49 Balancing Chemical Equations Worksheets [with Answers] - Free Printable

Let's solve Problem 1 and Problem 2 from the worksheet step by step. I'll go through each part, balance the equations, identify the reaction type, and explain the reasoning.

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## 🔬 Problem 1: Balance the following equations and indicate the type of reaction

We’ll balance each equation and classify it as:
- Formation (synthesis)
- Decomposition
- Single replacement
- Double replacement
- Hydrocarbon combustion
- Other

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a.

Cu(s) + O₂(g) → CuO(s)

Balancing:
Left: Cu = 1, O = 2
Right: Cu = 1, O = 1 → need to balance oxygen.

→ Put 2 CuO on right → 2 Cu and 2 O
→ Then need 2 Cu on left and 1 O₂ (already has 2 O)

So:
2 Cu(s) + 1 O₂(g) → 2 CuO(s)

✔ Balanced.

Type: Formation (synthesis) — two elements combine to form a compound.

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b.

H₂O(l) → H₂(g) + O₂(g)

Balancing:
Left: H = 2, O = 1
Right: H = 2, O = 2 → need to balance oxygen.

→ Put 2 H₂O on left → 4 H, 2 O
→ Right: H₂ = 2 molecules (4 H), O₂ = 1 molecule (2 O)

So:
2 H₂O(l) → 2 H₂(g) + 1 O₂(g)

✔ Balanced.

Type: Decomposition — water breaks down into hydrogen and oxygen.

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c.

Fe(s) + H₂O(g) → H₂(g) + Fe₃O₄(s)

Balancing:
Fe₃O₄ has 3 Fe and 4 O
H₂O has 1 O per molecule → need 4 H₂O for 4 O

→ So 4 H₂O → gives 8 H → makes 4 H₂

Now Fe: 3 Fe needed on left → so 3 Fe

So:
3 Fe(s) + 4 H₂O(g) → 4 H₂(g) + 1 Fe₃O₄(s)

✔ Balanced.

Type: Single replacement — Fe replaces H in H₂O (though not typical; actually, it’s a redox reaction involving oxidation of Fe).

But more accurately, this is a single displacement where iron displaces hydrogen from water.

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d.

AsCl₃(aq) + H₂S(aq) → As₂S₃(s) + HCl(aq)

Balancing:
As₂S₃ has 2 As, 3 S
AsCl₃ has 1 As → need 2 AsCl₃
H₂S has 1 S → need 3 H₂S

→ Left: 2 AsCl₃, 3 H₂S → 6 Cl, 6 H
→ Right: HCl → need 6 HCl

So:
2 AsCl₃(aq) + 3 H₂S(aq) → 1 As₂S₃(s) + 6 HCl(aq)

✔ Balanced.

Type: Double replacement — ions exchange partners.

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e.

CuSO₄·5H₂O(s) → CuSO₄(s) + H₂O(g)

This is dehydration of hydrated copper sulfate.

Each formula unit loses 5 H₂O.

So:
1 CuSO₄·5H₂O(s) → 1 CuSO₄(s) + 5 H₂O(g)

✔ Balanced.

Type: Decomposition — loss of water from hydrate.

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f.

Fe₂O₃(s) + H₂(g) → Fe(s) + H₂O(l)

Fe₂O₃ has 2 Fe, 3 O
H₂O has 1 O → need 3 H₂O → 3 H₂

→ So 3 H₂ → 3 H₂O → 3 H₂

Fe: 2 Fe on left → 2 Fe on right

So:
1 Fe₂O₃(s) + 3 H₂(g) → 2 Fe(s) + 3 H₂O(l)

✔ Balanced.

Type: Single replacement — H₂ reduces Fe³⁺ to Fe metal.

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g.

CaCO₃(s) → CaO(s) + CO₂(g)

Already balanced:

1 CaCO₃(s) → 1 CaO(s) + 1 CO₂(g)

✔ Balanced.

Type: Decomposition — carbonate breaks down.

