An ideal gaseous mixture of ethane `(C_(2)H_(6))` and ethene `(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 procedure:

### Step 1: Calculate the moles of the gas mixture using the ideal gas equation.

The ideal gas equation is given by:

\[ PV = nRT \]

Where:
- \( P \) = pressure (1 atm)
- \( V \) = volume (28 L)
- \( n \) = number of moles
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin (0°C = 273 K)

Substituting the values into the equation:

\[ n = \frac{PV}{RT} = \frac{(1 \text{ atm}) \times (28 \text{ L})}{(0.0821 \text{ L·atm/(K·mol)}) \times (273 \text{ K})} \]

Calculating the right side:

\[ n = \frac{28}{22.414} \approx 1.25 \text{ moles} \]

### Step 2: Set up the equations based on the reaction.

Let \( x \) be the number of moles of ethane \( (C_2H_6) \) and \( y \) be the number of moles of ethene \( (C_2H_4) \).

From the problem, we know:

1. The total moles of the mixture:
\[ x + y = 1.25 \quad \text{(Equation 1)} \]

2. The reaction with oxygen:
- For ethane \( (C_2H_6) \):
\[ C_2H_6 + \frac{7}{2} O_2 \rightarrow 2 CO_2 + 3 H_2O \]
This means \( x \) moles of \( C_2H_6 \) will require \( \frac{7}{2} x \) moles of \( O_2 \).

- For ethene \( (C_2H_4) \):
\[ C_2H_4 + 3 O_2 \rightarrow 2 CO_2 + 2 H_2O \]
This means \( y \) moles of \( C_2H_4 \) will require \( 3y \) moles of \( O_2 \).

The total moles of \( O_2 \) used is given as 128 g. The molar mass of \( O_2 \) is 32 g/mol, so the moles of \( O_2 \) is:

\[ \text{Moles of } O_2 = \frac{128 \text{ g}}{32 \text{ g/mol}} = 4 \text{ moles} \]

Thus, we have the second equation:

\[ \frac{7}{2} x + 3y = 4 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations.

From Equation 1:

\[ y = 1.25 - x \]

Substituting \( y \) in Equation 2:

\[ \frac{7}{2} x + 3(1.25 - x) = 4 \]

Expanding and simplifying:

\[ \frac{7}{2} x + 3.75 - 3x = 4 \]

Combining like terms:

\[ \frac{7}{2} x - 3x = 4 - 3.75 \]

\[ \frac{7}{2} x - \frac{6}{2} x = 0.25 \]

\[ \frac{1}{2} x = 0.25 \]

Thus,

\[ x = 0.5 \text{ moles of } C_2H_6 \]

Now substituting back to find \( y \):

\[ y = 1.25 - 0.5 = 0.75 \text{ moles of } C_2H_4 \]

### Step 4: Calculate the mole fraction of \( C_2H_6 \).

The mole fraction \( \chi_{C_2H_6} \) is given by:

\[ \chi_{C_2H_6} = \frac{x}{x + y} = \frac{0.5}{1.25} = 0.4 \]

### Final Answer:

The mole fraction of \( C_2H_6 \) in the mixture is **0.4**.

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