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

Let's solve this "Writing and Balancing Chemical Reactions" worksheet step by step. I'll go through each part of the problem, balancing the equations and identifying the reaction type.

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Part 1: Balance the following equations and indicate the type of reaction

We will balance each equation and classify it as:
- Formation
- Decomposition
- Single replacement
- Double replacement
- Hydrocarbon combustion
- Other

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#### a.
Cu(s) + O₂(g) → CuO(s)

Balance:
- Left: 1 Cu, 2 O
- Right: 1 Cu, 1 O → need 2 CuO on right → then 2 Cu on left

→ 2Cu + O₂ → 2CuO

✔ Balanced

Type: Formation (metal + oxygen → metal oxide)

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#### b.
H₂O(l) → H₂(g) + O₂(g)

Balance:
- Left: 2 H, 1 O
- Right: 2 H, 2 O → need 2 H₂O on left

→ 2H₂O → 2H₂ + O₂

✔ Balanced

Type: Decomposition (water breaks into hydrogen and oxygen)

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#### c.
Fe(s) + H₂O(g) → H₂(g) + Fe₃O₄(s)

Balance:
- Fe: 3 on right → 3 Fe on left
- H₂O: 4 H on right → 4 H₂O needed → 4 H₂ produced

→ 3Fe + 4H₂O → 4H₂ + Fe₃O₄

✔ Balanced

Type: Single replacement (Fe replaces H in water)

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#### d.
AsCl₃(aq) + H₂S(aq) → As₂S₃(s) + HCl(aq)

Balance:
- As: 2 on right → 2 AsCl₃ on left
- S: 3 on right → 3 H₂S on left
- H: 6 H from 3 H₂S → 6 HCl on right

→ 2AsCl₃ + 3H₂S → As₂S₃ + 6HCl

✔ Balanced

Type: Double replacement (ions swap partners)

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#### e.
CuSO₄·5H₂O(s) → CuSO₄(s) + H₂O(g)

Balance:
- 5 H₂O molecules → 5 H₂O on right

→ CuSO₄·5H₂O → CuSO₄ + 5H₂O

✔ Balanced

Type: Decomposition (loss of water from hydrate)

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#### f.
Fe₂O₃(s) + H₂(g) → Fe(s) + H₂O(l)

Balance:
- Fe: 2 on left → 2 Fe on right
- O: 3 → 3 H₂O → 3 H₂ needed

→ Fe₂O₃ + 3H₂ → 2Fe + 3H₂O

✔ Balanced

Type: Single replacement (H₂ reduces Fe³⁺ to Fe)

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#### g.
CaCO₃(s) → CaO(s) + CO₂(g)

Already balanced:

→ CaCO₃ → CaO + CO₂

✔ Balanced

Type: Decomposition (calcium carbonate decomposes)

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#### h.
Fe(s) + S₈(s) → FeS(s)

Balance:
- S₈ has 8 S atoms → need 8 FeS → 8 Fe on left

→ 8Fe + S₈ → 8FeS

✔ Balanced

Type: Formation (elemental metals and nonmetals form compound)

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#### i.
H₂S(aq) + KOH(aq) → H₂O(l) + K₂S(aq)

Balance:
- K: 2 on right → 2 KOH on left
- H: 2 from H₂S + 2 from 2KOH = 4 H → 2 H₂O on right

→ H₂S + 2KOH → 2H₂O + K₂S

✔ Balanced

Type: Acid-base neutralization (double replacement)

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#### j.
NaCl(l) → Na(l) + Cl₂(g)

Balance:
- NaCl → Na and Cl₂ → need 2 NaCl for Cl₂

→ 2NaCl → 2Na + Cl₂

✔ Balanced

Type: Decomposition (electrolysis of molten salt)

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

Balance:
- Al: 2 on right → 2 Al on left
- SO₄: 3 on right → 3 H₂SO₄ on left
- H: 6 H → 3 H₂

→ 2Al + 3H₂SO₄ → 3H₂ + Al₂(SO₄)₃

✔ Balanced

Type: Single replacement (Al displaces H⁺)

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

Balance:
- PO₄: 1 on both sides
- N: 3 on right → 3 NH₄OH on left
- H: left: 3(H₃PO₄) + 3(NH₄OH) = 3+3=6 H from acid + 3×5=15 H from base? Wait — better count:

Better:
- (NH₄)₃PO₄ needs 3 NH₄⁺ → 3 NH₄OH
- H₃PO₄ provides 3 H⁺
- 3 NH₄OH → 3 NH₄⁺ + 3 OH⁻ → forms 3 H₂O

So: H₃PO₄ + 3NH₄OH → 3H₂O + (NH₄)₃PO₄

✔ Balanced

Type: Double replacement / acid-base neutralization

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#### m.
C₃H₈(g) + O₂(g) → CO₂(g) + H₂O(l)

Combustion:
- C: 3 → 3 CO₂
- H: 8 → 4 H₂O
- O: right: 3×2 + 4×1 = 10 → 5 O₂

→ C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

✔ Balanced

Type: Hydrocarbon combustion

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#### n.
Al(s) + O₂(g) → Al₂O₃(s)

Balance:
- Al: 2 → 2 Al
- O: 3 → 3/2 O₂ → multiply all by 2:

→ 4Al + 3O₂ → 2Al₂O₃

✔ Balanced

Type: Formation (metal + oxygen → oxide)

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#### o.
CH₄(g) + O₂(g) → CO₂(g) + H₂O(l)

Combustion:
- C: 1 → 1 CO₂
- H: 4 → 2 H₂O
- O: 2 on right → 2 on left → but CO₂ has 2 O, H₂O has 1 O → total 4 O → need 2 O₂

→ CH₄ + 2O₂ → CO₂ + 2H₂O

✔ Balanced

Type: Hydrocarbon combustion

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#### p.
K₂SO₄(aq) + BaCl₂(aq) → KCl(aq) + BaSO₄(s)

Balance:
- Ba: 1 → 1 BaSO₄
- SO₄: 1 → 1 K₂SO₄
- K: 2 → 2 KCl
- Cl: 2 → 2 Cl → so 1 BaCl₂

→ K₂SO₄ + BaCl₂ → 2KCl + BaSO₄

✔ Balanced

Type: Double replacement (precipitation)

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#### q.
C₅H₁₂(l) + O₂(g) → CO₂(g) + H₂O(g)

Combustion:
- C: 5 → 5 CO₂
- H: 12 → 6 H₂O
- O: right: 5×2 + 6×1 = 16 → 8 O₂

→ C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O

✔ Balanced

Type: Hydrocarbon combustion

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#### r.
Ca(OH)₂(aq) + NH₄Cl(aq) → NH₄OH(aq) + CaCl₂(aq)

Balance:
- Ca: 1 → 1 CaCl₂ → 2 Cl → 2 NH₄Cl
- NH₄⁺: 2 → 2 NH₄OH
- OH: 2 from Ca(OH)₂ → 2 NH₄OH

→ Ca(OH)₂ + 2NH₄Cl → 2NH₄OH + CaCl₂

✔ Balanced

Type: Double replacement

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#### s.
V₂O₅(s) + Ca(s) → CaO(s) + V(s)

Balance:
- V: 2 → 2 V
- O: 5 → 5 CaO → 5 Ca
- Ca: 5 → 5 Ca

→ V₂O₅ + 5Ca → 5CaO + 2V

✔ Balanced

Type: Single replacement (Ca reduces V⁵⁺)

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#### t.
Na(s) + ZnI₂(aq) → NaI(aq) + Zn(s)

Balance:
- Zn: 1 → 1 Zn
- I: 2 → 2 NaI → 2 Na

→ 2Na + ZnI₂ → 2NaI + Zn

✔ Balanced

Type: Single replacement (Na displaces Zn)

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#### u.
C₇H₆O₃(l) + O₂(g) → CO₂(g) + H₂O(l)

This is salicylic acid, a hydrocarbon derivative with oxygen.

Balance:
- C: 7 → 7 CO₂
- H: 6 → 3 H₂O
- O: left: 3 (from C₇H₆O₃) + 2x (from O₂)
- right: 7×2 + 3×1 = 14 + 3 = 17 O

So: 3 + 2x = 17 → x = 7

→ C₇H₆O₃ + 7O₂ → 7CO₂ + 3H₂O

✔ Balanced

Type: Combustion (organic compound)

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#### v.
Ca(s) + N₂(g) → Ca₃N₂(s)

Balance:
- Ca: 3 → 3 Ca
- N: 2 → 1 N₂

→ 3Ca + N₂ → Ca₃N₂

✔ Balanced

Type: Formation (metal + nonmetal → ionic compound)

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#### w.
Fe₂O₃(s) + H₂(g) → Fe(s) + H₂O(l)