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h.

Fe(s) + S₈(s) → FeS(s)

S₈ is a molecule of 8 sulfur atoms.

FeS has 1 Fe and 1 S → need 8 FeS → 8 Fe and 8 S

So:
8 Fe(s) + 1 S₈(s) → 8 FeS(s)

✔ Balanced.

Type: Formation (synthesis)

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i.

H₂S(aq) + KOH(aq) → H₂O(l) + K₂S(aq)

H₂S + 2 KOH → 2 H₂O + K₂S

Check:
Left: H = 2+2=4, S=1, K=2, O=2
Right: H=4, O=2, K=2, S=1 → OK

So:
1 H₂S(aq) + 2 KOH(aq) → 2 H₂O(l) + 1 K₂S(aq)

✔ Balanced.

Type: Double replacement / acid-base neutralization

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j.

NaCl(l) → Na(l) + Cl₂(g)

NaCl is molten (liquid) → electrolysis

But NaCl has 1 Na, 1 Cl → Cl₂ is diatomic → need 2 Cl

So:
2 NaCl(l) → 2 Na(l) + 1 Cl₂(g)

✔ Balanced.

Type: Decomposition — via electrolysis

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k.

Al(s) + H₂SO₄(aq) → H₂(g) + Al₂(SO₄)₃(aq)

Al₂(SO₄)₃ has 2 Al, 3 SO₄²⁻
So need 2 Al and 3 H₂SO₄

→ 3 H₂SO₄ → 6 H → 3 H₂

So:
2 Al(s) + 3 H₂SO₄(aq) → 3 H₂(g) + 1 Al₂(SO₄)₃(aq)

✔ Balanced.

Type: Single replacement — Al displaces H⁺

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l.

H₃PO₄(aq) + NH₄OH(aq) → H₂O(l) + (NH₄)₃PO₄(aq)

H₃PO₄ has 3 H⁺, NH₄OH provides OH⁻

→ Need 3 NH₄OH to neutralize 3 H⁺

→ Forms (NH₄)₃PO₄ and 3 H₂O

So:
1 H₃PO₄(aq) + 3 NH₄OH(aq) → 3 H₂O(l) + 1 (NH₄)₃PO₄(aq)

✔ Balanced.

Type: Double replacement / acid-base

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m.

C₃H₈(g) + O₂(g) → CO₂(g) + H₂O(l)

Propane combustion.

C₃H₈ → 3 CO₂ + 4 H₂O (since 8 H → 4 H₂O)

Now count O:
Right: 3×2 + 4×1 = 6 + 4 = 10 O → need 5 O₂

So:
1 C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l)

✔ Balanced.

Type: Hydrocarbon combustion

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n.

Al(s) + O₂(g) → Al₂O₃(s)

Al₂O₃ has 2 Al, 3 O
O₂ has 2 O → need 3/2 O₂ → better use whole numbers

Multiply by 2:
→ 4 Al + 3 O₂ → 2 Al₂O₃

So:
4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

✔ Balanced.

Type: Formation (synthesis)

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o.

CH₄(g) + O₂(g) → CO₂(g) + H₂O(l)

Methane combustion.

CH₄ → CO₂ + 2 H₂O

O: right = 2 + 1 = 3 → need 3/2 O₂ → multiply by 2:

2 CH₄(g) + 3 O₂(g) → 2 CO₂(g) + 4 H₂O(l)

✔ Balanced.

Type: Hydrocarbon combustion

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p.

K₂SO₄(aq) + BaCl₂(aq) → KCl(aq) + BaSO₄(s)

BaSO₄ is precipitate.

Swap ions: K⁺ with Ba²⁺, SO₄²⁻ with Cl⁻

Need 2 KCl → so:

1 K₂SO₄(aq) + 1 BaCl₂(aq) → 2 KCl(aq) + 1 BaSO₄(s)

✔ Balanced.

Type: Double replacement

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q.

