Percentage yield from stoichiometric hydrogen

Steps:

• Molar mass of H₂ is 2 g/mol; moles of 60 kg H₂ = 60,000 g / 2 = 30,000 mol.
• Reaction: 3 mol H₂ produce 2 mol NH₃ (molar mass 17 g/mol), so 30,000 mol H₂ yield (2/3) × 30,000 = 20,000 mol NH₃.
• Theoretical mass of NH₃ = 20,000 mol × 17 g/mol = 340,000 g = 340 kg.
• Percentage yield = (actual yield / theoretical yield) × 100% = (60 / 340) × 100% ≈ 17.6%.

Why B is correct:

• 17% matches the percentage yield formula applied to the reaction stoichiometry, where actual yield is 60 kg against 340 kg theoretical.

Why the others are wrong:

• A: 5% underestimates yield by ignoring full stoichiometric ratio.
• C: 35% overestimates by misapplying mass ratios.
• D: 50% assumes 1:1 mass proportion, violating the 1:3 H₂:NH₃ mole ratio.

Final answer: B