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a mixture of 20ml of CO CH4 and N2 was burnt in excess of O2 resulting in the reduction of 13ml of volume. the residual gas was then treated withKOH solution to show a contraction of 14ml in volume. calculate volume of CO,CH4 and N2 misture. all measurements are made at constant pressure and temperature??
The gaseous mixture contains (10ml of CO+ 4 ml of CH4 + 6 ml of N2) = 20 ml
The solution is as follows:
Let x ml of CO, y ml of CH4 and Z ml of N2 be present in the mixture.
Then x + y+ z = 20 ml ----(1)
x ml of CO needs (x/2) ml of O2 ; y ml of CH4 needs 2y ml of O2
Total O2 used up during combustion is (x/2)+2y
It is to be noted that, 1 mole of CO gives 1 mole of CO2 and 1 mole of CH4 gives 1 mole of CO2. So, the overall reduction in volume of 13 ml is due to consumption of O2.
So, (x/2)+2y = 13 ml ----(2)
Solve for x and y using equations (1) and (2) . We get x = 10 ml and y= 4 ml.
So, volume of N2 = z= 6 ml.
On passing CO2 through KOH, it is absorbed to form solid salt of K2CO3. Volume of CO2 obtained is 14 ml from 10 ml CO and 4 ml CH4.
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