For a reaction,
${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)⟶2\mathrm{N}{\mathrm{H}}_3\left(\mathrm{g}\right)$, identify di-hydrogen $\left({\mathrm{H}}_2\right)$ as a limiting reagent in the following reaction mixtures.

${\text{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\to 2\mathrm{N}{\mathrm{H}}_3\left(\mathrm{g}\right)$

$\frac{\text{moles of }{\mathrm{N}}_2}{\text{stoichiometry of }{\text{N}}_2}\ne \frac{\text{mole of }{\text{H}}_2}{\text{stoichiometry of }{\text{H}}_2}$
Whose ratio is less called limiting reagent.

(a) mole of ${\mathrm{N}}_2=\frac{28}{28}=1$
$\frac{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{c}\mathrm{h}\mathrm{i}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{r}\mathrm{y}}$ ratio of ${\mathrm{N}}_2=\frac{1}{1}=1$
mole of ${\mathrm{H}}_2=\frac{6}{2}=3$
$\frac{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{c}\mathrm{h}\mathrm{i}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{r}\mathrm{y}}$ ratio of ${\mathrm{H}}_2=\frac{3}{3}=1$
there is no limiting reagent.

(b) mole of ${\mathrm{N}}_2=\frac{35}{28}=1.25$
$\frac{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{c}\mathrm{h}\mathrm{i}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{r}\mathrm{y}}$ ratio of ${\mathrm{N}}_2=\frac{1.25}{1}$ limiting reagent
mole of ${\mathrm{H}}_2=\frac{8}{2}=4$
$\frac{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{c}\mathrm{h}\mathrm{i}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{r}\mathrm{y}}$ ratio of ${\mathrm{H}}_2=\frac{4}{3}=1.33$

(c) mole of ${\mathrm{N}}_2=\frac{56}{28}=2$
$\frac{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{c}\mathrm{h}\mathrm{i}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{r}\mathrm{y}}$ ratio of ${\mathrm{N}}_2=\frac{2}{1}=2$
mole of ${\mathrm{H}}_2=\frac{10}{2}=5$
$\frac{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{c}\mathrm{h}\mathrm{i}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{r}\mathrm{y}}$ ratio of ${\mathrm{H}}_2=\frac{5}{3}=1.66$ limiting reagent.

(d) mole of ${\mathrm{N}}_2=\frac{14}{28}=0.5$
$\frac{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{c}\mathrm{h}\mathrm{i}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{r}\mathrm{y}}$ ratio of ${\mathrm{N}}_2=\frac{0.5}{1}=0.5$ limiting reagent
mole of ${\mathrm{H}}_2=\frac{4}{2}=2$
$\frac{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{c}\mathrm{h}\mathrm{i}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{r}\mathrm{y}}$ ratio of ${\mathrm{H}}_2=\frac{2}{3}=1.66$

$\therefore$ Answer is $\left(3\right)$ .