Question

The formula of a gaseous hydrocarbon which requires 6 times of its own volume of $O_2$ for complete oxidation and produces 4 times its own volume of $C{O}_2$ is $C_x{H}_y$. The value of $y$ is_________.

Answer 1

Answer: 8

Combustion $r{x}^n$ :

$C_x{H}_{y(g)}+\left(x+\frac{y}{4}\right)O_2(g)\to xC{O}_2(g)+\frac{y}{2}{H}_{2}O(\ell )$

$\underset{-}{V}\underset{-}{6V}\underset{Vx=4V}{-}$

$\Rightarrow x=4$

Sinc: (I) $V{o}_2=6\times V_{C_x{H}_y}$

$\Rightarrow V\left(x+\frac{y}{4}\right)=6V$

$\Rightarrow \left(x+\frac{y}{4}\right)=6$

$\Rightarrow 4+\frac{y}{4}=6$

$\Rightarrow y=8$