$$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$$
Consider the reaction: . If 20 g of dinitrogen reacts with 5 g of dihydrogen, then the limiting reagent of the reaction and number of moles of $$NH_3$$ formed respectively are

We have the reaction $$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$$. We are given 20 g of $$N_2$$ and 5 g of $$H_2$$.

The molar mass of $$N_2$$ is 28 g/mol, so the moles of $$N_2$$ are $$\dfrac{20}{28} = \dfrac{5}{7} \approx 0.714$$ mol. The molar mass of $$H_2$$ is 2 g/mol, so the moles of $$H_2$$ are $$\dfrac{5}{2} = 2.5$$ mol.

From the stoichiometry, 1 mole of $$N_2$$ requires 3 moles of $$H_2$$. So $$\dfrac{5}{7}$$ moles of $$N_2$$ would require $$3 \times \dfrac{5}{7} = \dfrac{15}{7} \approx 2.14$$ moles of $$H_2$$. We have 2.5 moles of $$H_2$$ available, which is more than enough. Therefore, $$N_2$$ is the limiting reagent because it will be consumed first.

Now, from the balanced equation, 1 mole of $$N_2$$ produces 2 moles of $$NH_3$$. So $$\dfrac{5}{7}$$ moles of $$N_2$$ will produce $$2 \times \dfrac{5}{7} = \dfrac{10}{7} \approx 1.42$$ moles of $$NH_3$$.

Hence, the correct answer is Option C.