Some Basic Concepts of Chemistry - Result Question 67
####5. For a reaction,
$N _2(g)+3 H _2(g) \longrightarrow 2 NH _3(g)$, identify dihydrogen $\left(H _2\right)$ as a limiting reagent in the following reaction mixtures.
(2019 Main, 9 April I)
(a) $56 g$ of $N _2+10 g$ of $H _2$
(b) $35 g$ of $N _2+8 g$ of $H _2$
(c) $14 g$ of $N _2+4 g$ of $H _2$
(d) $28 g$ of $N _2+6 g$ of $H _2$
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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 $56 g$ of $N _2+10 g$ of $H _2$ is taken as a combination then dihydrogen $\left(H _2\right)$ act as a limiting reagent in the reaction.
$$ \begin{array}{rrr} N _2(g)+3 H _2(g) \longrightarrow & NH _3(g) \ 2 \times 14 g & 3 \times 2 g & 2(14+3) g \ 28 g & 6 g & 34 g \end{array} $$
$28 g N _2$ requires $6 g H _2$ gas.
$56 g$ of $N _2$ requires $\frac{6 g}{28 g} \times 56 g=12 g$ of $H _2$ $12 g$ of $H _2$ gas is required for $56 g$ of $N _2$ gas but only $10 g$ of $H _2$ gas is present in option (a).
Hence, $H _2$ gas is the limiting reagent.
In option (b), i.e. $35 g$ of $N _2+8 g$ of $H _2$.
As $28 g N _2$ requires $6 g$ of $H _2$.
$35 g N _2$ requires $\frac{6 g}{28 g} \times 35 g H _2 \Rightarrow 7.5 g$ of $H _2$.
Here, $H _2$ gas does not act as limiting reagent since $7.5 g$ of $H _2$ gas is required for $35 g$ of $N _2$ and $8 g$ of $H _2$ is present in reaction mixture. Mass of $H _2$ left unreacted $=8-7.5 g$ of $H _2$.
$$ =0.5 \text { gof } H _2 \text {. } $$
Similarly, in option (c) and (d), $H _2$ does not act as limiting reagent.
For $14 g$ of $N _2+4 g$ of $H _2$.
As we know $28 g$ of $N _2$ reacts with $6 g$ of $H _2$.
$14 g$ of $N _2$ reacts with $\frac{6}{28} \times 14 g$ of $H _2 \Rightarrow 3 g$ of $H _2$.
For $28 g$ of $N _2+6 g$ of $H _2$, i.e. $28 g$ of $N _2$ reacts with $6 g$ of $H _2$ (by equation I).
