Question

200 ml of a gaseous mixture containing co and co2 and n2 on complete combustion in just sufficient amount of o2 showed contraction of 40 ml. when the resulting gases were passed through KOH solution it reduces by 50% then calculate the volume ration of VCo2: VCO: V N2 in the original mixture.

Answer 1

Volume of mixture of CO, CO2 and N2 = 200 ml 
Volume of CO = x ml
Volume of N2 = y ml
Volume of CO2 = 200 - x  - y ml

On combustion,  CO2 remains as it is.  Nitrogen burns only at very high temperatures. At low temperatures it does not form oxides.

    2 CO  +  O2  ==>  2 CO2
       x ml      x/2 ml      x ml

In the input the amount of O2 present = x/2 ml
Total volume of gas mixture + O2 = 200 + x/2 ml

Resulting mixture: total : 200 ml as:
   N2: y ml                   CO2:  x + (200 - x - y) = 200 - y  ml

Contraction (reduction) in volume of gases is
           40 ml = 200 + x/2 - 200 = x/2
            x = volume of CO in the mixture = 80 ml

Now the mixture of CO2 + N2 is passed through base KOH. All CO2 reacts and N2 is not reactive with   K OH.
          2 K OH + CO2 ==> K2 CO3 + H2 O

So volume of gas mixture reduces by 50% of 200 ml ie., 100 ml.
    =>          200 - y = 100 ml
    =>              y = 100 ml

So we had 80 ml of CO, 100 ml of N2, and 20 ml of CO2 in the mixture initially.

Ratio: of volumes of  CO2 : CO : N2  =  1 : 4 : 5