1. Balanced reaction

$N_2 + 3H_2 \rightarrow 2NH_3$

This means:

• $1$ mole of $N_2$ reacts with $3$ moles of $H_2$.

2. Molar masses

$M(N_2)=28\,\text{g mol}^{-1}, \qquad M(H_2)=2\,\text{g mol}^{-1}$

So for each option, convert grams to moles and compare with the stoichiometric ratio $1:3$.

3. Check each option

Option A: $56$ g of $N_2$ and $10$ g of $H_2$

Moles of $N_2$: $\frac{56}{28}=2\text{ mol}$

Moles of $H_2$: $\frac{10}{2}=5\text{ mol}$

For $2$ mol of $N_2$, required $H_2$ is: $2\times 3=6\text{ mol}$

Available $H_2 = 5$ mol, which is less than required.

So, $H_2$ is the limiting reagent.

Option B: $14$ g of $N_2$ and $4$ g of $H_2$

Moles of $N_2$: $\frac{14}{28}=0.5\text{ mol}$

Moles of $H_2$: $\frac{4}{2}=2\text{ mol}$

For $0.5$ mol of $N_2$, required $H_2$ is: $0.5\times 3=1.5\text{ mol}$

Available $H_2 = 2$ mol, which is more than required.

So, $H_2$ is not limiting.

Option C: $28$ g of $N_2$ and $6$ g of $H_2$

Moles of $N_2$: $\frac{28}{28}=1\text{ mol}$

Moles of $H_2$: $\frac{6}{2}=3\text{ mol}$

For $1$ mol of $N_2$, required $H_2$ is: $3\text{ mol}$

Available $H_2 = 3$ mol, exactly equal to required.

So, no limiting reagent among these two; mixture is stoichiometric.

Option D: $35$ g of $N_2$ and $8$ g of $H_2$

Moles of $N_2$: $\frac{35}{28}=1.25\text{ mol}$

Moles of $H_2$: $\frac{8}{2}=4\text{ mol}$

For $1.25$ mol of $N_2$, required $H_2$ is: $1.25\times 3=3.75\text{ mol}$

Available $H_2 = 4$ mol, which is more than required.

So, $H_2$ is not limiting.

4. Conclusion

Only in Option A, the available $H_2$ is less than the amount required.

Therefore, dihydrogen is the limiting reagent only in Option A.