JEE MAIN - Chemistry (2022 - 29th July Morning Shift - No. 2)
$$ \mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})} $$
$$20 \mathrm{~g} \quad ~~~5 \mathrm{~g}$$
Consider the above reaction, the limiting reagent of the reaction and number of moles of $$\mathrm{NH}_{3}$$ formed respectively are :
$$\mathrm{H}_{2}, 1.42$$ moles
$$\mathrm{H}_{2}, 0.71$$ moles
$$\mathrm{N}_{2}, 1.42$$ moles
$$\mathrm{N}_{2}, 0.71$$ moles
Explanation
$\underset{20 \,g}{\mathrm{N}_{2(\mathrm{~g})}}+\underset{5 \mathrm{~g}}{3 H_{2(g)}} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}$
Ideally $28 \mathrm{~g} \mathrm{~N}_{2}$ reacts with $6 \mathrm{~g} \,\mathrm{H}_{2}$ limiting reagent is $\mathrm{N}_{2}$
$\therefore$ Amount of $\mathrm{NH}_{3}$ formed on reacting $20 \mathrm{~g} \mathrm{~N}_2$ is,
$$ \begin{aligned} &=\frac{34 \times 20}{28}=24.28 \mathrm{~g} \\\\ &=1.42 \text { moles } \end{aligned} $$
Ideally $28 \mathrm{~g} \mathrm{~N}_{2}$ reacts with $6 \mathrm{~g} \,\mathrm{H}_{2}$ limiting reagent is $\mathrm{N}_{2}$
$\therefore$ Amount of $\mathrm{NH}_{3}$ formed on reacting $20 \mathrm{~g} \mathrm{~N}_2$ is,
$$ \begin{aligned} &=\frac{34 \times 20}{28}=24.28 \mathrm{~g} \\\\ &=1.42 \text { moles } \end{aligned} $$
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