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??

Question

Srinivas Rao · Answer

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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