1. **State the problem:** Calculate the theoretical yield of bromobenzene (C6H5Br) when 28.0 g of benzene (C6H6) reacts with 60.7 g of bromine (Br2). Then find the percentage yield if the actual yield is 51.2 g. 2. **Write the balanced chemical equation:** $$\mathrm{C_6H_6 + Br_2 \rightarrow C_6H_5Br + HBr}$$ 3. **Calculate molar masses:** - Benzene (C6H6): $6 \times 12.01 + 6 \times 1.008 = 78.11$ g/mol - Bromine (Br2): $2 \times 79.90 = 159.80$ g/mol - Bromobenzene (C6H5Br): $6 \times 12.01 + 5 \times 1.008 + 79.90 = 157.01$ g/mol 4. **Convert given masses to moles:** - Moles of benzene: $$\frac{28.0}{78.11} = 0.3586\text{ mol}$$ - Moles of bromine: $$\frac{60.7}{159.80} = 0.3799\text{ mol}$$ 5. **Determine the limiting reactant:** The reaction ratio is 1:1. - Benzene: 0.3586 mol - Bromine: 0.3799 mol Benzene is limiting because it has fewer moles. 6. **Calculate theoretical moles of bromobenzene produced:** Equal to moles of limiting reactant benzene: $$0.3586\text{ mol}$$ 7. **Calculate theoretical mass of bromobenzene:** $$0.3586 \times 157.01 = 56.3\text{ g}$$ 8. **Calculate percentage yield:** $$\text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 = \frac{51.2}{56.3} \times 100 = 90.9\%$$ **Final answers:** - Theoretical yield $m = 56.3$ g - Percentage yield $= 90.9$ %