A student reacts benzene,C6H6. with bromine. Br2. in an attempt to prepare bromobenzene.
C6HSBr :
C6H6 + Brl·---·-_·_-> C6HSBr + HBr
(a) What is the theoretical yield of bromobenzene in this reaction when 60.0 g of benzene reacts with 60.0 g Brl?
b) If the actual yield of bromobenzene was 56.7 g • what was the percentage yield?
This is a limiting reagent problem. Here is a worked example. Just follow the steps.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html
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To find the theoretical yield of bromobenzene (C6H5Br) in this reaction, we need to determine which reactant is limiting and calculate the amount of product that can be formed based on that reactant.
(a) The reaction equation given is:
C6H6 + Br2 → C6H5Br + HBr
1 mol of benzene (C6H6) reacts with 1 mol of Br2 to produce 1 mol of bromobenzene (C6H5Br).
First, we need to convert the given masses of benzene and Br2 into moles by using their molar masses. The molar mass of C6H6 is 78.11 g/mol, and the molar mass of Br2 is 159.81 g/mol.
Moles of benzene = (60.0 g) / (78.11 g/mol) = 0.767 mol
Moles of Br2 = (60.0 g) / (159.81 g/mol) = 0.375 mol
Since the stoichiometry of the reaction shows that the mole ratio between benzene and Br2 is 1:1, it means that the limiting reactant is Br2 since it has fewer moles.
To calculate the theoretical yield of bromobenzene, we need to find the mole ratio of bromobenzene to Br2, which is 1:1.
The theoretical yield of bromobenzene = 0.375 mol
To convert the theoretical yield from moles to grams, we use the molar mass of bromobenzene, which is 157.02 g/mol.
Theoretical yield of bromobenzene = (0.375 mol) × (157.02 g/mol) = 58.7625 g
Therefore, the theoretical yield of bromobenzene in this reaction is 58.8 g (rounded to one decimal place).
(b) The percentage yield can be calculated by dividing the actual yield by the theoretical yield and multiplying by 100.
Percentage yield = (actual yield / theoretical yield) × 100
Percentage yield = (56.7 g / 58.8 g) × 100
Percentage yield ≈ 96.4%
Therefore, the percentage yield of bromobenzene in this reaction is approximately 96.4%.
To calculate the theoretical yield of bromobenzene (C6H5Br) in this reaction, we need to determine which reactant is limiting and perform a stoichiometric calculation.
(a) Calculating the theoretical yield:
1. Start by determining the limiting reactant. To do this, compare the moles of benzene (C6H6) and bromine (Br2) using their respective molar masses.
- Molar mass of C6H6: 6 * atomic mass of C + 6 * atomic mass of H
= 6 * 12.01 g/mol + 6 * 1.01 g/mol
= 78.11 g/mol
- Molar mass of Br2: 2 * atomic mass of Br
= 2 * 79.90 g/mol
= 159.80 g/mol
Calculate the moles of benzene and bromine:
- Moles of C6H6 = mass of benzene / molar mass of C6H6
= 60.0 g / 78.11 g/mol
≈ 0.767 mol
- Moles of Br2 = mass of bromine / molar mass of Br2
= 60.0 g / 159.80 g/mol
≈ 0.375 mol
2. Next, use the stoichiometry of the balanced chemical equation to determine the moles of bromobenzene that can be produced.
The balanced chemical equation indicates that 1 mole of benzene reacts with 1 mole of bromine to produce 1 mole of bromobenzene.
Since the reaction is 1:1, the moles of bromobenzene that can be formed will be equal to the moles of the limiting reactant.
In this case, the limiting reactant is bromine (Br2), as it provides fewer moles compared to benzene (C6H6).
Therefore, the moles of bromobenzene formed = 0.375 mol.
3. Finally, calculate the theoretical yield of bromobenzene in grams using its molar mass.
- Molar mass of C6H5Br: 12.01 g/mol + 6 * 1.01 g/mol + 79.90 g/mol
= 157.02 g/mol
Theoretical yield = moles of bromobenzene * molar mass of C6H5Br
= 0.375 mol * 157.02 g/mol
≈ 58.88 g
Therefore, the theoretical yield of bromobenzene in this reaction is approximately 58.88 grams.
(b) Calculating the percentage yield:
The percentage yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100.
Given information:
- Actual yield = 56.7 g
- Theoretical yield = 58.88 g
Percentage yield = (actual yield / theoretical yield) * 100
= (56.7 g / 58.88 g) * 100
≈ 96.17%
Therefore, the percentage yield of bromobenzene in this reaction is approximately 96.17%.