Question

A mixture of ethane and ethene occupies $40litres$ at $1.00atm$ and at $400K$. The mixture reacts completely with $130g$ of $O_2$ to prduce $C{O}_2$ and $H_{2}O$. Assuming ideal gas behaviour, calculate the mole fractions of $C_2{H}_6$ and $C_2{H}_6$ in the mixture.

• A. The mole fraction $\left(\%\right)$ of ethene is $66.25$
• B. The mole fraction $\left(\%\right)$ of ethane is $66.25$
• C. The mole fraction $\left(\%\right)$ of ethene is $33.75$
• D. The mole fraction $\left(\%\right)$ of ethane is $33.75$

Answer 1

Answer: (B), (C)

The correct options are
B The mole fraction $\left(\%\right)$ of ethane is $66.25$
C The mole fraction $\left(\%\right)$ of ethene is $33.75$

$\text{Let the volume of ethane}=xlitre$
$\therefore$ $\text{Volume of ethene}=(40-x)litre$
Combustion reaction of ethane and ethene
$C_2{H}_6\left(g\right)+\frac{7}{2}{O}_2\left(g\right)\to 2C{O}_2\left(g\right)+3H_{2}O\left(l\right)or2C_6{H}_6\left(g\right)+7O_2\left(g\right)\to 4C{O}_2+H_{2}O\left(l\right)\left(ii\right)C_2{H}_4\left(g\right)+3O_2\left(g\right)\to 2C{O}_2\left(g\right)+2H_{2}O\left(l\right)$
Volume of $O_2$ required for complete combustion of ethane = $\frac{7x}{2}$
Volume of $O_2$ required for complete combustion of ethene = $40-x\times 3$
$\therefore$ Total volume of $O_2$ = $\frac{7x}{2}+(40-x)\times 3L$

$P=1atm;V=\frac{7x}{2}+(40-x)\times 3L;R=0.0821Latm{K}^{-1}mo{l}^{-1};T=400K$

We know $n=\frac{PV}{RT}=\frac{\frac{7x}{2}+(40-x)3}{0.082\times 400}=\frac{7x+(40-x)6}{2\times 0.082\times 400}$
The mass of n mole of $O_2$ =$\left[\frac{7x+(40-x)6}{2\times 0.082\times 400}\right]\times 32=130\Rightarrow \left[\frac{7x+240-6x}{65.6}\right]\times 32=130\Rightarrow 32x+240\times 32=8528\Rightarrow 32x=848\Rightarrow x=\frac{848}{32}=26.5$
Hence, mole fraction $\left(\%\right)$ of ethane = $\frac{26.5}{40}\times 100=66.25$
Mole fraction $\left(\%\right)$ of ethene = $33.75\%$