Question

A mixture of methane and ethyne in the volume ratio $x:y$ has a total volume of $30mL$. On complete combustion it gave 40 mL of $C{O}_2$ if the ratio had been $y:x$ what volume of $C{O}_2$ would have been obtained?

• A. $100mL$
• B. $60mL$
• C. $50mL$
• D. $150mL$

Answer 1

The correct option is C $50mL$

Let volume of methane = $xmL$
than volume of ethylene = $ymL$
given $x+y=30mL$ ...(1)
On complete combustion of mixture
$\underset{xmL}{C{H}_4}+O_2\to \underset{xmL}{C{O}_2}+2H_{2}O\underset{ymL}{C_2{H}_4}+3O_2\to \underset{2ymL}{2C{O}_2}+2H_{2}O$
Total vol. of $C{O}_2$ produced $=x+2ymL$
$\therefore$ $x+2y=\mathrm{40...}\left(2\right)$
from (1) and (2)
$x=20$ and $y=10mL$
If ratio had been $y:x$
than vol of $C{H}_4=y$ mL and vol of $C_2{H}_4=x$ mL
on combustion
$\underset{ymL}{C_2{H}_4}+O_2\to \underset{ymL}{C{O}_2}+2H_{2}O\underset{xmL}{C_2{H}_4}+3O_2\to \underset{2xmL}{2C{O}_2}+2H_{2}O$
$\therefore$ Total $C{O}_2$ formed $=(y+2x)mL=10+2\left(20\right)=50mL$