100 ml of a mixture of methane and acetylene was exploded with excess of oxygen. After cooling to room temperature the resulting gas mixture was passed through KOH when a reduction of 135 ml in volume was noted. The composition of the hydrocarbon mixture by volume at the same temperature and pressure will be

Options

A. 65% `CH_4, 35% C_2H_2`
B. `35% CH_4, 65% C_2H_2`
C. `25% CH_4,75% C_2H_2`
D. `85% CH_4, 15% C_2H_2`

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equations for the combustion of methane and acetylene.

1. **Combustion of Methane (CH₄)**:
\[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \]
From this equation, we can see that 1 volume of methane produces 1 volume of carbon dioxide.

2. **Combustion of Acetylene (C₂H₂)**:
\[ 2\text{C}_2\text{H}_2 + 5\text{O}_2 \rightarrow 4\text{CO}_2 + 2\text{H}_2\text{O} \]
This means that 1 volume of acetylene produces 2 volumes of carbon dioxide.

### Step 2: Define variables for the volumes of methane and acetylene in the mixture.

Let:
- \( x \) = volume of acetylene (C₂H₂) in mL
- \( 100 - x \) = volume of methane (CH₄) in mL

### Step 3: Calculate the total volume of carbon dioxide produced.

From the combustion reactions:
- The volume of CO₂ produced from methane is \( 100 - x \) mL.
- The volume of CO₂ produced from acetylene is \( 2x \) mL.

Thus, the total volume of CO₂ produced is:
\[ \text{Total CO}_2 = (100 - x) + 2x = 100 + x \text{ mL} \]

### Step 4: Set up the equation based on the volume reduction observed after passing through KOH.

According to the problem, after passing the gas mixture through KOH, the volume of gas decreased by 135 mL. This means:
\[ 100 + x = 135 \]

### Step 5: Solve for \( x \).

Rearranging the equation:
\[ x = 135 - 100 = 35 \text{ mL} \]

### Step 6: Calculate the volume of methane (CH₄).

Using the value of \( x \):
\[ \text{Volume of CH}_4 = 100 - x = 100 - 35 = 65 \text{ mL} \]

### Step 7: Calculate the percentage composition of the gases.

1. **Percentage of Methane (CH₄)**:
\[ \text{Percentage of CH}_4 = \left( \frac{65}{100} \right) \times 100 = 65\% \]

2. **Percentage of Acetylene (C₂H₂)**:
\[ \text{Percentage of C}_2\text{H}_2 = \left( \frac{35}{100} \right) \times 100 = 35\% \]

### Final Answer:
The composition of the hydrocarbon mixture by volume is:
- 65% CH₄
- 35% C₂H₂

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