What volume of air (in `m^(3)`) is needed for the combustion of `1m^(3)` of a gas having the following composition in percentage volume : `2%` of `C_(2)H_(2), 8%` of `CO, 35%` of `CH_(4), 50%` of `H_(2)` and `5%` of non-combustible gas. The air contains `20.8%` (by volume) of oxygen.

Explanation

To solve the problem of determining the volume of air needed for the combustion of 1 m³ of a gas mixture, we will follow these steps:

### Step 1: Identify the Composition of the Gas
The gas mixture consists of:
- 2% C₂H₂ (acetylene)
- 8% CO (carbon monoxide)
- 35% CH₄ (methane)
- 50% H₂ (hydrogen)
- 5% non-combustible gases

### Step 2: Calculate the Volume of Each Component in 1 m³
Given that the total volume of the gas is 1 m³, we can calculate the volume of each component:
- Volume of C₂H₂ = 2% of 1 m³ = 0.02 m³
- Volume of CO = 8% of 1 m³ = 0.08 m³
- Volume of CH₄ = 35% of 1 m³ = 0.35 m³
- Volume of H₂ = 50% of 1 m³ = 0.50 m³

### Step 3: Write the Combustion Reactions
We need to write the balanced chemical equations for the combustion of each component:
1. **For C₂H₂**:
\[ 2 \text{C}_2\text{H}_2 + 5 \text{O}_2 \rightarrow 4 \text{CO}_2 + 2 \text{H}_2\text{O} \]
- 1 m³ of C₂H₂ requires \( \frac{5}{2} \times 0.02 = 0.05 \) m³ of O₂.

2. **For CO**:
\[ 2 \text{CO} + \text{O}_2 \rightarrow 2 \text{CO}_2 \]
- 1 m³ of CO requires \( \frac{1}{2} \times 0.08 = 0.04 \) m³ of O₂.

3. **For CH₄**:
\[ \text{CH}_4 + 2 \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2\text{O} \]
- 1 m³ of CH₄ requires \( 2 \times 0.35 = 0.70 \) m³ of O₂.

4. **For H₂**:
\[ 2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O} \]
- 1 m³ of H₂ requires \( \frac{1}{2} \times 0.50 = 0.25 \) m³ of O₂.

### Step 4: Calculate the Total Volume of O₂ Required
Now, we sum the volumes of O₂ required for each component:
- Total O₂ required = 0.05 + 0.04 + 0.70 + 0.25 = 1.04 m³ of O₂.

### Step 5: Determine the Volume of Air Required
Given that air contains 20.8% O₂, we can calculate the volume of air needed using the formula:
\[ \text{Volume of air} = \frac{\text{Volume of O}_2}{\text{Percentage of O}_2 \text{ in air}} = \frac{1.04 \text{ m}^3}{0.208} \approx 5 \text{ m}^3 \]

### Final Answer
The volume of air needed for the combustion of 1 m³ of the gas mixture is approximately **5 m³**.

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