Question

A gas mixture of 3 $L$ containing propane and butane on complete combustion at ${25}^{o}C$ produced $10L$ of $C{O}_2$. The composition of gaseous mixture is:

• A. Propane = $2.5L$ & Butane = $0.5L$
• B. Propane = $1L$ & Butane = $2L$
• C. Propane = $2L$ & Butane = $1L$
• D. Propane = $1.5L$ & Butane = $1.5L$

Answer 1

The correct option is C Propane = $2L$ & Butane = $1L$

$C_3{H}_8+5O_2\to 3C{O}_2+4H_{2}O$
$C_4{H}_{10}+\frac{13}{2}{O}_2\to 4C{O}_2+5H_{2}O$
Let the volume of propane = $xL$
Then, volume of butane = $(3-x)L$
$1L$ of $C_3{H}_8$ or propane gives $3L$ of $C{O}_2$ and $1L$ of $C_4{H}_{10}$ or butane gives $4L$ of $C{O}_2$
$\Rightarrow C{O}_2$ produced by $xL$ of $C_3{H}_8$ = $3x$
$C{O}_2$ produced by $(3-x)L$ of $C_4{H}_{10}$ = $4\times (3-x)$
Hence, $C{O}_2$ produced :
= $3x+4(3-x)$
= $12-x$
And since, $12-x=10$ (given)
$x=2L$
Amount of propane in the mixture = $2L$
Amount of propane butane in the mixture = $1L$