For a reaction, N _ 2 (g) + 3 H _ 2 ( g ) ⟶ 2N H _ 3 ( g ) ,...

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

$\dfrac{\text{moles of}{N}_{2}}{\text{stoichiometry of}{\text{N}}_{2}} \ne \dfrac{\text{mole of}{\text{H}}_{2}}{\text{stoichiometry of}{\text{H}}_{2}}$
Whose ratio is less called limiting reagent.

(a) mole of ${N}_{2} = \dfrac{28}{28} = 1$
$\dfrac{moles}{stiochiometry}$ ratio of ${N}_{2} = \dfrac{1}{1} = 1$
mole of ${H}_{2} = \dfrac{6}{2} = 3$
$\dfrac{moles}{stiochiometry}$ ratio of ${H}_{2} = \dfrac{3}{3} = 1$
there is no limiting reagent.

(b) mole of ${N}_{2} = \dfrac{35}{28} = 1.25$
$\dfrac{moles}{stiochiometry}$ ratio of ${N}_{2} = \dfrac{1.25}{1}$ limiting reagent
mole of ${H}_{2} = \dfrac{8}{2} = 4$
$\dfrac{moles}{stiochiometry}$ ratio of ${H}_{2} = \dfrac{4}{3} = 1.33$

(c) mole of ${N}_{2} = \dfrac{56}{28} = 2$
$\dfrac{moles}{stiochiometry}$ ratio of ${N}_{2} = \dfrac{2}{1} = 2$
mole of ${H}_{2} = \dfrac{10}{2} = 5$
$\dfrac{moles}{stiochiometry}$ ratio of ${H}_{2} = \dfrac{5}{3} = 1.66$ limiting reagent.

(d) mole of ${N}_{2} = \dfrac{14}{28} = 0.5$
$\dfrac{moles}{stiochiometry}$ ratio of ${N}_{2} = \dfrac{0.5}{1} = 0.5$ limiting reagent
mole of ${H}_{2} = \dfrac{4}{2} = 2$
$\dfrac{moles}{stiochiometry}$ ratio of ${H}_{2} = \dfrac{2}{3} = 1.66$

$∴$ Answer is $\left(3\right)$ .