$$ \mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})} $$
$20 \mathrm{~g} \quad ~~~5 \mathrm{~g}$
Consider the above reaction, the limiting reagent of the reaction and number of moles of $\mathrm{NH}_{3}$ formed respectively are :
Solution
$\underset{20 \,g}{\mathrm{N}_{2(\mathrm{~g})}}+\underset{5 \mathrm{~g}}{3 H_{2(g)}} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}$<br/><br/> Ideally $28 \mathrm{~g} \mathrm{~N}_{2}$ reacts with $6 \mathrm{~g} \,\mathrm{H}_{2}$ limiting reagent is $\mathrm{N}_{2}$<br/><br/> $\therefore$ Amount of $\mathrm{NH}_{3}$ formed on reacting $20 \mathrm{~g} \mathrm{~N}_2$ is,
<br/><br/>
$$
\begin{aligned}
&=\frac{34 \times 20}{28}=24.28 \mathrm{~g} \\\\
&=1.42 \text { moles }
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
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