Question

$40mL$ of a mixture of hydrogen, $C{H}_4$ and $N_2$ was exploded with $10mL$ of oxygen. On cooling, the gases occupied $36.5mL$. After treatment with $KOH$, the volume reduced by $3mL$ and again on treatment with alkaline pyrogallol, the volume further decreased by $1.5mL$. Determine the composition of the mixture.

Answer 1

$V(H_2+C{H}_4+N_2)=40mL$

$V_{O_2}=10mL$

$H_2+\frac{1}{2}{O}_2\to H_{2}O\left(l\right)$

$C{H}_4+(1+\frac{4}{4})O_2\to C{O}_2+2H_{2}O\left(l\right)$

$N_2\to$ No reaction

Let volume of $H_2=xmL$

Volume of $C{H}_4=ymL$

volume of $N_2=\{40-(x+y\left)\right\}$

On cooling the gas remained

$=C{O}_2\left(g\right)$ produced $+N_2\left(g\right)+O_2$ (remained)

A/q, $y+\{40-(x+y\left)\right\}+10-(\frac{x}{2}+2y)=36.5$

(with x mL of $H_2,\frac{x}{2}$ mL of $O_2$ reacts & with y mL of $C{H}_{4}2ymL$ of $O_2$ reacts)

$40-x+10-\frac{x}{2}-2y=36.5$

$50-\frac{3x}{2}-2y=36.5$

$50-36.5=\frac{3x}{2}+2y$

$\therefore \frac{3x}{2}+2y=13.5$

$3x+4y=27$ ---------------(1)

After passing over KOH, $C{O}_2$ will be absorbed

$\therefore y=3ml$ -------------(2)

After passing over pyrogallol, $O_2$ will be absorbed.

$\therefore 10-(\frac{x}{2}+2y)=\frac{3}{2}$

$10-\frac{3}{2}=\frac{x}{2}+2y$

$17=x+4y$ ------------------(3)

from (2) & (3) $x=5$

$\therefore$ Volume of $H_2=5mL$

Volume of $N{H}_4=3mL$

Volume of $N_2=32mL$

$\therefore \%ofC{H}_4=\frac{3}{40}\times 100=7.5\%$

$\therefore \%of{H}_2=\frac{5}{40}\times 100=12.5\%$

$\therefore \%of{N}_2=\frac{32}{40}\times 100=80\%$