Problem 83 - Chemistry: The Central Science

Question

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-by-Step Solution

Answer

(a) The theoretical yield of bromobenzene is 63.9 g. (b) The percentage yield of bromobenzene is 66.2%.

Step 1: Determine the molar masses of benzene, bromine, and bromobenzene.

Calculate the molar mass of benzene (C6H6), bromine (Br2), and bromobenzene (C6H5Br) by adding up the molar masses of each element in the compounds according to the periodic table: Benzene (C6H6): \(6 \times 12.01 + 6 \times 1.01 = 72.06 \ \mathrm{g/mol}\) Bromine (Br2): \(2 \times 79.9 = 159.8 \ \mathrm{g/mol}\) Bromobenzene (C6H5Br): \(6 \times 12.01 + 5 \times 1.01 + 79.9 = 157.01 \ \mathrm{g/mol}\)

Step 2: Find the number of moles of reactants.

Divide the given masses by their molar masses: Benzene: \(30.0 \ \mathrm{g} \div 72.06 \ \mathrm{g/mol} = 0.416 \ \mathrm{mol}\) Bromine: \(65.0 \ \mathrm{g} \div 159.8 \ \mathrm{g/mol} = 0.407 \ \mathrm{mol}\)

Step 3: Find the limiting reactant.

Compare the molar ratio of the reactants to the coefficients in the balanced equation to determine which reactant is the limiting reactant: Benzene to Bromine: \(\frac{0.416}{1} \div \frac{0.407}{1} = 1.022\) Since the ratio is greater than the stoichiometric ratio (1:1), benzene is in excess and bromine is the limiting reactant.

Step 4: Calculate the theoretical yield.

Use the moles of the limiting reactant (bromine) and the balanced chemical equation to find the moles of bromobenzene produced: \(0.407 \ \mathrm{mol \ Bromine} \times \frac{1 \ \mathrm{mol \ Bromobenzene}}{1 \ \mathrm{mol \ Bromine}} = 0.407 \ \mathrm{mol \ Bromobenzene}\) Now, convert the moles of bromobenzene to grams using its molar mass: Theoretical yield: \(0.407 \ \mathrm{mol} \times 157.01 \ \mathrm{g/mol} = 63.9 \ \mathrm{g}\) (a) The theoretical yield of bromobenzene is 63.9 g.

Step 5: Determine the percentage yield.

Divide the actual yield by the theoretical yield and multiply by 100: Percentage yield: \(\frac{42.3 \ \mathrm{g}}{63.9 \ \mathrm{g}} \times 100 = 66.2 \% \) (b) The percentage yield of bromobenzene is 66.2%.