Answer: B. Calculate moles of starting ethanol and theoretical moles of product.
Moles of \(\text{C}_2\text{H}_5\text{OH}\) = 23/46 = 0.500 mol.
From the 1:1 stoichiometry of \(\text{C}_2\text{H}_5\text{OH} + \text{HBr} \rightarrow \text{C}_2\text{H}_5\text{Br} + \text{H}_2\text{O}\), the theoretical moles of bromoethane = 0.500 mol.
Theoretical mass of \(\text{C}_2\text{H}_5\text{Br}\) = \(0.500 \times 109 = 54.5\) g.
%yield = \(\frac{27}{54.5} \times 100\% \approx 49.5\%\), which rounds to 50%.
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