Question

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

• A. $C_4{H}_{10}$
• B. $C_4{H}_8$
• C. $C_4{H}_5$
• D. $C_8{H}_{16}$

Answer 1

Answer: (A)

$C_4{H}_{10}$

Given, $0.8$ mol of $C{O}_2$ & $1.0$ mol of $H_{2}O$ is produced by combustion of a hydrocarbon $C_x{H}_y$

We know, Reaction for combustion of hydrocarbons

i.e. $C_x{H}_y+(x+\frac{y}{4})O_2⟶xC{O}_2+\frac{y}{2}{H}_{2}O$

here $x=0.8$, $\frac{y}{2}=1.0$ [given]

$\Rightarrow x=0.8$ $y=2$

$\therefore$ Molar ratio of $C$ & $H=0.8:2$

Now dividing by smaller no, ratio= $\frac{0.8}{0.8}:\frac{2}{0.8}=1:2.5$

Whole number ratio of $C$ & $H$=$2:5$ or $4:10$

$\therefore$ The Hydrocarbon is $C_4{H}_{10}$