Worksheet for identifying and balancing chemical reactions.

Let’s go through each reaction one by one. For each, we’ll first figure out what type of reaction it is, then balance the equation.

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1. Zn + HCl → ZnCl₂ + H₂

- Type: Single replacement (Zn replaces H in HCl)
- Balance:
Left: Zn=1, H=1, Cl=1
Right: Zn=1, H=2, Cl=2
→ Put a 2 in front of HCl
Balanced: Zn + 2HCl → ZnCl₂ + H₂

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2. P + O₂ → P₄O₁₀

- Type: Synthesis (two elements combine to make one compound)
- Balance:
Right has 4 P and 10 O → need 4 P on left, 5 O₂ (since O₂ gives 2 oxygens)
Balanced: 4P + 5O₂ → P₄O₁₀

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3. NaBr + H₃PO₄ → Na₃PO₄ + HBr

- Type: Double replacement (ions swap partners)
- Balance:
Right has 3 Na → need 3 NaBr on left
That gives 3 Br → need 3 HBr on right
H: left has 3 from H₃PO₄, right has 3 from 3HBr → good
PO₄: 1 on each side → good
Balanced: 3NaBr + H₃PO₄ → Na₃PO₄ + 3HBr

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4. Ca(OH)₂ + Al₂(SO₄)₃ → CaSO₄ + Al(OH)₃

- Type: Double replacement
- Balance:
Look at SO₄: right has 1, left has 3 → put 3 in front of CaSO₄
Now Ca: left has 1, right has 3 → put 3 in front of Ca(OH)₂
Now OH: left has 6 (3×2), right has 3 per Al(OH)₃ → need 2 Al(OH)₃ to get 6 OH
Check Al: left has 2, right has 2 → good
Balanced: 3Ca(OH)₂ + Al₂(SO₄)₃ → 3CaSO₄ + 2Al(OH)₃

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5. Mg + Fe₂O₃ → Fe + MgO

- Type: Single replacement (Mg replaces Fe)
- Balance:
Fe: left has 2 → put 2 in front of Fe
O: left has 3 → need 3 MgO on right
Mg: now 3 on right → put 3 in front of Mg on left
Balanced: 3Mg + Fe₂O₃ → 2Fe + 3MgO

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6. C₂H₄ + O₂ → CO₂ + H₂O

- Type: Combustion (hydrocarbon + oxygen → CO₂ + H₂O)
- Balance:
C: 2 on left → 2 CO₂ on right
H: 4 on left → 2 H₂O on right (gives 4 H)
O: right = 2×2 + 2×1 = 6 → need 3 O₂ on left
Balanced: C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

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7. PbSO₄ → PbSO₃ + O₂

- Type: Decomposition (one compound breaks into two or more)
- Balance:
Left: Pb=1, S=1, O=4
Right: Pb=1, S=1, O=3+2=5 → not balanced
Try 2PbSO₄ → 2PbSO₃ + O₂
Left: O=8, Right: 2×3 + 2 = 8 → yes!
Balanced: 2PbSO₄ → 2PbSO₃ + O₂

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8. NH₃ + I₂ → NI₃ + H₂

- Type: Single replacement? Actually, this is a bit tricky — nitrogen is being combined with iodine, hydrogen is released. It’s often called a “displacement” or “replacement”, but technically it’s a redox reaction. For school level, call it single replacement (I replaces H).
- Balance:
N: 1 on each side
H: 3 on left, 2 on right → LCM is 6 → use 2NH₃ and 3H₂
Then I: right has 3 per NI₃ → if we have 2NI₃, that’s 6 I → need 3I₂ on left
So: 2NH₃ + 3I₂ → 2NI₃ + 3H₂
Check: N=2, H=6, I=6 on both sides → good
Balanced: 2NH₃ + 3I₂ → 2NI₃ + 3H₂

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9. H₂O + SO₃ → H₂SO₄

- Type: Synthesis (two compounds form one)
- Already balanced!
H=2, O=1+3=4, S=1 → same on right
Balanced: H₂O + SO₃ → H₂SO₄

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10. H₂SO₄ + NH₄OH → H₂O + (NH₄)₂SO₄

- Type: Acid-base neutralization (also double replacement)
- Balance:
Right has 2 NH₄ → need 2 NH₄OH on left
Then H: left = 2 (from acid) + 2×5 = 12? Wait — let’s count atoms carefully.

Actually:
Left: H₂SO₄ + NH₄OH → H=2+5=7? No — better to think ions.

