Balancing Chemical Equations Practice Sheet - Free Printable

Here is the fully balanced version of each chemical equation from your worksheet, along with a clear explanation for each.

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1. KOH + H₃PO₄ → K₃PO₄ + H₂O

This is an acid-base neutralization reaction. Phosphoric acid (H₃PO₄) has three acidic hydrogens, so it requires 3 KOH to fully neutralize it.

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

- K: 3 on left, 3 on right
- O: 3+4=7 on left, 4+3=7 on right
- H: 3+3=6 on left, 6 on right (3×2 in H₂O)
- P: 1 on both sides

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2. NH₃ + O₂ → NO + H₂O

This is the oxidation of ammonia to nitric oxide (Ostwald process).

Start by balancing N and H first.

NH₃ has 3 H, H₂O has 2 H → LCM of 3 and 2 is 6 → use 2 NH₃ and 3 H₂O.

Then balance N: 2 NH₃ → 2 NO

Now oxygen: Right side = 2 O (from 2 NO) + 3 O (from 3 H₂O) = 5 O atoms → need 5/2 O₂ → multiply entire equation by 2 to eliminate fraction.

✔ Balanced Equation:
4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

- N: 4 = 4
- H: 12 = 12
- O: 10 = 4 (NO) + 6 (H₂O) = 10

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3. Al + O₂ → Al₂O₃

Aluminum reacts with oxygen to form aluminum oxide.

Left: Al and O₂
Right: Al₂O₃ → 2 Al, 3 O

LCM of O: 2 and 3 → 6 O atoms → 3 O₂ and 2 Al₂O₃ → that gives 4 Al on right → so 4 Al on left.

✔ Balanced Equation:
4 Al + 3 O₂ → 2 Al₂O₃

- Al: 4 = 4
- O: 6 = 6

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4. CaS + H₂O → Ca(HS)₂ + Ca(OH)₂

This is a disproportionation reaction where sulfide reacts with water to form hydrosulfide and hydroxide.

Note: The products are Ca(HS)₂ and Ca(OH)₂, meaning 2 Ca atoms on right → need 2 CaS on left.

Also, left: 2 S, right: 2 S (in Ca(HS)₂)

Hydrogen: Right side: Ca(HS)₂ has 2H, Ca(OH)₂ has 2H → total 4H → need 2 H₂O on left.

✔ Balanced Equation:
2 CaS + 2 H₂O → 1 Ca(HS)₂ + 1 Ca(OH)₂

- Ca: 2 = 1+1 = 2
- S: 2 = 2
- H: 4 = 2 (from HS) + 2 (from OH) = 4
- O: 2 = 2 (from OH)

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5. Au₂S₃ + H₂ → Au + H₂S

Gold sulfide reduced by hydrogen gas.

Au₂S₃ has 2 Au and 3 S → need 2 Au and 3 H₂S on right.

H₂S has 2 H per molecule → 3 H₂S needs 6 H → so 3 H₂ on left.

✔ Balanced Equation:
1 Au₂S₃ + 3 H₂ → 2 Au + 3 H₂S

- Au: 2 = 2
- S: 3 = 3
- H: 6 = 6

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6. PCl₅ + H₂O → H₃PO₄ + HCl

Phosphorus pentachloride hydrolyzes to phosphoric acid and HCl.

PCl₅ → H₃PO₄ → need to account for 5 Cl → 5 HCl

Now H: H₃PO₄ has 3H, 5 HCl has 5H → total 8H → need 4 H₂O (since each has 2H)

Check O: 4 H₂O → 4 O → matches H₃PO₄ which has 4 O.

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

- P: 1 = 1
- Cl: 5 = 5
- H: 8 = 3 + 5 = 8
- O: 4 = 4

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7. Ba₃N₂ + HF → BaF₂ + NH₃

Barium nitride reacts with hydrofluoric acid.

Ba₃N₂ → 3 Ba, 2 N → need 3 BaF₂ and 2 NH₃

F: 3 BaF₂ → 6 F → need 6 HF

H: 6 HF → 6 H; 2 NH₃ → 6 H → matches.

✔ Balanced Equation:
1 Ba₃N₂ + 6 HF → 3 BaF₂ + 2 NH₃

- Ba: 3 = 3
- N: 2 = 2
- H: 6 = 6
- F: 6 = 6

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8. CaCl₂ + Na₃PO₄ → Ca₃(PO₄)₂ + NaCl

Double displacement reaction.

Ca₃(PO₄)₂ requires 3 Ca and 2 PO₄ → so need 3 CaCl₂ and 2 Na₃PO₄

That gives: 6 Cl and 6 Na → so 6 NaCl

✔ Balanced Equation:
3 CaCl₂ + 2 Na₃PO₄ → 1 Ca₃(PO₄)₂ + 6 NaCl

- Ca: 3 = 3
- Cl: 6 = 6
- Na: 6 = 6
- P: 2 = 2
- O: 8 = 8

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9. FeS + O₂ → Fe₂O₃ + SO₂

Iron sulfide oxidized to iron oxide and sulfur dioxide.

Fe₂O₃ → 2 Fe → need 2 FeS
SO₂ → 1 S per molecule → 2 FeS gives 2 S → need 2 SO₂

Now O: Fe₂O₃ has 3 O, 2 SO₂ has 4 O → total 7 O → need 7/2 O₂ → multiply by 2.

So: 4 FeS + 7 O₂ → 2 Fe₂O₃ + 4 SO₂

✔ Balanced Equation:
4 FeS + 7 O₂ → 2 Fe₂O₃ + 4 SO₂

- Fe: 4 = 4
- S: 4 = 4
- O: 14 = 6 (from 2 Fe₂O₃) + 8 (from 4 SO₂) = 14

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10. HClO₄ + P₄O₁₀ → H₃PO₄ + Cl₂O₇

Perchloric acid reacts with phosphorus pentoxide to form phosphoric acid and dichlorine heptoxide.

P₄O₁₀ → 4 P → need 4 H₃PO₄

Each H₃PO₄ has 3 H → total 12 H → need 12 HClO₄

Cl: 12 Cl → need 6 Cl₂O₇ (since each has 2 Cl)

Check O:

Left: 12 HClO₄ → 12×4 = 48 O; P₄O₁₀ → 10 O → total 58 O

Right: 4 H₃PO₄ → 4×4 = 16 O; 6 Cl₂O₇ → 6×7 = 42 O → total 58 O ✔

✔ Balanced Equation:
12 HClO₄ + 1 P₄O₁₀ → 4 H₃PO₄ + 6 Cl₂O₇

- H: 12 = 12
- Cl: 12 = 12
- P: 4 = 4
- O: 48+10=58 = 16+42=58

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## ✔ Final Answer Summary:

1. 3 KOH + 1 H₃PO₄ → 1 K₃PO₄ + 3 H₂O
2. 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O
3. 4 Al + 3 O₂ → 2 Al₂O₃
4. 2 CaS + 2 H₂O → 1 Ca(HS)₂ + 1 Ca(OH)₂
5. 1 Au₂S₃ + 3 H₂ → 2 Au + 3 H₂S
6. 1 PCl₅ + 4 H₂O → 1 H₃PO₄ + 5 HCl
7. 1 Ba₃N₂ + 6 HF → 3 BaF₂ + 2 NH₃
8. 3 CaCl₂ + 2 Na₃PO₄ → 1 Ca₃(PO₄)₂ + 6 NaCl
9. 4 FeS + 7 O₂ → 2 Fe₂O₃ + 4 SO₂
10. 12 HClO₄ + 1 P₄O₁₀ → 4 H₃PO₄ + 6 Cl₂O₇

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