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Mole Fraction

An ideal gase...

Question

An ideal gaseous mixture of ethane $(C_{2} H_{6})$ and ethene $(C_{2} H_{4})$ occupies $28 litres$ at 1 atm and $273 K$ . The mixture reacts completely with 128 g of $O_{2}$ to produce $C O_{2}$ and $H_{2} O$ . The mole fraction of $C_{2} H_{6}$ in the mixture is:

A

0.6

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B

0.4

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C

0.5

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D

0.8

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Solution

The correct option is B 0.4
Balanced reactions when they react with $O_{2}$ are as follows:
$C_{2} H_{6} + 3.5 O_{2} \to 2 C O_{2} + 3 H_{2} O$
$C_{2} H_{4} + 3 O_{2} \to 2 C O_{2} + 2 H_{2} O$
Let volume of ethane be `x`.
$22.4 \times 4 = 3.5 x + 3 (28 - x)$
$x = 11.2$
at constant T and P ,
$V \propto n$
Therefore, mole fraction of ethane in mixture becomes
$= \frac{11.2}{28} = 0.4$

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