Standard way:
H₂SO₄ has 2 H⁺, NH₄OH provides OH⁻ → needs 2 OH⁻ to neutralize → so 2 NH₄OH

Then products: 2 H₂O and (NH₄)₂SO₄

Check:
Left: H₂SO₄ + 2NH₄OH → H: 2 + 2×5 = 12? Wait — NH₄OH is NH₄⁺ and OH → formula unit has 5 H? Let's write molecularly:

NH₄OH = N H₅ O → actually, it’s better to treat as NH₃·H₂O, but for balancing:

Use:
H₂SO₄ + 2NH₄OH → 2H₂O + (NH₄)₂SO₄

Count atoms:

Left:
H: 2 (from H₂SO₄) + 2×(4+1)=2×5=10 → total H=12?
Wait — H₂SO₄ has 2H, each NH₄OH has 5H? That can’t be right.

Actually, NH₄OH is ammonium hydroxide: NH₄⁺ and OH⁻ → so formula is often written as NH₄OH meaning one N, five H, one O? But in reality, when reacting:

H₂SO₄ + 2NH₃ → (NH₄)₂SO₄ — but here it’s NH₄OH.

Better approach:
The reaction is:
H₂SO₄ + 2NH₄OH → (NH₄)₂SO₄ + 2H₂O

Now check atoms:

Left:
H: from H₂SO₄ = 2; from 2NH₄OH = 2*(4+1) = 10? Wait — NH₄OH has N, 5H, O? Actually, standard atomic count:

NH₄OH: N=1, H=5, O=1? But that’s not accurate — in chemical equations, we treat NH₄OH as providing NH₄⁺ and OH⁻, so when writing molecular equation, it’s fine.

Let’s count properly:

Reactants:
H₂SO₄: H=2, S=1, O=4
2NH₄OH: N=2, H=2*(4+1)=10? Wait — NH₄OH is typically considered as having 5 H atoms? Actually, no — NH₄ is 4H, OH is 1H and 1O, so total H per NH₄OH is 5? Yes.

But in product:
(NH₄)₂SO₄: N=2, H=8, S=1, O=4
2H₂O: H=4, O=2

Total right: H=8+4=12, O=4+2=6, etc.

Left: H₂SO₄ (H=2) + 2NH₄OH (each has H=5 → 10) → total H=12 → matches
O: H₂SO₄ has 4, 2NH₄OH has 2×1=2 → total 6 → matches right (4 from sulfate + 2 from water)
N: 2=2, S:1=1 → perfect.

So:
Balanced: H₂SO₄ + 2NH₄OH → 2H₂O + (NH₄)₂SO₄

Type: Acid-base neutralization (or double replacement)

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11. H₂SO₄ + Fe → H₂ + FeSO₄

- Type: Single replacement (Fe replaces H)
- Balance:
Already balanced!
H: 2=2, S:1=1, O:4=4, Fe:1=1
Balanced: H₂SO₄ + Fe → H₂ + FeSO₄

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12. HCl + NH₃ → NH₄Cl

- Type: Synthesis (or acid-base forming salt)
- Already balanced!
H:1+3=4, Cl:1, N:1 → right: NH₄Cl has N=1, H=4, Cl=1
Balanced: HCl + NH₃ → NH₄Cl

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Final Answer:

1. Type: Single replacement; Balanced: Zn + 2HCl → ZnCl₂ + H₂
2. Type: Synthesis; Balanced: 4P + 5O₂ → P₄O₁₀
3. Type: Double replacement; Balanced: 3NaBr + H₃PO₄ → Na₃PO₄ + 3HBr
4. Type: Double replacement; Balanced: 3Ca(OH)₂ + Al₂(SO₄)₃ → 3CaSO₄ + 2Al(OH)₃
5. Type: Single replacement; Balanced: 3Mg + Fe₂O₃ → 2Fe + 3MgO
6. Type: Combustion; Balanced: C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
7. Type: Decomposition; Balanced: 2PbSO₄ → 2PbSO₃ + O₂
8. Type: Single replacement; Balanced: 2NH₃ + 3I₂ → 2NI₃ + 3H₂
9. Type: Synthesis; Balanced: H₂O + SO₃ → H₂SO₄
10. Type: Acid-base neutralization; Balanced: H₂SO₄ + 2NH₄OH → 2H₂O + (NH₄)₂SO₄
11. Type: Single replacement; Balanced: H₂SO₄ + Fe → H₂ + FeSO₄
12. Type: Synthesis; Balanced: HCl + NH₃ → NHCl