Chemistry practice problems worksheet focusing on balancing chemical equations.

Let’s solve each problem one by one. We’ll write the unbalanced equation first, then balance it step by step.

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Problem 11: Dinitrogen Pentoxide + Water → Nitric Acid

Unbalanced:
N₂O₅ + H₂O → HNO₃

Count atoms:
Left: N=2, O=5+1=6, H=2
Right: H=1, N=1, O=3

We need 2 HNO₃ on right to match 2 N from left:
N₂O₅ + H₂O → 2HNO₃

Now check:
Left: N=2, O=6, H=2
Right: H=2, N=2, O=6 → Balanced!

✔ Answer: N₂O₅ + H₂O → 2HNO₃

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Problem 12: Magnesium + Titanium(IV) Chloride → Magnesium Chloride + Titanium

Titanium(IV) means Ti⁴⁺, so chloride is TiCl₄
Magnesium is Mg, forms Mg²⁺ → MgCl₂
Titanium product is just Ti (elemental)

Unbalanced:
Mg + TiCl₄ → MgCl₂ + Ti

Balance Cl: 4 on left, 2 on right → put 2 in front of MgCl₂
Mg + TiCl₄ → 2MgCl₂ + Ti

Now Mg: 1 on left, 2 on right → put 2 in front of Mg
2Mg + TiCl₄ → 2MgCl₂ + Ti

Check:
Left: Mg=2, Ti=1, Cl=4
Right: Mg=2, Cl=4, Ti=1 → Balanced!

✔ Answer: 2Mg + TiCl₄ → 2MgCl₂ + Ti

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Problem 13: Carbon + Zinc Oxide → Zinc + Carbon Dioxide

Unbalanced:
C + ZnO → Zn + CO₂

Balance O: 1 on left, 2 on right → put 2 in front of ZnO
C + 2ZnO → Zn + CO₂

Now Zn: 2 on left, 1 on right → put 2 in front of Zn
C + 2ZnO → 2Zn + CO₂

Check:
Left: C=1, Zn=2, O=2
Right: Zn=2, C=1, O=2 → Balanced!

✔ Answer: C + 2ZnO → 2Zn + CO₂

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Problem 14: Bromine + Sodium Iodide → Sodium Bromide + Iodine

Bromine is Br₂ (diatomic), Iodine is I₂ (diatomic)
Sodium iodide = NaI, sodium bromide = NaBr

Unbalanced:
Br₂ + NaI → NaBr + I₂

Balance I: 1 on left, 2 on right → put 2 in front of NaI
Br₂ + 2NaI → NaBr + I₂

Now Na: 2 on left, 1 on right → put 2 in front of NaBr
Br₂ + 2NaI → 2NaBr + I₂

Check:
Left: Br=2, Na=2, I=2
Right: Na=2, Br=2, I=2 → Balanced!

✔ Answer: Br₂ + 2NaI → 2NaBr + I₂

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Problem 15: Phosphorus Trichloride + Chlorine gas → Phosphorus Pentachloride

Phosphorus trichloride = PCl₃
Chlorine gas = Cl₂
Phosphorus pentachloride = PCl₅

Unbalanced:
PCl₃ + Cl₂ → PCl₅

Balance Cl: Left has 3+2=5? Wait — let’s count properly.

Actually, PCl₃ has 3 Cl, Cl₂ has 2 Cl → total 5 Cl on left if we use 1 each? But PCl₅ has 5 Cl.

Try:
PCl₃ + Cl₂ → PCl₅

Left: P=1, Cl=3+2=5
Right: P=1, Cl=5 → Already balanced!

✔ Answer: PCl₃ + Cl₂ → PCl₅

Wait — that seems too easy. Let me double-check.

Yes! One molecule of PCl₃ and one of Cl₂ give exactly PCl₅. Perfect.

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Problem 16: Phosphorus + Bromine → Phosphorus Tribromide

Phosphorus is usually P₄ (but sometimes written as P — but in reactions like this, often simplified to P). However, standard practice for balancing uses P₄.

But looking at common textbook problems, they often simplify phosphorus to “P” unless specified.

Phosphorus tribromide = PBr₃

If we use P:
P + Br₂ → PBr₃

Balance Br: 2 on left, 3 on right → LCM is 6 → use 3 Br₂ and 2 PBr₃
P + 3Br₂ → 2PBr₃

Now P: 1 on left, 2 on right → put 2 in front of P
2P + 3Br₂ → 2PBr₃

Check:
Left: P=2, Br=6
Right: P=2, Br=6 → Balanced!

Some textbooks might expect P₄, but since the problem says “Phosphorus” without specifying, and given other problems are simple, we’ll go with P.

