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 $CO _{2}$ is $C _{ x } H _{ y }$. The value of $y$ is_________.

Answer 1

Combustion $rx ^{ n}$ :

$C _{ x } H _{ y ( g )}+\left( x +\frac{ y }{4}\right) O _{2}( g ) \rightarrow x CO _{2}( g )+\frac{ y }{2} H _{2} O (\ell)$

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

$\Rightarrow x =4$

Sinc: (I) $Vo _{2}=6 \times V _{ C _{ x } H _{y}}$

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

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

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

$\Rightarrow y =8$