For the reaction 3 H2 + N2 → 2 NH3, If you have 4.0 g H2 and 28.0 g N2, which is limiting and what is the theoretical NH3 yield (in grams)?

• A. N2 is limiting; NH3 ≈ 44.0 g
• B. H2 is excess; NH3 ≈ 10 g
• C. N2 is limiting; NH3 ≈ 15 g
• D. H2 is limiting; NH3 ≈ 22.5 g

Answer: (D)

The main idea is identifying the limiting reactant by comparing how much of each reactant is needed versus how much is actually available, using the stoichiometric ratio.

Convert masses to moles: H2 ≈ 4.0 g / 2.02 g/mol ≈ 1.98 mol; N2 ≈ 28.0 g / 28.02 g/mol ≈ 1.00 mol. The reaction requires 3 mol H2 for every 1 mol N2. To use all of the N2 you’d need 3.00 mol H2, but only about 1.98 mol H2 are available. So hydrogen runs out first and is the limiting reactant.

From the ratio 3 H2 → 2 NH3, the amount of NH3 produced is (2/3) times the amount of limiting H2: NH3 moles ≈ (2/3) × 1.98 ≈ 1.32 mol. Converting to grams: 1.32 mol × 17.03 g/mol ≈ 22.5 g.

The theoretical yield of NH3 is about 22.5 g.