✔ Answer: 2P + 3Br₂ → 2PBr₃

*(Note: If using P₄, it would be P₄ + 6Br₂ → 4PBr₃ — also correct, but more complex. Since problem doesn’t specify, simpler version is acceptable.)*

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Problem 17: Calcium Hydride + Water → Calcium Hydroxide + Hydrogen gas

Calcium hydride = CaH₂
Water = H₂O
Calcium hydroxide = Ca(OH)₂
Hydrogen gas = H₂

Unbalanced:
CaH₂ + H₂O → Ca(OH)₂ + H₂

Count atoms:

Left: Ca=1, H=2+2=4, O=1
Right: Ca=1, O=2, H=2+2=4? Wait — Ca(OH)₂ has 2 O and 2 H from OH, plus H₂ has 2 H → total H=4, O=2

But left has only 1 O → need 2 H₂O

Try:
CaH₂ + 2H₂O → Ca(OH)₂ + H₂

Now left: Ca=1, H=2+4=6, O=2
Right: Ca=1, O=2, H=2 (from OH) + 2 (from H₂) = 4 → Not enough H on right.

We need more H₂. Try putting 2 in front of H₂:

CaH₂ + 2H₂O → Ca(OH)₂ + 2H₂

Now right: H = 2 (from OH) + 4 (from 2H₂) = 6 → matches left!

Check all:
Left: Ca=1, H=2+4=6, O=2
Right: Ca=1, O=2, H=2+4=6 → Balanced!

✔ Answer: CaH₂ + 2H₂O → Ca(OH)₂ + 2H₂

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Problem 18: Sulfuric Acid + Potassium Hydroxide → Potassium Sulfate + Water

Sulfuric acid = H₂SO₄
Potassium hydroxide = KOH
Potassium sulfate = K₂SO₄
Water = H₂O

Unbalanced:
H₂SO₄ + KOH → K₂SO₄ + H₂O

Balance K: 1 on left, 2 on right → put 2 in front of KOH
H₂SO₄ + 2KOH → K₂SO₄ + H₂O

Now H: Left = 2 (from acid) + 2 (from 2KOH) = 4 H
Right: H₂O has 2 H → need 2 H₂O

So:
H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

Check:
Left: H=4, S=1, O=4+2=6, K=2
Right: K=2, S=1, O=4+2=6, H=4 → Balanced!

✔ Answer: H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

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Problem 19: Propane (C₃H₈) + Oxygen → Carbon Dioxide + Water

Combustion reaction.

Unbalanced:
C₃H₈ + O₂ → CO₂ + H₂O

Balance C: 3 on left → 3 CO₂
C₃H₈ + O₂ → 3CO₂ + H₂O

Balance H: 8 on left → 4 H₂O (since each has 2 H)
C₃H₈ + O₂ → 3CO₂ + 4H₂O

Now balance O:
Right: 3×2 + 4×1 = 6 + 4 = 10 O atoms
Left: O₂ → need 5 O₂ molecules

So:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Check:
Left: C=3, H=8, O=10
Right: C=3, H=8, O=6+4=10 → Balanced!

✔ Answer: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

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Problem 20: Benzene (C₆H₆) + Oxygen → Carbon Dioxide + Water

Another combustion.

Unbalanced:
C₆H₆ + O₂ → CO₂ + H₂O

Balance C: 6 on left → 6 CO₂
C₆H₆ + O₂ → 6CO₂ + H₂O

Balance H: 6 on left → 3 H₂O (each has 2 H)
C₆H₆ + O₂ → 6CO₂ + 3H₂O

Now O: Right = 6×2 + 3×1 = 12 + 3 = 15 O atoms
Left: O₂ → need 15/2 = 7.5 → not whole number.

Multiply entire equation by 2 to eliminate fraction:

Start over with coefficients doubled:

2C₆H₆ + ? O₂ → 12CO₂ + 6H₂O

Now O on right: 12×2 + 6×1 = 24 + 6 = 30 → need 15 O₂

So:
2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O

Check:
Left: C=12, H=12, O=30
Right: C=12, H=12, O=24+6=30 → Balanced!

✔ Answer: 2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O

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Final Answers:

11. N₂O₅ + H₂O → 2HNO₃
12. 2Mg + TiCl₄ → 2MgCl₂ + Ti
13. C + 2ZnO → 2Zn + CO₂
14. Br₂ + 2NaI → 2NaBr + I₂
15. PCl₃ + Cl₂ → PCl₅
16. 2P + 3Br₂ → 2PBr₃
17. CaH₂ + 2H₂O → Ca(OH)₂ + 2H₂
18. H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
19. C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
20. 2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O