A 20 mL mixture of CO, `CH_4`, and Helium (He) gases is exploded by an electric discharge at room temperature with excess of oxygen. The volume contraction is found to be 13 mL. A further contraction of 14 mL occurs when the residual gas is treated wityh KOH solution. Find out the composition of the gaseous mixture in terms of volume percentage.

Explanation

Let `CO=amlCH_4=bmL`
Helium`=[20-(a+b)]mL`
(a). `CO+(1)/(2)O_2toCO_2`
(b). `CH_4+2O_2toCO_2+2H_2O`
(c). `Heimplies[20-(a+b)]`
(Helium does not react, it will remain same.)
Initial volume is
`a+b+[20-(a+b)]`
Contraction `=13mL`
Initial volume`=`Final volume
`a+(a)/(2)+b+2b+[20-(a+b)]-13`
`=(a+b)+[20-(a+b)]`
`implies(a+(a)/(2)+b+2b)-13=a+b`
`impliesa+b=14`
Solving, `a=10,b=4`
Percentage of `CO=(10)/(20)xx100=50%`
Percentage of `CH_4=(4)/(20)xx100=20%`
Percentage of `He=100-(50+20)=30%`