Question

A mixture of ethane $(C_2{H}_6)$ and ethene $(C_2{H}_4)$ occupies $40L$ at $1.00$ atm and at $400K$. The mixture reacts completely with $130g$ of $O_2$, to produce $C{O}_2$ and $H_{2}O$. Assuming ideal gas behaviour, calculate the mole fractions of $C_2{H}_4$ and $C_2{H}_6$ in the mixture.

Answer 1

The total moles of gaseous mixture $=\frac{pV}{RT}=\frac{1\times 40}{0.082\times 400}$
$=1.22$
Let the mixture contain $x$ mole of ethane.
Therefore,
$\underset{x}{C_2{H}_6}+\frac{7}{2}{O}_2⟶2C{O}_2+3H_{2}O$
$\underset{1.22-x}{C_2{H}_4}+3O_2⟶2C{O}_2+2H_{2}O$
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 $=1-0.66=0.34$