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

Let `a mL CO, b mL CH_(4)` and `C mL N_(2)` be present in mixture,
Then `a+b+c=20`….(1)
`CO+(1)/(2)O_(2)rarrCO_(2)`
`:.`Volume of `CO=a, :.` Volume of `CO_(2)=a`
`CH_(4)+2O_(2)rarrCO_(2)+2H_(2)O_((l))`
`:.` Volume of `CH_(4)=b, :.` Volume of `CO_(2)=b`
`N_(2)+O_(2)rarr"No reaction"`
Volume of `CO_(2)` formed `=` Volume absorbed by `KOH`
`a+b=14 mL`....(2)
Now initial volume of `CO+CH_(4)+N_(2)+"Vol. of" O_(2)` taken-volume of `CO_(2)` formed `-` volume of `N_(2)-` volume of `O_(2)` left `=13` (the contraction)
`:. a+b+c+Vol.O_(2)` taken- vol.of `O_(2)` left `-(a+b)-c=13`
`:.` Vol. of `O_(2)` used `=13`
`:. (a)/(2)+2b=13`
`( :' "volume of" O_(2) "used" =(a)/(2)+2b)`....(3)
Solving Eqa. (1),(2) and (3), we get
`a=10 mL, b=4 mL, c=6 mL`