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In Haber process 3 litres of dihydrgen and 30 litres of dinitrogen were taken for reaction which yielded only $50 \%$ of the expected product. What will be the composition of gaseous mixture under the above said condition in the end?
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The correct answer is:
10 litres ammonia, 25 litres nitrogen, 15 litres hydrogen
$\mathrm{N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3$ $1 \mathrm{~mol} \quad 3 \mathrm{~mol} 2 \mathrm{~mol}$ 2 vol $3 \mathrm{~mol} 2 \mathrm{vol}$
$(\operatorname{mol} \%=\operatorname{vol} \%)$
$$
\begin{aligned}
& \begin{array}{l}
10 \mathrm{vol} \quad 30 \mathrm{vol} \quad 20 \mathrm{vol} \\
\text { (expected = product) }
\end{array} \\
& (30-x)(30-x) 2 x \\
& \text { Given that } 2 x=20 \times \frac{50}{100}=10 \\
& \therefore \quad x=5
\end{aligned}
$$
Composition of gaseous mixture
$$
\begin{aligned}
\mathrm{N}_2 & =(30-x)=25 \text { litre } \\
\mathrm{H}_2 & =(30-3 x)=15 \text { litre } \\
\mathrm{NH}_3 & =2 x=10 \text { litre }
\end{aligned}
$$
$(\operatorname{mol} \%=\operatorname{vol} \%)$
$$
\begin{aligned}
& \begin{array}{l}
10 \mathrm{vol} \quad 30 \mathrm{vol} \quad 20 \mathrm{vol} \\
\text { (expected = product) }
\end{array} \\
& (30-x)(30-x) 2 x \\
& \text { Given that } 2 x=20 \times \frac{50}{100}=10 \\
& \therefore \quad x=5
\end{aligned}
$$
Composition of gaseous mixture
$$
\begin{aligned}
\mathrm{N}_2 & =(30-x)=25 \text { litre } \\
\mathrm{H}_2 & =(30-3 x)=15 \text { litre } \\
\mathrm{NH}_3 & =2 x=10 \text { litre }
\end{aligned}
$$
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