Question

A mixture of $N{H}_3(g)$ and $N_2{H}_4(g)$ is placed in a sealed container at $300K$. The total pressure is $0.5$ atm. The container is heated to $1200K$ at which both substances decompose completely according to the equations
$2N{H}_3(g)\to N2(g)+3H_2(g)$
and $N_2{H}_4(g)\to N_2(g)+2H_2(g)$
After decomposition is complete, the total pressure at $1200K$ is found to be $4.5$ atm. Find the mole $\%$ of $N_2{H}_4$ in the original mixture.

Answer 1

Answer: 25

Let initial mixture contains $n_1$ and $n_2$ moles of $N{H}_3$ and $N_2{H}_4$ respectively.
Total moles of gases originally present $=n_1+n_2$
Total moles of gases after decomposition of gases
$=2n_1+3n_2$
$0.5\times V=(n_1+n_2)R\times 300$
$4.5\times V=(2n_1+3n_2)R\times 1200$
$\frac{2n_1+3n_2}{n_1+n_2}=\frac{9}{4}$
$\frac{n_2}{n_1}=\frac{1}{3}$
$\frac{n_2}{n_1+n_2}\times 100=25\%$