Question

A mixture of ethane and ethene occupies $41\,L$ at $1\,atm$ and $500\, K$. The mixture reacts completely with $\frac{10}{3}$ mole of ${{O}_{2}}$ to produce $C{{O}_{2}}$ and ${{H}_{2}}O$ . The mole fraction of ethane and ethene in the mixture are respectively ($R\, =\, 0.082\, L \,atm$ ${{K}^{-1}}mo{{l}^{-1}}$ )

• A. $0.50, 0.50$
• B. $0.75, 0.25$
• C. $0.67, 0.33$
• D. $0.25, 0.75$

Answer 1

Answer: (C)

$0.67, 0.33$

Answer 2

For a gaseous mixture of
${{C}_{2}}{{H}_{6}}$ and ${{C}_{2}}{{H}_{4}}$
$pV=nRT$ or $n=\frac{PV}{RT}=\frac{1\times 41}{0.082\times 500}$
or $n=1$
$\therefore$ Total mole of ${{C}_{2}}{{H}_{6}}+{{C}_{2}}{{H}_{4}}=1$
mole Let the mole of
${{C}_{2}}{{H}_{6}}=x$
then mole of ${{C}_{2}}{{H}_{4}}=1-x$
${{C}_{2}}{{H}_{6}}+\frac{7}{2}{{O}_{2}}\xrightarrow{{}}2C{{O}_{2}}+3{{H}_{2}}O$
${{C}_{2}}{{H}_{4}}+3{{O}_{2}}\xrightarrow{{}}2C{{O}_{2}}+2{{H}_{2}}O$
$\therefore$ Mole of ${{O}_{2}}$
needed for complete reaction of mixture
$=\frac{7}{2}x+3(1-x)$
$\therefore$ $\frac{7}{2}x+3(1-x)=\frac{10}{3}$
Or $x=\frac{2}{3}$
Thus, mole fraction of
${{C}_{2}}{{H}_{6}}=\frac{2}{3}=0.67$
and mole fraction of
${{C}_{2}}{{H}_{4}}=1-\frac{2}{3}=0.33$