SATHEE: Some Basic Concepts of Chemistry

Some Basic Concepts of Chemistry - Result Question 5

####5. For a reaction,

$N_{2} (g) + 3 H_{2} (g) ⟶ 2 N H_{3} (g)$ , identify dihydrogen $(H_{2})$ 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 $H_{2}$

(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 $(H_{2})$ act as a limiting reagent in the reaction.

$\begin{array}{rrr} N_{2} (g) + 3 H_{2} (g) ⟶ & N H_{3} (g) 2 \times 14 g & 3 \times 2 g & 2 (14 + 3) g 28 g & 6 g & 34 g \end{array}$

$28 g N_{2}$ 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 $H_{2}$ .