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

The Correct Option is C

To determine if H_2 is the limiting reagent in the given reaction mixtures, we need to calculate the moles of each reactant and compare the ratio to the stoichiometric ratio in the balanced chemical equation. The balanced equation for the reaction is:

N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)

The molar masses of the reactants are approximately:
N_2: 28 \, \text{g/mol}
H_2: 2 \, \text{g/mol}

We need to use these to calculate the moles of each reactant for the options and see if H_2 is the limiting reagent:

1. Option 1: 14 \, \text{g of} \, N_2 + 4 \, \text{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}
   • The stoichiometric requirement is 1 mole of N_2 for every 3 moles of H_2. Here, for 0.5 mole of N_2, we need 1.5 moles of H_2, but we have 2 moles of H_2.
   • Dihydrogen is not the limiting reagent.

2. Option 2: 28 \, \text{g of} \, N_2 + 6 \, \text{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 mole of N_2, we need 3 moles of H_2, and we have exactly 3 moles.
   • Dihydrogen is not the limiting reagent.

3. Option 3: 56 \, \text{g of} \, N_2 + 10 \, \text{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 moles of N_2, we need 6 moles of H_2 (2 moles of N_2 \times 3), but we only have 5 moles.
   • Dihydrogen is the limiting reagent.

4. Option 4: 35 \, \text{g of} \, N_2 + 8 \, \text{g of} \, H_2

   • Moles of N_2 = \frac{35}{28} \approx 1.25 \, \text{mol}
   • Moles of H_2 = \frac{8}{2} = 4 \, \text{mol}
   • For 1.25 moles of N_2, we need 3.75 moles of H_2 (1.25 moles of N_2 \times 3), and we have 4 moles of H_2.
   • Dihydrogen is not the limiting reagent.

Thus, in option 3, H_2 is the limiting reagent as it is insufficient to fully react with the available N_2.