Question

An ideal gaseous mixture of ethane $\left(C_2{H}_6\right)$ and ethene $\left(C_2{H}_4\right)$ occupies $28$ litre at $1atm$ and $0^{o}C$. The mixture reacts completely with $128gm$ $O_2$ to produce ${CO}_2$ and $H_{2}O$. Mole fraction at $C_2{H}_4$ in the mixture is :

• A. $0.6$
• B. $0.4$
• C. $0.5$
• D. $0.8$

Answer 1

The correct option is A $0.6$

As we know that,

Total amount of mixture $\left(n\right)=\frac{PV}{RT}$

Given :

$P=1atm$

$V=28L$

$R=0.082Latm{K}^{-1}{mol}^{-1}$

$T=273K$

$\therefore n=\frac{1\times 28}{0.082\times 273}=\frac{28}{22.386}=1.25\text{moles}$

The combustion reactions of ethane and ethene are :

$\underset{Ethane}{C_2{H}_6}+\frac{7}{2}{O}_2⟶2C{O}_2+3H_{2}O$

$\underset{Ethene}{C_2{H}_4}+3O_2⟶2C{O}_2+2H_{2}O$

Let the amount of ethane in the mixture be x.

$\therefore$ the amount of oxygen consumed $=\frac{7}{2}x+3\times (1.25-x)=3.75+\frac{1}{2}x$

The amount of oxygen $(i.e.3.75+\frac{1}{2}x)$ will be equal to the given amount $(i.e.128gper32g{mol}^{-1}=4\text{moles})$

$\therefore 3.75+\frac{1}{2}x=4$

$\Rightarrow \frac{1}{2}x=4-3.75$

$\Rightarrow x=0.5$

$\therefore$ Amount of ethene in the mixture $=1.25-0.5=0.75\text{moles}$

$\therefore$ Mole fraction of $C_2{H}_6=\frac{\text{Moles of}{C}_2{H}_6}{\text{Total moles}}=\frac{0.75}{1.25}=\frac{3}{5}=0.6$

Hence the correct option is A.