Balancing Chemical Equations. CA Standards Students know how to ... - Free Printable

Let's solve each of these combustion reactions step by step.

A combustion reaction is a chemical reaction where a hydrocarbon (or organic compound) reacts with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O). The general form is:

> CₓHᵧ + O₂ → CO₂ + H₂O

We need to balance each equation using the law of conservation of mass — atoms of each element must be equal on both sides.

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🔧 Strategy for Balancing:

1. Balance carbon (C) first.
2. Then balance hydrogen (H).
3. Finally, balance oxygen (O) — usually from O₂.

Let’s go through each one.

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48. CH₄ + O₂ → CO₂ + H₂O

- Carbon: 1 on left → 1 CO₂ on right ✔
- Hydrogen: 4 H → need 2 H₂O (since each has 2 H) → 2 H₂O
- Now check oxygen:
- Right: 1 CO₂ (2 O) + 2 H₂O (2 O) = 4 O atoms
- Left: O₂ → so need 2 O₂ molecules (4 O atoms)

✔ Balanced:
> CH₄ + 2O₂ → CO₂ + 2H₂O

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49. C₂H₆ + O₂ → CO₂ + H₂O

- C: 2 → 2 CO₂
- H: 6 → 3 H₂O (each has 2 H)
- Now O:
- Right: 2 CO₂ → 4 O; 3 H₂O → 3 O → total 7 O
- So need 7/2 O₂ → but we want whole numbers → multiply entire equation by 2

Start over:
> C₂H₆ + O₂ → 2CO₂ + 3H₂O

Oxygen on right: 2×2 + 3×1 = 4 + 3 = 7 → need 7/2 O₂ → so use fractions and double:

Multiply all by 2:
> 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

✔ Balanced:
> 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

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50. C₃H₈ + O₂ → CO₂ + H₂O

- C: 3 → 3 CO₂
- H: 8 → 4 H₂O
- O: right side: 3×2 + 4×1 = 6 + 4 = 10 O → so need 5 O₂

✔ Balanced:
> C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

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51. C₄H₁₀ + O₂ → CO₂ + H₂O

- C: 4 → 4 CO₂
- H: 10 → 5 H₂O
- O: right: 4×2 + 5×1 = 8 + 5 = 13 O → need 13/2 O₂ → so multiply by 2

Double everything:
> 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

✔ Balanced:
> 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

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52. C₅H₁₂ + O₂ → CO₂ + H₂O

- C: 5 → 5 CO₂
- H: 12 → 6 H₂O
- O: 5×2 + 6×1 = 10 + 6 = 16 O → need 8 O₂

✔ Balanced:
> C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O

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53. C₆H₁₄ + O₂ → CO₂ + H₂O

- C: 6 → 6 CO₂
- H: 14 → 7 H₂O
- O: 6×2 + 7×1 = 12 + 7 = 19 O → need 19/2 O₂ → multiply by 2

→ 2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

✔ Balanced:
> 2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

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54. C₂H₄ + O₂ → CO₂ + H₂O

- C: 2 → 2 CO₂
- H: 4 → 2 H₂O
- O: 2×2 + 2×1 = 4 + 2 = 6 O → need 3 O₂

✔ Balanced:
> C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

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55. C₂H₂ + O₂ → CO₂ + H₂O

- C: 2 → 2 CO₂
- H: 2 → 1 H₂O
- O: 2×2 + 1×1 = 4 + 1 = 5 O → need 5/2 O₂ → multiply by 2

→ 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O

✔ Balanced:
> 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O

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56. C₆H₆ + O₂ → CO₂ + H₂O

- C: 6 → 6 CO₂
- H: 6 → 3 H₂O
- O: 6×2 + 3×1 = 12 + 3 = 15 O → need 15/2 O₂ → multiply by 2

→ 2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O

✔ Balanced:
> 2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O

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✔ Final Answers:

| Problem | Balanced Equation |
|--------|-------------------|
| 48. | CH₄ + 2O₂ → CO₂ + 2H₂O |
| 49. | 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O |
| 50. | C₃H₈ + 5O₂ → 3CO₂ + 4H₂O |
| 51. | 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O |
| 52. | C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O |
| 53. | 2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O |
| 54. | C₂H₄ + 3O₂ → 2CO₂ + 2H₂O |
| 55. | 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O |
| 56. | 2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O |

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📌 Summary Tips:

- Always balance C, then H, then O.
- If you get fractional O₂, multiply the entire equation by 2 to eliminate fractions.
- Check atom counts after balancing.

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