Percentage Yield from Hydrogen Limiting Reagent

Steps:

• Calculate moles of H₂ in 60 kg: 60,000 g / 2 g/mol = 30,000 mol.
• Determine theoretical moles of NH₃ using stoichiometry: (2/3) × 30,000 mol = 20,000 mol.
• Compute theoretical mass of NH₃: 20,000 mol × 17 g/mol = 340,000 g = 340 kg.
• Apply percentage yield formula: (actual yield / theoretical yield) × 100% = (60 kg / 340 kg) × 100% ≈ 17.6%.

Why B is correct:

• Percentage yield = (actual / theoretical) × 100% directly computes to ~17%, per standard yield definition.

Why the others are wrong:

• A: Far too low; ignores proper stoichiometric ratio, underestimating theoretical yield by ~3.5 times.
• C: Overestimates yield; likely from assuming 1:2 H₂:NH₃ ratio instead of 3:2.
• D: Assumes 50% as simple mass halving, disregarding molar masses and reaction proportions.

Final answer: B