1.80 g of E-Stilbene (molar mass: 180 g/mol) reacts with 3.25 g of pyridinium tribromide (molar mass: 320 g/mol) in acetic acid to give 2.55 g of pure meso-1,2-dibromo-1,2-diphenylethane (molar mass: 340 g/mol). What is the percentage yield of this reaction?
A 50.0%
B 65.0%
C 75.0%
D 80.0%
The correct answer and explanation is :
To determine the percentage yield of this reaction, we need to calculate the theoretical yield of meso-1,2-dibromo-1,2-diphenylethane based on the limiting reagent and compare it to the actual yield.
Step 1: Identify the limiting reagent
First, we need to calculate the number of moles of both E-Stilbene and pyridinium tribromide.
- E-Stilbene:
The molar mass of E-Stilbene is given as 180 g/mol. The number of moles of E-Stilbene is calculated as:
[
\text{moles of E-Stilbene} = \frac{\text{mass of E-Stilbene}}{\text{molar mass of E-Stilbene}} = \frac{1.80 \, \text{g}}{180 \, \text{g/mol}} = 0.010 \, \text{mol}
] - Pyridinium tribromide:
The molar mass of pyridinium tribromide is 320 g/mol. The number of moles of pyridinium tribromide is calculated as:
[
\text{moles of pyridinium tribromide} = \frac{\text{mass of pyridinium tribromide}}{\text{molar mass of pyridinium tribromide}} = \frac{3.25 \, \text{g}}{320 \, \text{g/mol}} = 0.01016 \, \text{mol}
]
Since the reaction involves a 1:1 molar ratio between E-Stilbene and pyridinium tribromide, E-Stilbene is the limiting reagent because it has fewer moles (0.010 mol).
Step 2: Calculate the theoretical yield
In this reaction, 1 mole of E-Stilbene reacts with 1 mole of pyridinium tribromide to give 1 mole of meso-1,2-dibromo-1,2-diphenylethane. Since E-Stilbene is the limiting reagent, the theoretical yield of meso-1,2-dibromo-1,2-diphenylethane is based on the number of moles of E-Stilbene:
[
\text{moles of meso-1,2-dibromo-1,2-diphenylethane} = \text{moles of E-Stilbene} = 0.010 \, \text{mol}
]
The molar mass of meso-1,2-dibromo-1,2-diphenylethane is 340 g/mol. Therefore, the theoretical yield is:
[
\text{theoretical yield} = \text{moles of meso-1,2-dibromo-1,2-diphenylethane} \times \text{molar mass of meso-1,2-dibromo-1,2-diphenylethane}
]
[
\text{theoretical yield} = 0.010 \, \text{mol} \times 340 \, \text{g/mol} = 3.40 \, \text{g}
]
Step 3: Calculate the percentage yield
The actual yield is given as 2.55 g. The percentage yield is calculated as:
[
\text{percentage yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100
]
[
\text{percentage yield} = \left( \frac{2.55 \, \text{g}}{3.40 \, \text{g}} \right) \times 100 = 75.0\%
]
Conclusion:
The percentage yield of the reaction is 75.0%, which corresponds to answer choice C.