Question

$34$ gm of a mixture containing $N_2$ and $H_2$ in $1:3$ by mole is partially converted into ${NH}_3$. Calculate the vapour density of the mixture (containing remaining $N_2,H_2$ and ${NH}_3$ formed) after reaction if it has been found that the ${NH}_3$ formed required $0.5$ moles of $H_3{PO}_4$ for complete neutralization.

Answer 1

$N_2+3H_2\rightleftharpoons 2N{H}_3$

Let mass of $N_2$ be $x$ $g$

Then mass of $H_2$ will be $(34-x)g$

moles of $N_2=\frac{x}{28}$, moles of $H_2=\frac{34-x}{2}$

as their molar ratio is $1:3$

So, $\frac{\frac{x}{28}}{\frac{(34-x)}{2}}=\frac{1}{3}\Rightarrow \frac{6x}{28}=34-x$

$\Rightarrow 6x=34\times 28-28x$

$\Rightarrow 34x=34\times 28$

$\Rightarrow x=28\Rightarrow$ Mass of $N_2$

$\Rightarrow 34-x=34-28=6⟶$ Mass of $H_2$

Number of moles of $N_2=\frac{28}{28}=1$

Number of moles of $H_2=\frac{6}{2}=3$

$\underset{1}{N_2}+\underset{3}{3H_2}\rightleftharpoons \underset{-}{2N{H}_3}$

$1-y$ $3-3y$ $2y$

As $N{H}_3$ required $0.5$ moles of $H_{3}P{O}_4$

So, $\underset{\left(H_{3}P{O}_4\right)}{n\times n_f}=\underset{\left(N{H}_3\right)}{n\times n_f}$

(as $n_f$ for $H_{3}P{O}_4$ is $3$)

$\Rightarrow 0.5\times 3=n\times 1$

$n=1.5$ moles of $N{H}_3$

Thus $2y=1.5$

$\Rightarrow y=\frac{1.5}{2}=0.75$ moles of $N{H}_3$

Number of moles remain in solution Mass remain in solution

$N_2⟶1-0.75=0.25$ $0.25\times 28=7g$

$H_2⟶3-3\times 0.75=0.75$ $0.75\times 2=1.5g$

$N{H}_3⟶2\times 0.75=1.5$ $1.5\times 31=46.5g$

Vapour density=$\frac{EffectiveMolarmass\left(EMM\right)}{2}$

$\Rightarrow EMM=\frac{m_1+m_2+m_3}{n_1+n_2+n_3}=\frac{7+1.5+46.5}{0.25+0.75+1.5}$

$\Rightarrow EMM=22$

Vapour density of mixture= $\frac{22}{2}=11$ .