A mixture of `C_(2)H_(2)` and `C_(3)H_(8)` occupied a certain volume at 80 mm Hg. The mixture was completely burnt to `CO_(2)` and `H_(2)O(l)`. When the pressure of `CO_(2)` was found to be 230 mm Hg at the same temperature and volume, the mole fraction of `C_(3)H_(8)` in the mixture is :

Options

A. 0.125
B. 0.875
C. 0.6
D. 0.8

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the combustion reactions for both hydrocarbons
1. **Combustion of Acetylene (C₂H₂)**:
\[ C_2H_2 + \frac{5}{2} O_2 \rightarrow 2 CO_2 + H_2O \]
This means 1 mole of C₂H₂ produces 2 moles of CO₂.

2. **Combustion of Propane (C₃H₈)**:
\[ C_3H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2O \]
This means 1 mole of C₃H₈ produces 3 moles of CO₂.

### Step 2: Define variables for the moles of each component
Let:
- \( x \) = moles of C₂H₂
- \( y \) = moles of C₃H₈

### Step 3: Set up the equations based on the problem statement
1. **Total pressure before combustion**:
\[ P_{total} = P_{C_2H_2} + P_{C_3H_8} = 80 \, \text{mm Hg} \]
This gives us the first equation:
\[ x + y = 80 \quad (1) \]

2. **Total pressure after combustion**:
The total moles of CO₂ produced after combustion is:
\[ 2x + 3y \]
Given that the pressure of CO₂ is 230 mm Hg, we have the second equation:
\[ 2x + 3y = 230 \quad (2) \]

### Step 4: Solve the equations
From equation (1):
\[ y = 80 - x \quad (3) \]

Substituting equation (3) into equation (2):
\[ 2x + 3(80 - x) = 230 \]
Expanding this gives:
\[ 2x + 240 - 3x = 230 \]
Combining like terms:
\[ -x + 240 = 230 \]
Solving for \( x \):
\[ -x = 230 - 240 \]
\[ -x = -10 \implies x = 10 \]

Now substituting \( x \) back into equation (3) to find \( y \):
\[ y = 80 - 10 = 70 \]

### Step 5: Calculate the mole fraction of C₃H₈
The mole fraction of C₃H₈ is given by:
\[ \text{Mole fraction of } C_3H_8 = \frac{y}{x + y} = \frac{70}{10 + 70} = \frac{70}{80} = 0.875 \]

### Final Answer
The mole fraction of C₃H₈ in the mixture is **0.875**.

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