A 20.0 cm 3 mixture of CO , CH 4 and He gases is exploded by an electric discharge at room temperature with excess of oxygen. The volume contraction is found to be 13.0 cm 3 . A further contraction of 14.0 cm 3 occurs when the residual gas is treated with KOH solution. Find out the composition of the gaseous mixture in terms of volume percentage.

Question

A 20.0 cm 3 mixture of CO , CH 4 and He gases is exploded by an electric discharge at room temperature with excess of oxygen. The volume contraction is found to be 13.0 cm 3 . A further contraction of 14.0 cm 3 occurs when the residual gas is treated with KOH solution. Find out the composition of the gaseous mixture in terms of volume percentage. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

The reaction involved in the explosion process is

x m L CO (g) + \frac{x}{2} m L \frac{1}{2} O_{2} (g) \to x m L C O_{2} (g)

y m L C H_{4} (g) + 2 y m L 2 O_{2} (g) \to y m L C O_{2} (g) + 2 H_{2} O (l)

The first step volume contraction can be calculated as :

\Rightarrow (x + \frac{x}{2} + y + 2 y) - (x + y) = 13

x + 4 y = 26....

(i)
The second volume contraction is due to absorption of

C O_{2}

.
Hence,

x + y = 14.....

(ii)
Now, solving equations (i) and (ii),

x = 10 m L, y = 4 m L

and volume of

He = 20 - 14 = 6 m L

\Rightarrow V o l %

of

CO = \frac{10}{20} \times 100 = 50%

V o l %

of

C H_{4} = \frac{4}{20} \times 100 = 20%

V o l %

of

He = 30%