Question

38 ml mixture was mixed with 31 ml O2 & exploded, volume of residual gas was 29ml which reduce to 12 ml when passed in KOH solution determine composition of given mixture

Answer 1

Answer:

6 ml.

Explanation:

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 CH4needs 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.