Question

For a reaction,
$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$; identify dihydrogen $(H_2)$ as a limiting reagent in the following reaction mixtures.

• A. $14g$ of $N_2$ + $4g$ of $H_2$
• B. $28g$ of $N_2$ + $6g$ of $H_2$
• C. $56g$ of $N_2$ + $10g$ of $H_2$
• D. $35g$ of $N_2$ + $8g$ of $H_2$

Answer 1

Answer: (C)

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.
...(I)
$28\, g\, N _{2}$ 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 $28\, g\, N _{2}$ required $6\, g$ of $H _{2}$.
$35\, g\, N _{2}$ required $\frac{6 g }{28 g } \times 35\, g\, H _{2}$
$\Rightarrow 7.5\, g$ 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 $H _{2}=0.5\, g$ of $H _{2}$
Similarly, in option (c) and (d), $H _{2}$ does not act as limiting reagent.
For $14\, g$ of $N _{2}+4\, g$ or $H _{2}$.
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 $N _{2}+6\, g$ or $H _{2}$,
i.e. $28\, g$ of $N _{2}$ reacts with $6\, g$ of $H _{2}$ (by equation $I$ ).