An ideal gaseous mixture of ethane `(C_(2)H_(6))` and ethane `(C_(2)H_(4))` occupies `28` litre at `1atm` `0^(@)C`. The mixture reacts completely with `128 gm O_(2)` to produce `CO_(2)` and `H_(2)O`. Mole of fraction at `C_(2)H_(6)` in the mixtture is-

Options

A. `0.6`
B. `0.4`
C. `0.5`
D. `0.8`

Explanation

To solve the problem step by step, we will follow the outlined process to find the mole fraction of ethane (C₂H₆) in the mixture.

### Step 1: Determine the total moles of the gas mixture
Given that the mixture occupies 28 liters at 1 atm and 0°C, we can use the ideal gas law to find the total number of moles of the gas mixture.

Using the ideal gas equation:
\[ PV = nRT \]

Where:
- \( P = 1 \, \text{atm} \)
- \( V = 28 \, \text{L} \)
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 273 \, \text{K} \) (0°C)

Rearranging for \( n \):
\[ n = \frac{PV}{RT} \]
Substituting the values:
\[ n = \frac{1 \times 28}{0.0821 \times 273} \]
Calculating:
\[ n \approx \frac{28}{22.414} \approx 1.25 \, \text{moles} \]

### Step 2: Set up the reactions for ethane and ethene
Let:
- \( A \) = moles of ethane (C₂H₆)
- \( B \) = moles of ethene (C₂H₄)

From the problem, we know:
\[ A + B = 1.25 \]

The reactions for the combustion of ethane and ethene are:
1. For ethane:
\[ C_2H_6 + 3.5 O_2 \rightarrow 2 CO_2 + 3 H_2O \]
This means that 1 mole of ethane requires 3.5 moles of O₂.

2. For ethene:
\[ C_2H_4 + 3 O_2 \rightarrow 2 CO_2 + 2 H_2O \]
This means that 1 mole of ethene requires 3 moles of O₂.

### Step 3: Calculate the total moles of O₂ used
From the problem, we know that 128 g of O₂ is used. The molar mass of O₂ is approximately 32 g/mol, so:
\[ \text{Moles of } O_2 = \frac{128}{32} = 4 \, \text{moles} \]

### Step 4: Set up the equation for O₂ consumption
The total moles of O₂ consumed can be expressed as:
\[ 3.5A + 3B = 4 \]

### Step 5: Solve the system of equations
Now we have two equations:
1. \( A + B = 1.25 \)
2. \( 3.5A + 3B = 4 \)

We can solve these equations simultaneously.

From equation 1:
\[ B = 1.25 - A \]

Substituting \( B \) into equation 2:
\[ 3.5A + 3(1.25 - A) = 4 \]
Expanding:
\[ 3.5A + 3.75 - 3A = 4 \]
Combining like terms:
\[ 0.5A + 3.75 = 4 \]
Subtracting 3.75 from both sides:
\[ 0.5A = 0.25 \]
Dividing by 0.5:
\[ A = 0.5 \]

Now substituting back to find \( B \):
\[ B = 1.25 - 0.5 = 0.75 \]

### Step 6: Calculate the mole fraction of ethane
The mole fraction of ethane (C₂H₆) is given by:
\[ X_{C_2H_6} = \frac{A}{A + B} = \frac{0.5}{0.5 + 0.75} = \frac{0.5}{1.25} = 0.4 \]

### Final Answer
The mole fraction of ethane in the mixture is \( 0.4 \).