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Question

A mixture of 20 mL CO, CH4 and N2 was burnt in excess of O2, 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. The amount of volume (in mL) of N2 in the mixture is: (assume H2O is in the liquid state and all measurements are made at constant P and T)

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Solution

Let the mixture contain a,b and c mL of CO,CH4 and N2 in it. Therefore, a+b+c=20...(i)
The reactions during combustion of mixture are as follows:
CO+12O2CO2
CH4+2O2CO2+2H2O(l)
N2+O2No reaction
Thus, reduction in volume is 2b due to 2nd reaction and a2 due to 1st reaction.
Total CO2 formed =a+b= volume absorbed by KOH
a+b=14...(ii)
and, (a/2)+2b = 13...(iii)
By equations (i),(ii) and (iii), a=10 mL, b=4 mL, c=6 mL

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