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 with a specific composition, we will follow these steps:

### Step 1: Identify the composition of the gas
The gas composition by volume is:
- 2% C₂H₂
- 8% CO
- 35% CH₄
- 50% H₂
- 5% Non-combustible gas

### Step 2: Calculate the moles of each component in 1 m³ of gas
Since the total volume is 1 m³, the moles of each component can be calculated as follows:
- Moles of C₂H₂ = 0.02 m³
- Moles of CO = 0.08 m³
- Moles of CH₄ = 0.35 m³
- Moles of H₂ = 0.50 m³
- Non-combustible gas does not participate in combustion.

### Step 3: Write the combustion reactions
1. For C₂H₂:
\[ C_2H_2 + \frac{5}{2} O_2 \rightarrow 2 CO_2 + H_2O \]
1 mole of C₂H₂ requires 2.5 moles of O₂.

2. For CO:
\[ 2 CO + O_2 \rightarrow 2 CO_2 \]
2 moles of CO require 1 mole of O₂, thus 0.08 m³ of CO requires 0.04 m³ of O₂.

3. For CH₄:
\[ CH_4 + 2 O_2 \rightarrow CO_2 + 2 H_2O \]
1 mole of CH₄ requires 2 moles of O₂, thus 0.35 m³ of CH₄ requires 0.70 m³ of O₂.

4. For H₂:
\[ 2 H_2 + O_2 \rightarrow 2 H_2O \]
2 moles of H₂ require 1 mole of O₂, thus 0.50 m³ of H₂ requires 0.25 m³ of O₂.

### Step 4: Calculate the total moles of O₂ required
Now we can sum the moles of O₂ required for each component:
- From C₂H₂: \(0.02 \times 2.5 = 0.05\) m³ O₂
- From CO: \(0.08 \times 0.5 = 0.04\) m³ O₂
- From CH₄: \(0.35 \times 2 = 0.70\) m³ O₂
- From H₂: \(0.50 \times 0.5 = 0.25\) m³ O₂

Total O₂ required:
\[ 0.05 + 0.04 + 0.70 + 0.25 = 1.04 \text{ m³ O₂} \]

### Step 5: Calculate the volume of air needed
Given that the air contains 20.8% O₂, we can calculate the volume of air required:
\[ \text{Volume of air} = \frac{\text{Volume of O₂ required}}{\text{Percentage of O₂ in air}} = \frac{1.04 \text{ m³}}{0.208} \approx 5 \text{ m³} \]

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

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