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. `(19)/(20)`
C. `(17)/(18)`
D. `(15)/(16)`

Answer

To find the mole fraction of `CH4` in the original gas mixture of `CH4` and `C2H2`, we can follow these steps:

### Step 1: Define Variables
Let:
- \( P_{CH4} \) = partial pressure of `CH4`
- \( P_{C2H2} \) = partial pressure of `C2H2`
- Total pressure \( P_{total} = P_{CH4} + P_{C2H2} = 63 \, \text{torr} \)

### Step 2: Express Partial Pressures
From the total pressure, we can express the partial pressures as:
- \( P_{C2H2} = 63 - P_{CH4} \)

### Step 3: Write the Combustion Reactions
The combustion reactions for `CH4` and `C2H2` are:
1. For `CH4`:
\[ CH4 + 2O2 \rightarrow CO2 + 2H2O \]
This means 1 mole of `CH4` produces 1 mole of `CO2`.

2. For `C2H2`:
\[ C2H2 + 5/2 O2 \rightarrow 2CO2 + H2O \]
This means 1 mole of `C2H2` produces 2 moles of `CO2`.

### Step 4: Calculate Total CO2 Produced
Let \( x \) be the number of moles of `CH4` and \( y \) be the number of moles of `C2H2`. The total moles of `CO2` produced from the combustion will be:
\[ \text{Total } CO2 = x + 2y \]

### Step 5: Relate CO2 Pressure to Original Gases
The pressure of `CO2` collected after combustion is given as \( 99 \, \text{torr} \). According to the ideal gas law, the pressure of a gas is proportional to the number of moles. Thus:
\[ P_{CO2} = x + 2y \]
And since we know the total pressure of `CO2` is \( 99 \, \text{torr} \):
\[ x + 2y = 99 \]

### Step 6: Set Up the System of Equations
Now we have a system of equations:
1. \( P_{CH4} + P_{C2H2} = 63 \)
2. \( x + 2y = 99 \)

From \( P_{CH4} = x \) and \( P_{C2H2} = 63 - x \):
\[ x + 2(63 - x) = 99 \]

### Step 7: Solve for x
Expanding and simplifying:
\[ x + 126 - 2x = 99 \]
\[ 126 - x = 99 \]
\[ x = 126 - 99 = 27 \]

### Step 8: Calculate y
Substituting \( x \) back into the equation for \( P_{C2H2} \):
\[ P_{C2H2} = 63 - 27 = 36 \]
Now, using the first equation:
\[ 27 + 2y = 99 \]
\[ 2y = 99 - 27 = 72 \]
\[ y = 36 \]

### Step 9: Calculate Mole Fraction of CH4
The mole fraction of `CH4` is given by:
\[ \text{Mole fraction of } CH4 = \frac{x}{x + y} = \frac{27}{27 + 36} = \frac{27}{63} = \frac{9}{21} = \frac{3}{7} \]

### Final Answer
The mole fraction of `CH4` in the original gas mixture is \( \frac{3}{7} \).

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