A mixture of `20 mL` of `CO, CH_(4)` and `N_(2)` was burnt in excess of `O_(2)` resulting in reduction of `13 mL` of volume. The residual gas was then treated with `KOH` solution to show a contraction of `14 mL` in volume. Calculate volume of `Co, CH_(4)` and `N_(2)` in mixture. All measurements are made at constant pressure and temperature.

Explanation

`"Suppose volume of CO = a mL," CH_(4)="b mL and N"_(2)="c mL. Then a + b + c = 20 mL …(i)"`
The combustion reaction will be
`CO+(1)/(2)O_(2)rarr CO_(2)`
`CH_(4)+2O_(2) rarr CO_(2)+2H_(2)O(l)`
`N_(2)+O_(2) rarr "No reaction"`
Thus, a mL of CO will produce `CO_(2)=amL`
`"b mL of CH"_(4)" will produce CO"_(2)=bmL`
`N_(2)` will remain as such, i.e. = c mL
As `CO_(2)` is absorbed by KOH, decrease in volume on treating with KOH will be `=a+b=14mL` (Given) .....(ii)
The first given decrease is due to `O_(2)` consumed.
a mL of CO will consume `O_(2)=(a)/(2)mL`
b mL of `CH_(4)` will consume `O_(2)=2bmL`
`therefore O_(2)` consumed `=(1)/(2)+2b=13mL" (Given ) ...(iii)"`
From eqns. (i) and (ii), `c = 20 - 14 = 6 mL`
From Eqns. (ii) and (iii),
`(a)/(2)+2(14-a)=13 " or "(3)/(2)a=15" or "a=10mL`
`therefore " From eqn. (ii), "10+b=14" or "b=4mL`
`CO=10mL, CH_(4)=4mL, N_(2)=6mL`