A mixture of 20 mL of CO, CH_(4) and N_(2) was burnt in excess of O_(2) resulting in the 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 the mixture. All measurements are made at constant pressure and temperature.

A mixture of 20 mL of CO, CH_(4) and N_(2) was burnt in excess of O_(2

1.	A mixture of 20 mL of CO, CH_(4) and N_(2) was burnt in excess of O_(2) resulting in the 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 the mixture. All measurements are made at constant pressure and temperature.
Answer» SOLUTION :`"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`