A mixture of `CH_(4)` and `C_(2)H_(2)` occupied a certain volume at a total pressure equal to 63 torr. The same gas mixture was burnt to `CO_(2) and H_(2)O(l).CO_(2)(g)` alone was collected in the same volume and at the same temperature, the pressure was found to be 99 torr.
What was the mole fraction of `CH_(4)` in the original gas mixture?

Options

A. `(19)/(21)`
B. `(3)/(7)`
C. `(7)/(8)`
D. `(15)/(16)`

Answer

To solve the problem, we need to find the mole fraction of `CH4` in the original gas mixture of `CH4` and `C2H2`. Here’s a step-by-step solution:

### Step 1: Define the Variables
Let:
- \( P \) = partial pressure of \( CH_4 \)
- \( 63 - P \) = partial pressure of \( C_2H_2 \)

### Step 2: Write the Combustion Reactions
1. For methane (\( CH_4 \)):
\[ CH_4 + 2 O_2 \rightarrow CO_2 + 2 H_2O \]
From this reaction, 1 mole of \( CH_4 \) produces 1 mole of \( CO_2 \).

2. For ethyne (\( C_2H_2 \)):
\[ C_2H_2 + \frac{5}{2} O_2 \rightarrow 2 CO_2 + H_2O \]
From this reaction, 1 mole of \( C_2H_2 \) produces 2 moles of \( CO_2 \).

### Step 3: Calculate the Total Pressure of \( CO_2 \)
The total pressure of \( CO_2 \) produced from both reactions is:
\[ P_{CO_2} = P + 2(63 - P) \]
This simplifies to:
\[ P_{CO_2} = P + 126 - 2P = 126 - P \]

### Step 4: Set Up the Equation
According to the problem, the total pressure of \( CO_2 \) collected is 99 torr:
\[ 126 - P = 99 \]

### Step 5: Solve for \( P \)
Rearranging the equation gives:
\[ P = 126 - 99 = 27 \text{ torr} \]

### Step 6: Calculate the Mole Fraction of \( CH_4 \)
The mole fraction of \( CH_4 \) can be calculated using its partial pressure:
\[ \text{Mole fraction of } CH_4 = \frac{P_{CH_4}}{P_{total}} = \frac{P}{63} = \frac{27}{63} \]
This simplifies to:
\[ \text{Mole fraction of } CH_4 = \frac{3}{7} \]

### Final Answer
The mole fraction of \( CH_4 \) in the original gas mixture is \( \frac{3}{7} \).

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