Same as f. → already done

→ Fe₂O₃ + 3H₂ → 2Fe + 3H₂O

✔ Balanced

Type: Single replacement

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#### x.
C₁₅H₃₀(l) + O₂(g) → CO₂(g) + H₂O(g)

Combustion:
- C: 15 → 15 CO₂
- H: 30 → 15 H₂O
- O: right: 15×2 + 15×1 = 30 + 15 = 45 → 45/2 = 22.5 → ×2

Multiply all by 2:
→ 2C₁₅H₃₀ + 45O₂ → 30CO₂ + 30H₂O

✔ Balanced

Type: Hydrocarbon combustion

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#### y.
BN(s) + F₂(g) → BF₃(s) + N₂(g)

Balance:
- B: 1 → 1 BF₃
- N: 2 → 2 BN → 1 N₂
- F: 3 → 3 F₂ → 3/2 F₂ → ×2

→ 2BN + 3F₂ → 2BF₃ + N₂

✔ Balanced

Type: Other (not standard category — possibly combination or redox)

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#### z.
C₁₂H₂₆(l) + O₂(g) → CO₂(g) + H₂O(g)

Combustion:
- C: 12 → 12 CO₂
- H: 26 → 13 H₂O
- O: right: 12×2 + 13×1 = 24 + 13 = 37 → 37/2 = 18.5 → ×2

→ 2C₁₂H₂₆ + 37O₂ → 24CO₂ + 26H₂O

✔ Balanced

Type: Hydrocarbon combustion

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✔ Part 1 Summary Table

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

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Part 2: Predict products, states, type, and balance

Now let’s predict and balance each.

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#### a. Lead(II) nitrate + sodium iodide

Reaction: Double replacement
Precipitate: PbI₂ (yellow solid)

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

✔ Balanced
Type: Double replacement (precipitation)

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#### b. Zinc sulfide reacts with oxygen

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

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

✔ Balanced
Type: Combustion / oxidation

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#### c. Butane (C₄H₁₀) as fuel → combustion

Equation:
2C₄H₁₀(l) + 13O₂(g) → 8CO₂(g) + 10H₂O(g)

Balanced:
- C: 8 → 8 CO₂
- H: 20 → 10 H₂O
- O: 26 → 13 O₂

✔ Balanced
Type: Hydrocarbon combustion

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#### d. Barium hydroxide + HCl → neutralization

Products: BaCl₂ and H₂O

Equation:
Ba(OH)₂(aq) + 2HCl(aq) → BaCl₂(aq) + 2H₂O(l)

✔ Balanced
Type: Acid-base neutralization (double replacement)

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#### e. Copper + silver nitrate

Reaction: Single replacement (Cu displaces Ag)

Equation:
Cu(s) + 2AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)

✔ Balanced
Type: Single replacement

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#### f. Sulfur burns in oxygen → sulfur dioxide

Equation:
S(s) + O₂(g) → SO₂(g)

✔ Balanced
Type: Combination / combustion

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#### g. Aluminum sulfate + calcium hydroxide

Reaction: Double replacement → precipitate of Al(OH)₃ and CaSO₄

But Al(OH)₃ is insoluble; CaSO₄ is slightly soluble.

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

Note: CaSO₄ may be aqueous if dilute, but often considered precipitate.

✔ Balanced
Type: Double replacement (precipitation)

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#### h. Zinc metal + sulfuric acid

Reaction: Single replacement → H₂ gas

Equation:
Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)

✔ Balanced
Type: Single replacement

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#### i. Aluminum powder + chlorine gas

Reaction: Formation of aluminum chloride

Equation:
2Al(s) + 3Cl₂(g) → 2AlCl₃(s)

✔ Balanced
Type: Formation / combination

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#### j. Sucrose undergoes cellular respiration

Sucrose: C₁₂H₂₂O₁₁

Respiration: C₁₂H₂₂O₁₁ + O₂ → CO₂ + H₂O

Balance:
- C: 12 → 12 CO₂
- H: 22 → 11 H₂O
- O: left: 11 + 2x, right: 12×2 + 11×1 = 24 + 11 = 35

So: 11 + 2x = 35 → x = 12

→ C₁₂H₂₂O₁₁ + 12O₂ → 12CO₂ + 11H₂O

✔ Balanced
Type: Combustion (biological oxidation)

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✔ Part 2 Summary

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

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✔ Final Notes

All equations are now balanced and classified correctly.

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