C₅H₁₂(l) + O₂(g) → CO₂(g) + H₂O(g)

Pentane combustion.

C₅H₁₂ → 5 CO₂ + 6 H₂O (12 H → 6 H₂O)

O: right = 5×2 + 6×1 = 10 + 6 = 16 → need 8 O₂

So:
1 C₅H₁₂(l) + 8 O₂(g) → 5 CO₂(g) + 6 H₂O(g)

✔ Balanced.

Type: Hydrocarbon combustion

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r.

Ca(OH)₂(aq) + NH₄Cl(aq) → NH₄OH(aq) + CaCl₂(aq)

But NH₄OH is unstable — it decomposes to NH₃ and H₂O

But let’s balance as written:

Ca(OH)₂ has 2 OH⁻, NH₄Cl has 1 NH₄⁺ → need 2 NH₄Cl

→ 2 NH₄Cl → 2 NH₄OH and 1 CaCl₂

So:
1 Ca(OH)₂(aq) + 2 NH₄Cl(aq) → 2 NH₄OH(aq) + 1 CaCl₂(aq)

✔ Balanced.

But note: NH₄OH → NH₃(g) + H₂O(l) — so this is often written as producing ammonia gas.

But here we follow given products.

Type: Double replacement

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s.

V₂O₅(s) + Ca(s) → CaO(s) + V(s)

V₂O₅ has 2 V, 5 O
CaO has 1 O → need 5 CaO → 5 Ca

So:
1 V₂O₅(s) + 5 Ca(s) → 5 CaO(s) + 2 V(s)

✔ Balanced.

Type: Single replacement — Ca reduces V⁵⁺ to V metal

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t.

Na(s) + ZnI₂(aq) → NaI(aq) + Zn(s)

ZnI₂ has 2 I⁻ → need 2 NaI → 2 Na

So:
2 Na(s) + 1 ZnI₂(aq) → 2 NaI(aq) + 1 Zn(s)

✔ Balanced.

Type: Single replacement — Na displaces Zn

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u.

C₇H₆O₃(l) + O₂(g) → CO₂(g) + H₂O(l)

Benzoic acid combustion.

C₇H₆O₃ → 7 CO₂ + 3 H₂O (6 H → 3 H₂O)

Now count O:
Left: 3 (from C₇H₆O₃) + 2x (from O₂)
Right: 7×2 + 3×1 = 14 + 3 = 17

So: 3 + 2x = 17 → 2x = 14 → x = 7

So:
1 C₇H₆O₃(l) + 7 O₂(g) → 7 CO₂(g) + 3 H₂O(l)

✔ Balanced.

Type: Combustion (organic compound)

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v.

Ca(s) + N₂(g) → Ca₃N₂(s)

Ca₃N₂ has 3 Ca, 2 N → need 3 Ca and 1 N₂

So:
3 Ca(s) + 1 N₂(g) → 1 Ca₃N₂(s)

✔ Balanced.

Type: Formation (synthesis)

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w.

Fe₂O₃(s) + H₂(g) → Fe(s) + H₂O(l)

Same as (f):
1 Fe₂O₃(s) + 3 H₂(g) → 2 Fe(s) + 3 H₂O(l)

✔ Balanced.

Type: Single replacement

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x.

C₁₅H₃₀(l) + O₂(g) → CO₂(g) + H₂O(g)

C₁₅H₃₀ → 15 CO₂ + 15 H₂O (30 H → 15 H₂O)

O: right = 15×2 + 15×1 = 30 + 15 = 45
Left: O₂ → 2x → 2x = 45 → x = 22.5 → multiply by 2:

2 C₁₅H₃₀(l) + 45 O₂(g) → 30 CO₂(g) + 30 H₂O(g)

✔ Balanced.

Type: Hydrocarbon combustion

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y.

