Question

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

• A. $x_{C_2{H}_6}=0.3293$, $x_{C_2{H}_4}=0.6707$,
• B. $x_{C_2{H}_6}=0.6707$, $x_{C_2{H}_4}=0.3293$,
• C. $x_{C_2{H}_6}=0.5$, $x_{C_2{H}_4}=0.5$,
• D. none of these

Answer 1

Answer: (B)

$x_{C_2{H}_6}=0.6707$, $x_{C_2{H}_4}=0.3293$,

For $O_2$, $PV=nRT$
or, $1\times 40=n\times 0.0821\times 400$
or, $n=1.218$
$C_2{H}_6\left(g\right)+\frac{7}{2}{O}_2\left(g\right)\to 2C{O}_2\left(g\right)+3H_{2}O\left(l\right)$
$C_2{H}_4\left(g\right)+3O_2\left(g\right)\to 2C{O}_2\left(g\right)+3H_{2}O\left(l\right)$
Let the moles of ethane be $a$.
Volume of $O_2$ required by ethane $=7/2a$
Volume of ethane $=(1.218-a)$
Volume of $O_2$ required by ethene$=3(1.218-a)$
Given,
$\frac{7}{2}a+3(1.218-a)=\frac{130}{32}$ (moles of oxygen)
or $\frac{7}{2}a+3.654-3a=4.0625$
or $a=0.817$(moles of ethane)
$\therefore 1.218-a=1.218-0.817$ $=0.401$(moles of ethene)
Mole fraction of ethane $=\frac{0.817}{1.218}=0.6707$
Mole fraction of ethene$=1-0.6707=0.3293$