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Chapter 3: Problem 81

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?

Short Answer

Expert verified
(a) The theoretical yield of bromobenzene is 60.3 g. (b) The percentage yield of bromobenzene is 70.2%.

Step by step solution

01

1. Write the balanced chemical equation

The balanced chemical equation is: \[ C_6H_6 + Br_2 \longrightarrow C_6H_5Br + HBr \]
02

2. Calculate the number of moles of reactants

We need the molar masses of benzene and bromine to convert the given mass to moles. Molar mass of benzene, \(C_6H_6 = 6 \times 12.01 + 6 \times 1.01 = 78.12\,\text{g/mol}\) Molar mass of bromine, \(Br_2 = 2 \times 79.90 = 159.80\,\text{g/mol}\) Now, find the moles of benzene and bromine. Moles of benzene = \(\frac{30.0\,\text{g}}{78.12\,\text{g/mol}} = 0.384\,\text{mol}\) Moles of bromine = \(\frac{65.0\,\text{g}}{159.80\,\text{g/mol}} = 0.407\,\text{mol}\)
03

3. Determine the limiting reactant

To determine the limiting reactant, we will compare the mole ratios of the reactants to the balanced equation. Mole ratio of benzene to bromine: \(\frac{0.384\,\text{mol}}{0.407\,\text{mol}} = 0.943\) From the balanced equation, the mole ratio of benzene to bromine should be 1:1. Since the ratio is less than 1 in our case, benzene is the limiting reactant.
04

4. Calculate the theoretical yield of bromobenzene

The balanced equation tells us that one mole of benzene reacts to produce one mole of bromobenzene. Thus, the theoretical yield of bromobenzene is equal to the moles of the limiting reactant (benzene). With the molar mass of bromobenzene, we can calculate its theoretical yield: Molar mass of bromobenzene, \(C_6H_5Br = 12.01\times 6+ 1.01\times 5 + 79.90 = 157.03\,\text{g/mol}\) Theoretical yield = moles of benzene \(\times\) molar mass of bromobenzene Theoretical yield = \(0.384\,\text{mol} \times 157.03\,\text{g/mol} = 60.3\,\text{g}\)
05

5. Calculate the percentage yield

Now we can calculate the percentage yield using the theoretical yield (60.3 g) and the actual yield (42.3 g): Percentage yield = \(\frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100\) Percentage yield = \(\frac{42.3\,\text{g}}{60.3\,\text{g}} \times 100 = 70.2\%\) The percentage yield of bromobenzene is 70.2%.

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Most popular questions from this chapter

When ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) reacts with chlorine \(\left(\mathrm{Cl}_{2}\right)\), the main product is \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\), but other products containing \(\mathrm{Cl}\), such as \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2},\) are also obtained in small quantities. The formation of these other products reduces the yield of \(\mathrm{C}_{2} \mathrm{H}_{5}\) Cl. (a) Calculate the theoretical yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) when \(125 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{6}\) reacts with \(255 \mathrm{~g}\) of \(\mathrm{Cl}_{2}\), assuming that \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{Cl}_{2}\) react only to form \(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}\) and HCl. (b) Calculate the percent yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) if the reaction produces \(206 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\).

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