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Answer :
Final answer:
To find the theoretical yield of bromobenzene, calculate the moles of benzene (45.0 g) and bromine (97.5 g), identify the limiting reagent, and use its moles to calculate the moles of bromobenzene that can be theoretically produced. Multiply by the molar mass of bromobenzene to get the theoretical yield in grams, and use this to find the percent yield based on the actual product obtained.
Explanation:
To calculate the theoretical yield of bromobenzene, we need to determine the limiting reagent and use it to calculate the moles of bromobenzene that can be formed. Benzene (C6H6) reacts with bromine (Br2) in a 1:1 molar ratio to produce bromobenzene (C6H5Br). First, we convert the masses of benzene and bromine to moles:
- For benzene: molecular weight of C6H6 = 78.11 g/mol, so moles of benzene = 45.0 g / 78.11 g/mol.
- For bromine: molecular weight of Br2 = 159.808 g/mol, so moles of bromine = 97.5 g / 159.808 g/mol.
After calculating the moles of each reactant, we identify the limiting reagent, which is the one present in the lower molar amount. The theoretical yield is then calculated based on the moles of the limiting reagent, using the corresponding molar ratio. If benzene is the limiting reagent, the theoretical yield of bromobenzene would be the moles of benzene used multiplied by the molecular weight of bromobenzene (C6H5Br), 157.01 g/mol.
The percent yield is then calculated by taking the ratio of the actual yield (mass of product isolated) to the theoretical yield and multiplying by 100. The percent yield could vary, for example, a yield of 40% would mean that if 0.08 moles of product were obtained from a possible 0.2 moles based on the limiting reagent, the yield would be (0.08/0.2) imes 100.
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