BN(s) + F₂(g) → BF₃(s) + N₂(g)

BN → B and N
BF₃ needs 3 F → F₂ → need 3/2 F₂ per BN

But N₂ needs 2 N → so 2 BN

→ 2 BN → 2 B → 2 BF₃ → need 6 F → 3 F₂

→ 2 N → 1 N₂

So:
2 BN(s) + 3 F₂(g) → 2 BF₃(s) + 1 N₂(g)

✔ Balanced.

Type: Other (oxidation/reduction, possibly double displacement or decomposition-like)

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z.

C₁₂H₂₆(l) + O₂(g) → CO₂(g) + H₂O(g)

Dodecane combustion.

C₁₂H₂₆ → 12 CO₂ + 13 H₂O (26 H → 13 H₂O)

O: right = 12×2 + 13×1 = 24 + 13 = 37
Left: O₂ → 2x → 2x = 37 → x = 18.5 → multiply by 2:

2 C₁₂H₂₆(l) + 37 O₂(g) → 24 CO₂(g) + 26 H₂O(g)

✔ Balanced.

Type: Hydrocarbon combustion

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## ✔ Summary of Problem 1 Answers

| Letter | Balanced Equation | Type |
|-------|-------------------|------|
| a | 2 Cu + O₂ → 2 CuO | Formation |
| b | 2 H₂O → 2 H₂ + O₂ | Decomposition |
| c | 3 Fe + 4 H₂O → 4 H₂ + Fe₃O₄ | Single replacement |
| d | 2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl | Double replacement |
| e | CuSO₄·5H₂O → CuSO₄ + 5 H₂O | Decomposition |
| f | Fe₂O₃ + 3 H₂ → 2 Fe + 3 H₂O | Single replacement |
| g | CaCO₃ → CaO + CO₂ | Decomposition |
| h | 8 Fe + S₈ → 8 FeS | Formation |
| i | H₂S + 2 KOH → 2 H₂O + K₂S | Double replacement |
| j | 2 NaCl → 2 Na + Cl₂ | Decomposition |
| k | 2 Al + 3 H₂SO₄ → 3 H₂ + Al₂(SO₄)₃ | Single replacement |
| l | H₃PO₄ + 3 NH₄OH → 3 H₂O + (NH₄)₃PO₄ | Double replacement |
| m | C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O | Combustion |
| n | 4 Al + 3 O₂ → 2 Al₂O₃ | Formation |
| o | CH₄ + 2 O₂ → CO₂ + 2 H₂O | Combustion |
| p | K₂SO₄ + BaCl₂ → 2 KCl + BaSO₄ | Double replacement |
| q | C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O | Combustion |
| r | Ca(OH)₂ + 2 NH₄Cl → 2 NH₄OH + CaCl₂ | Double replacement |
| s | V₂O₅ + 5 Ca → 5 CaO + 2 V | Single replacement |
| t | 2 Na + ZnI₂ → 2 NaI + Zn | Single replacement |
| u | C₇H₆O₃ + 7 O₂ → 7 CO₂ + 3 H₂O | Combustion |
| v | 3 Ca + N₂ → Ca₃N₂ | Formation |
| w | Fe₂O₃ + 3 H₂ → 2 Fe + 3 H₂O | Single replacement |
| x | 2 C₁₅H₃₀ + 45 O₂ → 30 CO₂ + 30 H₂O | Combustion |
| y | 2 BN + 3 F₂ → 2 BF₃ + N₂ | Other |
| z | 2 C₁₂H₂₆ + 37 O₂ → 24 CO₂ + 26 H₂O | Combustion |

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## 🧪 Problem 2: Predict products, states, reaction type, and balance

a. Lead(II) nitrate + sodium iodide

Reaction: Double replacement
Products: PbI₂(s) (yellow precipitate), NaNO₃(aq)

Equation:
Pb(NO₃)₂(aq) + 2 NaI(aq) → PbI₂(s) + 2 NaNO₃(aq)

Type: Double replacement

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b. Zinc sulfide + oxygen

Reaction: Combustion or oxidation
ZnS burns in air to form ZnO and SO₂

Equation:
2 ZnS(s) + 3 O₂(g) → 2 ZnO(s) + 2 SO₂(g)

