Percentage yield from experimental data

Steps:

• Molar mass of C₂H₅Br (bromoethane) is 109 g/mol; moles = 10.9 g / 109 g/mol = 0.1 mol.
• Reaction is 1:1, so theoretical moles of C₂H₅OH (ethanol) = 0.1 mol.
• Molar mass of C₂H₅OH is 46 g/mol; theoretical yield = 0.1 mol × 46 g/mol = 4.6 g.
• Percentage yield = (actual yield / theoretical yield) × 100% = (3.45 g / 4.6 g) × 100% = 75%.

Why C is correct:

• Percentage yield formula gives exactly 75% using stoichiometric 1:1 ratio and standard molar masses matching the masses provided.

Why the others are wrong:

• A. 32%: Too low; ignores proper molar mass ratio (46/109 ≈ 0.422, but misapplied).
• B. 42%: Incorrect; possibly from erroneous moles or mass ratio without 1:1 stoichiometry.
• D. 100%: Assumes perfect reaction; actual yield (3.45 g) is less than theoretical (4.6 g).

Final answer: C