Balancing Equations Worksheet featuring ten chemical reactions to balance, with blank spaces for coefficients.

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 to make 4 H
- Oxygen: 2 on left, 2 (in CO₂) + 1 (in H₂O) = 3 → not balanced

Try:
CH₄ + 2O₂ → CO₂ + 2H₂O

Now check:
- C: 1 = 1 ✔
- H: 4 = 4 ✔
- O: 4 (left) = 2 (CO₂) + 2 (H₂O) = 4 ✔

✔ Balanced:
CH₄ + 2O₂ → CO₂ + 2H₂O

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2. Na⁺ + Cl⁻ → NaCl

This is an ionic combination forming a compound.

Na⁺ + Cl⁻ → NaCl

Each ion combines in a 1:1 ratio.

✔ Balanced as written:
Na⁺ + Cl⁻ → NaCl

(Though often written as Na + ½Cl₂ → NaCl for elemental form, here it's ions.)

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3. Al + O₂ → Al₂O₃

Aluminum reacts with oxygen to form aluminum oxide.

Unbalanced:
Al + O₂ → Al₂O₃

- Al: 1 left, 2 right → need 2 Al
- O: 2 left, 3 right → LCM of 2 and 3 is 6 → use 3O₂ and 2Al₂O₃

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

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 synthesis of octane from CO and H₂ — but this isn't a standard reaction. However, let’s balance it as given.

C₈H₁₈ has 8 C and 18 H.

Left side: CO and H₂

We need 8 carbon atoms → 8 CO
Hydrogen: 18 H → 9 H₂

But also, CO has oxygen → will produce water.

Each CO contributes one O → 8 O atoms → forms 8 H₂O (since each H₂O has one O)

But H₂O needs 2 H per molecule → 8 H₂O needs 16 H

Total H needed: 18 (for C₈H₁₈) + 16 (for H₂O) = 34 H → 17 H₂ molecules

So:

8CO + 17H₂ → C₈H₁₈ + 8H₂O

Check:
- C: 8 = 8 ✔
- O: 8 = 8 (in H₂O) ✔
- H: 34 = 18 (C₈H₁₈) + 16 (8 H₂O) = 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 count
- Each CO turns into CO₂ → so 1 CO → 1 CO₂

To remove 3 O from Fe₂O₃, need 3 CO → produces 3 CO₂

So:
Fe₂O₃ + 3CO → 2Fe + 3CO₂

Check:
- Fe: 2 = 2 ✔
- O: 3 + 3 = 6; right: 3×2 = 6 ✔
- C: 3 = 3 ✔

✔ Balanced:
Fe₂O₃ + 3CO → 2Fe + 3CO₂

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7. H₂SO₄ + Pb(OH)₄ → Pb(SO₄)₂ + H₂O

Acid-base neutralization.

Pb(OH)₄ is lead(IV) hydroxide, H₂SO₄ is sulfuric acid.

Pb(SO₄)₂ has 2 SO₄²⁻ → so need 2 H₂SO₄

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

Now H: left: 2×2 + 4 = 8 H
Right: 2 H in H₂O → need 4 H₂O

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

Check:
- H: 4 (from 2H₂SO₄) + 4 (from Pb(OH)₄) = 8 H → 4 H₂O = 8 H ✔
- S: 2 = 2 ✔
- O: 8 (H₂SO₄) + 4 (Pb(OH)₄) = 12 → right: 8 (SO₄) + 4 (H₂O) = 12 ✔
- 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 left, 3 right → need 3 HCl
- H: 3 left → H₂ needs 2 H → need 3/2 H₂

So:
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 or acid reaction.

Ca₃(PO₄)₂ contains 3 Ca and 2 PO₄³⁻

Products: CaSO₄ and Ca(H₂PO₄)₂

Note: Ca(H₂PO₄)₂ uses 2 H₂PO₄⁻ → so takes 2 PO₄ from original

So:
- One Ca₃(PO₄)₂ gives 3 Ca and 2 PO₄
- We can make:
- One Ca(H₂PO₄)₂ → uses 2 PO₄ and 2 Ca
- Remaining 1 Ca → makes 1 CaSO₄

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

But we need to balance H₂SO₄ and H₂PO₄

In Ca(H₂PO₄)₂, there are 4 H and 2 P — but H comes from H₂SO₄

H₂SO₄ provides 2 H and 1 SO₄

Each H₂SO₄ → 2 H → can make one H₂PO₄⁻? But H₂PO₄⁻ has 2 H already

Actually, H₂PO₄⁻ comes from PO₄³⁻ + 2H⁺ → H₂PO₄⁻

So each PO₄³⁻ needs 2 H⁺ → 2 H⁺ from H₂SO₄

For two PO₄³⁻ → need 4 H⁺ → 2 H₂SO₄

And 2 H₂SO₄ → 2 SO₄²⁻ → makes 2 CaSO₄?

Wait — earlier we said only one CaSO₄?

But we have 3 Ca atoms.

Let’s try:

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

But that uses only 1 Ca in CaSO₄ and 2 in Ca(H₂PO₄)₂ → total 3 Ca ✔

But H₂SO₄: 2 → gives 2 SO₄ → but only 1 CaSO₄ formed → missing one SO₄

Wait: Ca(H₂PO₄)₂ does not contain sulfate.

So if we use 2 H₂SO₄ → 2 SO₄²⁻ → need 2 CaSO₄

But we only have 3 Ca.

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

Now check:
- Ca: 3 = 2 + 1 = 3 ✔
- P: 2 = 2 ✔
- O: too many — better count elements.

But wait: Ca(H₂PO₄)₂ has 2 H₂PO₄ → each has 2 H → total 4 H

From 2 H₂SO₄ → 4 H → good

S: 2 = 2 ✔

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

Check all:
- Ca: 3 = 2 + 1 = 3 ✔
- P: 2 = 2 ✔
- O: Left: Ca₃(PO₄)₂ has 8 O from PO₄ (each PO₄ has 4 O → 2×4=8), plus 2H₂SO₄ → 8 O → total 16 O
Right: 2CaSO₄ → 2×4 = 8 O, Ca(H₂PO₄)₂ → 2×4 = 8 O → total 16 ✔
- H: 4 = 4 ✔
- S: 2 = 2 ✔

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

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10. H₃PO₄ + HCl → PCl₅ + H₂O

Phosphoric acid reacting with HCl to form phosphorus pentachloride.

H₃PO₄ + HCl → PCl₅ + H₂O

- P: 1 = 1 ✔
- Cl: 1 left, 5 right → need 5 HCl
- H: left: 3 (H₃PO₄) + 5 (HCl) = 8 H → right: H₂O → need 4 H₂O
- O: left: 4 (H₃PO₄) → right: 4 H₂O → 4 O ✔

So:
H₃PO₄ + 5HCl → PCl₅ + 4H₂O

Check:
- P: 1 = 1 ✔
- Cl: 5 = 5 ✔
- H: 3 + 5 = 8 → 4×2 = 8 ✔
- 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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Let me know if you'd like explanations for any specific step!