Question

A mixture of ethane $\left(C_2{H}_6\right)$ and ethene $\left(C_2{H}_4\right)$ occupies 40 L at 1.00 atm and at 400 K.The mixture reacts completely with 130 g of $O_2$ to produce $C{O}_2$ and $H_{2}O$. Assuming ideal gas behaviour. If the value mole fraction of $C_2{H}_4$ in the mixture is $33\times {10}^{-x}$, then what is the value of x?

Answer 1

For $O_2$
$PV=nRT$
or $1\times 40=n\times 0.0821\times 400$
or $n=1.218$
$C_2{H}_6\left(g\right)+\frac{7}{2}{O}_2\left(g\right)⟶2C{O}_2\left(g\right)+3H_{2}O\left(l\right)$
$C_2{H}_4\left(g\right)+3O_2\left(g\right)⟶2C{O}_2\left(g\right)+2H_{2}O\left(l\right)$
Let the moles of ethane be $a$.
$Volumeof{O}_{2}requiredbyethane=\frac{7}{2}a$
$Volumeofethene=1.218-a$
$Volumeof{O}_{2}requiredbyethene=3(1.218-a)$
Given
$\frac{7}{2}a+3(1.218-a)=\frac{130}{32}$(moles of oxygen)
or $\frac{7}{2}a+3.654-3a=4.0625$
$(1.218-a)=(1.218-0.817)=0.401$ (moles of ethene)
$Molefractionofethane=\frac{0.817}{1.218}=0.6707$
$Molefractionofethene=(1-0.6707)=0.3293$