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: $${\mathrm{C}}_6{\mathrm{H}}_6+{\mathrm{Br}}_2⟶{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{Br}+\mathrm{HBr}$$ (a) What is the theoretical yield of bromobenzene in this reaction when $30.0\mathrm{g}$ of benzene reacts with $65.0\mathrm{g}$ of bromine? (b) If the actual yield of bromobenzene was $42.3\mathrm{g}$, what was the percentage yield?

Short answer

The theoretical yield of bromobenzene in this reaction is $60.30g$, and the percentage yield is $70.15\mathrm{\%}$.

Step-by-step solution

Convert mass to moles

Moles = (Mass) / (Molar Mass) The molar mass of benzene, bromine, and bromobenzene can be calculated as: $C_6{H}_6$ (benzene): $6\times 12.01+6\times 1.01=78.12g/mol$ $B{r}_2$ (bromine): $2\times 79.90=159.8g/mol$ $C_6{H}_{5}Br$ (bromobenzene): $6\times 12.01+5\times 1.01+79.90=157.03g/mol$ Now, we can convert the mass of benzene and bromine to moles: Moles of benzene: $30.0g/78.12g/mol=0.384mol$ Moles of bromine: $65.0g/159.8g/mol=0.407mol$

Determine the limiting reactant

In the balanced chemical equation, we can see that one mole of benzene reacts with one mole of bromine to produce one mole of bromobenzene. Now, we have to compare the mole ratio of benzene and bromine to determine the limiting reactant. Mole ratio: $0.384molbenzene/0.407molbromine=0.943$ Since the mole ratio is less than 1, it indicates that benzene is the limiting reactant in this reaction.

Calculate the theoretical yield of bromobenzene

As benzene is the limiting reactant, the theoretical yield of bromobenzene can be calculated using the moles of benzene: Theoretical yield = moles of benzene * molar mass of bromobenzene Theoretical yield = $0.384mol\times 157.03g/mol=60.30g$

Calculate the percentage yield

Now that we have the theoretical and actual yields, we can calculate the percentage yield using the formula: Percentage yield = (Actual yield / Theoretical yield) * 100 Percentage yield = $\frac{42.3g}{60.30g}$ * 100 = 70.15 % The percentage yield of this reaction is 70.15%.