Chemistry, asked by tin8i1palvive, 1 year ago


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?

Answers

Answered by kvnmurty
189
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.

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