Percentage yield from iron production reaction

Steps:

• Molar mass of Fe₂O₃ is 160 g/mol; 2Fe is 112 g/mol.
• Theoretical yield of Fe: (112/160) × 80 tonnes = 56 tonnes.
• Actual yield is 50 tonnes.
• Percentage yield = (50/56) × 100% ≈ 89%.

Why D is correct:

• Percentage yield formula gives (actual/theoretical) × 100% = 89%, matching option D.

Why the others are wrong:

• A: 45% from incorrect ratio like 50/112 × 100%.
• B: 63% from mistaking input mass for theoretical yield (50/80 × 100% ≈ 63%).
• C: 68% from approximate mass ratio error (e.g., 112/165 × 100% ≈ 68%).

Final answer: D