When benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) reacts with bromine \(\left(\mathrm{Br}_{2}\right),\) bromobenzene $\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}\right)$ is obtained: $$ \mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{Br}_{2} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}+\mathrm{HBr} $$ (a) When \(30.0 \mathrm{~g}\) of benzene reacts with \(65.0 \mathrm{~g}\) of bromine, what is the theoretical yield of bromobenzene? (b) If the actual yield of bromobenzene is \(42.3 \mathrm{~g}\), what is the percentage yield?

Step 1: Determine the moles of each reactant First, we need to find the number of moles of benzene and bromine. To do this, divide the mass by the molar mass of each substance. Number of moles of benzene = \(\frac{30.0 \ g}{78.11 \ g/mol} \approx0.384 \ mol\) Number of moles of bromine = \(\frac{65.0 \ g}{159.81 \ g/mol} \approx0.407 \ mol\) Step 2: Determine the limiting reactant For each reactant, calculate the mole ratio with respect to the balanced chemical equation: Mole ratio of benzene = \(\frac{0.384 \ mol}{1}\) Mole ratio of bromine = \(\frac{0.407 \ mol}{1}\) Since the mole ratio of benzene is smaller, it is the limiting reactant. Step 3: Calculate the theoretical yield of bromobenzene Use the given balanced chemical equation to calculate the number of moles of bromobenzene produced by 0.384 moles of benzene: \[\frac{\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br} \ moles}{\mathrm{C}_{6} \mathrm{H}_{6} \ moles} = \frac{1}{1}\] Number of moles of bromobenzene produced = \(0.384 \ mol\) Next, multiply the number of moles of bromobenzene produced by its molar mass to find the theoretical yield in grams. Theoretical Yield of bromobenzene = 0.384 mol × 157.01 g/mol = \(60.2 \ g\) So, the theoretical yield of bromobenzene is 60.2 g.

Step 1: Calculate the percentage yield Now, we'll use the actual yield and theoretical yield to calculate the percentage yield. Percentage Yield = \(\frac{\textrm{Actual Yield}}{\textrm{Theoretical Yield}}\times 100\%\) Percentage Yield = \(\frac{42.3 \ g}{60.2 \ g}\times 100\% \approx 70.3\%\) So, the percentage yield of bromobenzene is approximately 70.3%.