Question

$0.2$ mol sample of hydrocarbon, $C_X{H}_Y$ yields after complete combustion with excess $O_2$ gas, $0.8$ mol of $C{O}_2$ and $1.0$ mol of $H_{2}O$. Hence, the hydrocarbon is :

• A. $C_4{H}_{10}$
• B. $C_4{H}_8$
• C. $C_4{H}_6$
• D. $C_8{H}_{16}$

Answer 1

The correct option is A $C_4{H}_{10}$

The balanced reaction is as follows:
$C_X{H}_Y+[X+\frac{Y}{4}]O_2\to XC{O}_2+\frac{Y}{2}{H}_{2}O$
$0.2$ excess $0.2X$ $\frac{0.2Y}{2}$ (Moles before reaction)
Therefore, $0.2X=0.8$ and $\frac{0.2}{Y}=1.0$
$X=4$ and $Y=10$

Hence, the gas is $C_4{H}_{10}$.