Question Detail
Question
Benzene reacts with bromine to produce bromobenzene and hydrobromic acid. If 30.0g of benzene reacts with 65.0g of bromine and produces 56.7g of bromobenzene, what is the percent yield of the reaction? [Br = 80, C = 12, H = 1]
Correct Answer
Option B is the correct answer.
Detailed Explanation
The balanced equation is C₆H₆ + Br₂ → C₆H₅Br + HBr. Molar mass C₆H₆ = 78 g/mol, Br₂ = 160 g/mol, C₆H₅Br = 157 g/mol. Moles C₆H₆ = 30.0/78 = 0.3846 mol. Moles Br₂ = 65.0/160 = 0.40625 mol. Benzene is the limiting reagent. Theoretical yield of C₆H₅Br from benzene = 0.3846 mol × 157 g/mol = 60.38 g. Percent yield = (actual yield / theoretical yield) × 100 = (56.7 / 60.38) × 100 = 93.9%. The closest answer provided is 94.5%, likely due to rounding differences in molar masses (e.g., using Br=80, C=12, H=1 gives C₆H₆=78, Br₂=160, C₆H₅Br=12*6+5+80=157).
Hint
Determine the limiting reagent first, then calculate the theoretical yield of bromobenzene. Percent yield = (actual yield / theoretical yield) × 100.