Some Basic Concepts of Chemistry 1 Question 5
5. For a reaction,
$\mathrm{N}{2}(g)+3 \mathrm{H}{2}(g) \longrightarrow 2 \mathrm{NH}{3}(g)$, identify dihydrogen $\left(\mathrm{H}{2}\right)$ as a limiting reagent in the following reaction mixtures.
(2019 Main, 9 April I)
(a) $56 \mathrm{~g}$ of $\mathrm{N}{2}+10 \mathrm{~g}$ of $\mathrm{H}{2}$
(b) $35 \mathrm{~g}$ of $\mathrm{N}{2}+8 \mathrm{~g}$ of $\mathrm{H}{2}$
(c) $14 \mathrm{~g}$ of $\mathrm{N}{2}+4 \mathrm{~g}$ of $\mathrm{H}{2}$
(d) $28 \mathrm{~g}$ of $\mathrm{N}{2}+6 \mathrm{~g}$ of $\mathrm{H}{2}$
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Solution:
- Key Idea The reactant which is present in the lesser amount, i.e. which limits the amount of product formed is called limiting reagent.
When $56 \mathrm{~g}$ of $\mathrm{N}{2}+10 \mathrm{~g}$ of $\mathrm{H}{2}$ is taken as a combination then dihydrogen $\left(\mathrm{H}_{2}\right)$ act as a limiting reagent in the reaction.
$$ \begin{array}{rrr} \mathrm{N}{2}(g)+3 \mathrm{H}{2}(g) \longrightarrow & \mathrm{NH}_{3}(g) \ 2 \times 14 \mathrm{~g} & 3 \times 2 \mathrm{~g} & 2(14+3) \mathrm{g} \ 28 \mathrm{~g} & 6 \mathrm{~g} & 34 \mathrm{~g} \end{array} $$
$28 \mathrm{~g} \mathrm{~N}{2}$ requires $6 \mathrm{~g} \mathrm{H}{2}$ gas.
$56 \mathrm{~g}$ of $\mathrm{N}{2}$ requires $\frac{6 \mathrm{~g}}{28 \mathrm{~g}} \times 56 \mathrm{~g}=12 \mathrm{~g}$ of $\mathrm{H}{2}$ $12 \mathrm{~g}$ of $\mathrm{H}{2}$ gas is required for $56 \mathrm{~g}$ of $\mathrm{N}{2}$ gas but only $10 \mathrm{~g}$ of $\mathrm{H}_{2}$ gas is present in option (a).
Hence, $\mathrm{H}_{2}$ gas is the limiting reagent.
In option (b), i.e. $35 \mathrm{~g}$ of $\mathrm{N}{2}+8 \mathrm{~g}$ of $\mathrm{H}{2}$.
As $28 \mathrm{~g} \mathrm{~N}{2}$ requires $6 \mathrm{~g}$ of $\mathrm{H}{2}$.
$35 \mathrm{~g} \mathrm{~N}{2}$ requires $\frac{6 \mathrm{~g}}{28 \mathrm{~g}} \times 35 \mathrm{~g} \mathrm{H}{2} \Rightarrow 7.5 \mathrm{~g}$ of $\mathrm{H}_{2}$.
Here, $\mathrm{H}{2}$ gas does not act as limiting reagent since $7.5 \mathrm{~g}$ of $\mathrm{H}{2}$ gas is required for $35 \mathrm{~g}$ of $\mathrm{N}{2}$ and $8 \mathrm{~g}$ of $\mathrm{H}{2}$ is present in reaction mixture. Mass of $\mathrm{H}{2}$ left unreacted $=8-7.5 \mathrm{~g}$ of $\mathrm{H}{2}$.
$$ =0.5 \text { gof } \mathrm{H}_{2} \text {. } $$
Similarly, in option (c) and (d), $\mathrm{H}_{2}$ does not act as limiting reagent.
For $14 \mathrm{~g}$ of $\mathrm{N}{2}+4 \mathrm{~g}$ of $\mathrm{H}{2}$.
As we know $28 \mathrm{~g}$ of $\mathrm{N}{2}$ reacts with $6 \mathrm{~g}$ of $\mathrm{H}{2}$.
$14 \mathrm{~g}$ of $\mathrm{N}{2}$ reacts with $\frac{6}{28} \times 14 \mathrm{~g}$ of $\mathrm{H}{2} \Rightarrow 3 \mathrm{~g}$ of $\mathrm{H}_{2}$.
For $28 \mathrm{~g}$ of $\mathrm{N}{2}+6 \mathrm{~g}$ of $\mathrm{H}{2}$, i.e. $28 \mathrm{~g}$ of $\mathrm{N}{2}$ reacts with $6 \mathrm{~g}$ of $\mathrm{H}{2}$ (by equation I).
