Question

200 ml of gaseous mixture containing $CO$, $C{O}_2$ and $N_2$ on complete combustion in just sufficient amount of $O_2$ showed a contraction of 40 ml . When the resulting gases were passed through $KOH$ solution it reduces by 50 ml, then calculate the volume of $V_{CO}:V_{C{O}_2}:V_{N_2}$ in the original mixture.

• A. 4 : 1 : 5
• B. 2 : 3 : 5
• C. 1 : 4 : 5
• D. 1 : 3 : 5

Answer 1

The correct option is A 4 : 1 : 5

Given, Volume of $CO$, $C{O}_2,N_2$ $=200$ ml

On combustion, $C{O}_2$ remains intact and nitrogen does not react at low temperatures.

$2CO$ +$O_2\to 2C{O}_2$

x x/2 x

Total Volume = 200 + $\frac{x}{2}$

Resulting mixture = $N_2$ = y ml

$C{O}_2$ $=200-yml$

Contradiction = Final - Initial = 200 +$\frac{x}{2}-200=$$\frac{x}{2}$

$40=\frac{x}{2}$

Hence $x=80$ ml

When mixture passes through $KOH$, only $C{O}_2$ reacts.

As volume reduces by 50% ,

$200-y=100$ ml

$y=100$ ml

Hence $CO=80$ ml, $y=100$ ml and $C{O}_2$ = 20 ml

$V_{CO}:V_{C{O}_2}:V_{N_2}$ $=4:1:5$

Therefore, option $A$ is correct.