Question

A mixture of ethane $(C_2H_6 )$ and ethene $(C_2H_4)$ occupies $40\, L$ at $1.00$ atm and at $400\, K$. The mixture reacts completely with $130 \,g$ of $O_2$, to produce $CO_2$ and $H_2O$. Assuming ideal gas behaviour, calculate the mole fractions of $C_2H_4\,$ and $\, C_2H_6$ in the mixture.

Answer 1

The total moles of gaseous mixture $ = \frac{pV}{RT} = \frac{1\times 40}{0.082\times400}$
$= 1.22$
Let the mixture contain $x$ mole of ethane.
Therefore,
$\underset{x}{C _{2} H _{6}}+\frac{7}{2} O _{2} \longrightarrow 2 CO _{2}+3 H _{2} O $
$\underset{1.22 - x}{C _{2} H _{4}}+3 O _{2} \longrightarrow 2 CO _{2}+2 H _{2} O$
Total moles of $ O _{2} $ required $=\frac{7}{2} x+3(1.22-x)=\frac{x}{2}+3.66 $
$\frac{130}{32}=\frac{x}{2}+3.66 $
$\Rightarrow x=0.805 $ mole ethane and $0.415$ mole ethene.
$\Rightarrow $ Mole fraction of ethane $=\frac{0.805}{1.22}=0.66 $
Mole fraction of ethene $=1-0.66=0.34$