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

The reaction involved in the explosion process is
`{:(CO(g) ,+, (1)/(2)O_(2)(g), rarr ,CO_(2)(g)),("x mL" ,,(x)/(2)mL,,"x mL"),(CH_(4)(g),+,2O_(2)(g),rarr,CO_(2)(g)+2H_(2)O(l)),("y mL",,"2y mL",,"y mL"):}`
The first step volume contraction can be calculated as :
`(x+(x)/(2)+y+2y)-(x+y) = 13`
`rArr x+4y = 26` ....(i)
The second volume contraction is due to absorption of `CO_(2)`.
Hence, `x+y = 14` ....(ii)
Now, solving equations (i) and (ii),
`x = 10 mL, y = 4 mL` and volume of `He = 20 - 14 = 6 mL`
`rArr Vol %` of `CO = (10)/(20) xx 100 = 50%`
Vol `%` of `CH_(4) = (4)/(20) xx 100 = 20%`
Vol `%` of `He = 30%`