Question

A mixture of ethane and ethene occupies $20$ L at $1.0$ atm and $400$ K. The mixture reacts completely with $65$ g of $O_2$ to produce $C{O}_2$ and water. Assuming ideal behaviour, the ratio of mole fractions of ethane and ethene in the mixture is approximately:

• A. $1:2$
• B. $2:1$
• C. $1:1$
• D. $2:3$

Answer 1

Answer: (B)

$2:1$

The ideal gas equation is $n=\frac{P\times V}{R\times T}$.
Here, $R$ is the ideal gas constant with value of $0.0821$ $Latmmo{l}^{-1}{K}^{-1}$.
Substituting values in the ideal gas equation, we get
$n=\frac{1\times 20}{0.082\times 400}=0.61mol=a+b$.
Here, $a$ and $b$ are number of moles of ethane and ethene respectively.
The reactions are as follows:
$C_2{H}_6+3.5O_2\to 2C{O}_2+3H_{2}O$
$C_2{H}_4+3O_2\to 2C{O}_2+2H_{2}O$
$65$ g of oxygen corresponds to $\frac{65}{32}=2.031$ mol.
This oxygen reacts with a mixture which contains $a$ moles of ethane, requiring $3.5a$ moles of oxygen and $b$ moles of ethene, requiring $3b$ moles of oxygen.
Thus, $3.5a+3b=2.031$.
Substituting $a+b=0.61$ in the above equation, the values of $a$ and $b$ are calculated as $0.402$ and $0.208$ respectively.
The mole fraction of ethane is $\frac{0.402}{0.402+0.208}=0.66$.
The mole fraction of ethene is $1-0.66=0.34$.
Hence, the ratio of mole fractions of ethane and ethene is $2:1$.