:
- Al: 2 = 2 ✔
- Cl: 6 = 6 ✔
- H: 6 = 6 ✔

✔ Balanced: 2Al + 6HCl → 2AlCl₃ + 3H₂

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

This is a double displacement reaction.

Ca₃(PO₄)₂ contains:
- 3 Ca
- 2 PO₄³⁻

H₂SO₄ provides SO₄²⁻

Products:
- CaSO₄: 1 Ca and 1 SO₄
- Ca(H₂PO₄)₂: 1 Ca and 2 H₂PO₄⁻

We need to distribute Ca and PO₄.

Total Ca: 3 → so need 1 CaSO₄ and 2 Ca(H₂PO₄)₂? But that would be 3 Ca → yes.

But PO₄: 2 PO₄ → each Ca(H₂PO₄)₂ has 2 H₂PO₄ → so 2 × 2 = 4 H₂PO₄ → but only 2 PO₄ available → no.

Wait: PO₄²⁻ becomes H₂PO₄⁻ → each PO₄²⁻ makes one H₂PO₄⁻

So Ca₃(PO₄)₂ → can give 2 PO₄²⁻ → which become 2 H₂PO₄⁻

But Ca(H₂PO₄)₂ needs 2 H₂PO₄⁻ → so one molecule of Ca(H₂PO₄)₂ requires 2 H₂PO₄⁻ → so one Ca₃(PO₄)₂ can make one Ca(H₂PO₄)₂

Then leftover Ca: 3 - 1 = 2 → need 2 CaSO₄

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

Now check:

Left:
- Ca: 3
- P: 2
- O: many
- H: 4
- S: 2

Right:
- Ca: 2 (CaSO₄) + 1 (Ca(H₂PO₄)₂) = 3 ✔
- S: 2 ✔
- P: 2 ✔
- H: 2 × 2 = 4 ✔ (from Ca(H₂PO₄)₂)
- O: balanced by inspection

Yes, this works.

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

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10. H₃PO₄ + HCl → PCl₅ + H₂O

Phosphoric acid reacts with HCl to form phosphorus pentachloride?

But this is not a typical reaction — likely a redox or substitution.

But let’s try balancing.

PCl₅ has:
- 1 P
- 5 Cl

H₃PO₄ has 1 P

So need 1 H₃PO₄

To get 5 Cl → need 5 HCl

Now:
H₃PO₄ + 5HCl → PCl₅ + H₂O

Now H: 3 + 5 = 8 H → need 4 H₂O

O: 4 from H₃PO₄ → 4 H₂O has 4 O → good

So:
H₃PO₄ + 5HCl → PCl₅ + 4H₂O

Check:
- P: 1 = 1 ✔
- Cl: 5 = 5 ✔
- H: 3 + 5 = 8 → 4 H₂O has 8 H ✔
- O: 4 = 4 ✔

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

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## ✔ Final Balanced Equations:

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

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