Balancing Equations Practice Worksheet featuring chemical reactions like CH₄ + O₂ → CO₂ + H₂O.

Let's solve each of the combustion reactions in the worksheet by balancing the chemical equations. These are all hydrocarbon combustion reactions, where a hydrocarbon reacts with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O).

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🔹 1. Methane Combustion:

Unbalanced:
$$
\_\_ \text{CH}_4 + \_\_ \text{O}_2 \rightarrow \_\_ \text{CO}_2 + \_\_ \text{H}_2\text{O}
$$

#### Step-by-step balancing:

1. Carbon (C): 1 C on left → 1 C on right → so CO₂ coefficient = 1
$$
\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}
$$

2. Hydrogen (H): 4 H on left → 2 H per H₂O → need 2 H₂O to get 4 H
$$
\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}
$$

3. Oxygen (O):
Right side: 2 O from CO₂ + 2×1 = 2 O from 2 H₂O → total 4 O atoms
Left: O₂ has 2 O per molecule → need 2 O₂ to get 4 O
$$
\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}
$$

✔ Balanced!

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🔹 2. Propane Combustion:

Unbalanced:
$$
\_\_ \text{C}_3\text{H}_8 + \_\_ \text{O}_2 \rightarrow \_\_ \text{CO}_2 + \_\_ \text{H}_2\text{O}
$$

#### Step-by-step:

1. Carbon (C): 3 C → need 3 CO₂
$$
\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3\text{CO}_2 + \text{H}_2\text{O}
$$

2. Hydrogen (H): 8 H → each H₂O has 2 H → need 4 H₂O
$$
\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}
$$

3. Oxygen (O):
Right: 3 CO₂ → 3×2 = 6 O, 4 H₂O → 4×1 = 4 O → total 10 O atoms
Left: O₂ → each has 2 O → need 5 O₂ to give 10 O
$$
\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}
$$

✔ Balanced!

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🔹 3. Octane Combustion:

Unbalanced:
$$
\_\_ \text{C}_8\text{H}_{18} + \_\_ \text{O}_2 \rightarrow \_\_ \text{CO}_2 + \_\_ \text{H}_2\text{O}
$$

#### Step-by-step:

1. Carbon (C): 8 C → need 8 CO₂
$$
\text{C}_8\text{H}_{18} + \text{O}_2 \rightarrow 8\text{CO}_2 + \text{H}_2\text{O}
$$

2. Hydrogen (H): 18 H → each H₂O has 2 H → need 9 H₂O
$$
\text{C}_8\text{H}_{18} + \text{O}_2 \rightarrow 8\text{CO}_2 + 9\text{H}_2\text{O}
$$

3. Oxygen (O):
Right: 8 CO₂ → 8×2 = 16 O, 9 H₂O → 9×1 = 9 O → total 25 O atoms
Left: O₂ → each has 2 O → need 25/2 = 12.5 O₂ → but we can't have half molecules!

So multiply entire equation by 2 to eliminate fraction.

Start over with coefficients:
- C₈H₁₈ → 1
- CO₂ → 8
- H₂O → 9

Now O atoms needed: 8×2 + 9×1 = 16 + 9 = 25

So O₂ needed: 25/2 → write as 25/2 O₂

Multiply whole equation by 2:
- 2 C₈H₁₈
- 25 O₂
- 16 CO₂
- 18 H₂O

$$
2\text{C}_8\text{H}_{18} + 25\text{O}_2 \rightarrow 16\text{CO}_2 + 18\text{H}_2\text{O}
$$

✔ Balanced!

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✔ Final Balanced Equations:

1. CH₄ + 2O₂ → CO₂ + 2H₂O
2. C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
3. 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

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🧠 Tips for Balancing Combustion Reactions:

- Balance C first → CO₂
- Then H → H₂O
- Finally O → O₂
- If you get fractions, multiply entire equation by denominator to eliminate them.
- Always check atom counts on both sides!

Let me know if you'd like a printable version or more practice!