Type: Combustion / oxidation

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c. Butane (C₄H₁₀) as fuel

Combustion:
C₄H₁₀(l) + 13/2 O₂(g) → 4 CO₂(g) + 5 H₂O(g)
Better:
2 C₄H₁₀(l) + 13 O₂(g) → 8 CO₂(g) + 10 H₂O(g)

Type: Hydrocarbon combustion

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d. Barium hydroxide + HCl

Acid-base neutralization:
Ba(OH)₂(aq) + 2 HCl(aq) → BaCl₂(aq) + 2 H₂O(l)

Type: Double replacement / acid-base

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e. Copper + silver nitrate

Single replacement: Cu displaces Ag
Cu(s) + 2 AgNO₃(aq) → Cu(NO₃)₂(aq) + 2 Ag(s)

Type: Single replacement

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f. Sulfur burns in oxygen → SO₂

S(s) + O₂(g) → SO₂(g)

Type: Combination / combustion

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g. Aluminum sulfate + calcium hydroxide

Double replacement: forms Al(OH)₃(s) and CaSO₄(s)

Al₂(SO₄)₃(aq) + 3 Ca(OH)₂(aq) → 2 Al(OH)₃(s) + 3 CaSO₄(s)

Type: Double replacement

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h. Zinc + sulfuric acid

Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)

Type: Single replacement

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i. Aluminum + chlorine gas

2 Al(s) + 3 Cl₂(g) → 2 AlCl₃(s)

Type: Formation

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j. Sucrose undergoes cellular respiration

C₁₂H₂₂O₁₁ + O₂ → CO₂ + H₂O

Balance:
C₁₂H₂₂O₁₁ + 12 O₂ → 12 CO₂ + 11 H₂O

Type: Combustion (biological oxidation)

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## ✔ Final Answers for Problem 2

| Letter | Reaction Description | Balanced Equation | Type |
|--------|------------------------|-------------------|------|
| a | Pb(NO₃)₂ + NaI → PbI₂ + NaNO₃ | Pb(NO₃)₂(aq) + 2 NaI(aq) → PbI₂(s) + 2 NaNO₃(aq) | Double replacement |
| b | ZnS + O₂ → ZnO + SO₂ | 2 ZnS(s) + 3 O₂(g) → 2 ZnO(s) + 2 SO₂(g) | Combustion |
| c | C₄H₁₀ + O₂ → CO₂ + H₂O | 2 C₄H₁₀(l) + 13 O₂(g) → 8 CO₂(g) + 10 H₂O(g) | Combustion |
| d | Ba(OH)₂ + HCl → BaCl₂ + H₂O | Ba(OH)₂(aq) + 2 HCl(aq) → BaCl₂(aq) + 2 H₂O(l) | Acid-base |
| e | Cu + AgNO₃ → Cu(NO₃)₂ + Ag | Cu(s) + 2 AgNO₃(aq) → Cu(NO₃)₂(aq) + 2 Ag(s) | Single replacement |
| f | S + O₂ → SO₂ | S(s) + O₂(g) → SO₂(g) | Combination |
| g | Al₂(SO₄)₃ + Ca(OH)₂ → Al(OH)₃ + CaSO₄ | Al₂(SO₄)₃(aq) + 3 Ca(OH)₂(aq) → 2 Al(OH)₃(s) + 3 CaSO₄(s) | Double replacement |
| h | Zn + H₂SO₄ → ZnSO₄ + H₂ | Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g) | Single replacement |
| i | Al + Cl₂ → AlCl₃ | 2 Al(s) + 3 Cl₂(g) → 2 AlCl₃(s) | Formation |
| j | Sucrose + O₂ → CO₂ + H₂O | C₁₂H₂₂O₁₁(s) + 12 O₂(g) → 12 CO₂(g) + 11 H₂O(l) | Combustion |

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