A mixture of methane and ethyne in the volume ratio \(x: y\) has a total volume of \(30 ~mL\). On complete combustion it gave 40 mL of \(CO_2\) if the ratio had been \(y : x\) what volume of \(CO_2\) would have been obtained?

Let volume of methane = \(x~ mL\)
than volume of ethylene = \(y ~mL\)
given \(x + y = 30~mL\) ...(1)
On complete combustion of mixture
\(\underset {x~mL}{CH_4} + O_2 \rightarrow \underset{x~~mL}{CO_2} + 2H_2O \\
\underset{y~mL}{C_2H_4} + 3O_2 \rightarrow \underset{2y~mL}{2CO_2} + 2H_2O\)
Total vol. of \(CO_2\) produced \(= x + 2y~mL\)
\(\therefore\) \(x + 2y = 40...(2)\)
from (1) and (2)
\(x = 20\) and \(y = 10~ mL\)
If ratio had been \(y:x\)
than vol of \(CH_4 = y\) mL and vol of \(C_2H_4 = x\) mL
on combustion
\(\underset{y~mL}{C_2H_4 } + O_2 \rightarrow \underset{y~mL}{CO_2} + 2H_2O\\
\underset{x~mL}{C_2H_4} + 3O_2 \rightarrow \underset{2x~mL}{2CO_2} + 2H_2O\)
\(\therefore\) Total \(CO_2\) formed \(= (y + 2x) ~mL = 10 + 2(20) = 50~ mL\)