Molar percentage yield from reaction stoichiometry

Steps:

• Moles of C₂H₅Br = 5.0 g / 109 g/mol = 0.0459 mol.
• Theoretical moles of C₂H₅OH = 0.0459 mol (1:1 ratio).
• Actual moles of C₂H₅OH = 1.59 g / 46 g/mol = 0.0346 mol.
• Molar % yield = (0.0346 mol / 0.0459 mol) × 100% = 75%.

Why D is correct:

• Percentage yield formula is (actual moles of product / theoretical moles of product) × 100%, yielding exactly 75% here.

Why the others are wrong:

• A. 13%: From dividing actual mass by reactant molar mass (1.59 / 109 × 100%).
• B. 32%: From using ethanol molar mass for reactant (5.0 / 46 × 100%, inverted yield).
• C. 42%: From mass yield without molar conversion (1.59 / 5.0 × 100%, ignoring molar masses).

Final answer: D