Question

10ml of a mixture of CO, CH4 and N2 explodes with excess of oxygen gave a contraction pf 6.5ml. there was a further contraction of 7ml, when the residual gas treated with KOH. Volume of CO, CH4 and N2 respectively is?

Answer 1

Let the volumes of CO, CH4 and N2  in the mixture be x ml, y ml and (10-x-y) ml respectively.  We ignore the volume of water formed in the combustion as it is a liquid and its volume is small. Nitrogen combusts only at very very high temperatures. So it remains unreactive.

   2 CO  +  O2  ==>  2 CO2          CH4  +  2 O2  ==>  CO2  + 2 H2O
   x ml      x/2 ml        x ml              y ml     2 y ml          y ml      (liquid)

Volume of oxygen used: 2y+x/2  ml
Total volume of all gases before combustion = 10 + 2 y + x/2  ml

After combustion the total volume is: 10 ml 
    as  CO2 : x + y         N2:  10 - x- y

Reduction in volume:  10 + 2y + x/2 - 10 = 2 y + x/2 = 6.5  ml
   =>   4 y + x = 13 ml  --- (1)

When the mixture of CO2 + N2 passes over KOH, all of CO2 is absorbed. Only N2 remains. Then reduction of 7 ml corresponds to that of CO2.
   =>  x + y = 7 ml      --- (2)

Solving the two equations,   x = 5 ml   and   y = 2 ml

Volumes of CO = 5 ml,   CH4 = 2 ml ,     N2 = 3 ml.