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`

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the total number of moles of the gas mixture.
Using the ideal gas equation, \( PV = nRT \), we can rearrange it to find the number of moles \( n \):

\[ n = \frac{PV}{RT} \]

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

Substituting the values:

\[ n = \frac{1 \times 28}{0.0821 \times 273} \approx \frac{28}{22.414} \approx 1.25 \, \text{moles} \]

### Step 2: Write the balanced chemical reactions for the combustion of ethane and ethene.
The combustion reactions are:

1. For ethane \( (C_2H_6) \):
\[ C_2H_6 + \frac{7}{2} O_2 \rightarrow 2 CO_2 + 3 H_2O \]

2. For ethene \( (C_2H_4) \):
\[ C_2H_4 + 3 O_2 \rightarrow 2 CO_2 + 2 H_2O \]

### Step 3: Define variables for the moles of each component.
Let:
- \( x \) = moles of \( C_2H_6 \)
- \( 1.25 - x \) = moles of \( C_2H_4 \)

### Step 4: Calculate the total moles of \( O_2 \) required for the reaction.
From the balanced equations:
- Moles of \( O_2 \) required for \( C_2H_6 \): \( \frac{7}{2} x \)
- Moles of \( O_2 \) required for \( C_2H_4 \): \( 3(1.25 - x) \)

Total moles of \( O_2 \):

\[ \text{Total } O_2 = \frac{7}{2} x + 3(1.25 - x) = \frac{7}{2} x + 3.75 - 3x \]

Combining terms:

\[ \text{Total } O_2 = \left(\frac{7}{2} - 3\right)x + 3.75 = \left(\frac{1}{2}\right)x + 3.75 \]

### Step 5: Set up the equation based on the given mass of \( O_2 \).
Given that \( 128 \, \text{g} \) of \( O_2 \) is used, we can convert this mass to moles:

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

Setting the total moles of \( O_2 \) equal to 4:

\[ \left(\frac{1}{2}\right)x + 3.75 = 4 \]

### Step 6: Solve for \( x \).
Rearranging the equation:

\[ \frac{1}{2}x = 4 - 3.75 = 0.25 \]

Multiplying both sides by 2:

\[ x = 0.5 \]

### Step 7: Calculate the moles of \( C_2H_4 \).
Using \( x \):

\[ \text{Moles of } C_2H_4 = 1.25 - x = 1.25 - 0.5 = 0.75 \]

### Step 8: Calculate the mole fraction of \( C_2H_6 \).
The mole fraction \( \chi \) of \( C_2H_6 \) is given by:

\[ \chi_{C_2H_6} = \frac{\text{moles of } C_2H_6}{\text{total moles}} = \frac{x}{1.25} = \frac{0.5}{1.25} = 0.4 \]

Thus, the mole fraction of ethane \( (C_2H_6) \) in the mixture is **0.4**.

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