a) To determine the maximum amount of C6H5Br (bromobenzene) that could be obtained from 15.0 g of benzene, we need to use the concept of stoichiometry and compare the ratios of the substances involved in the reaction.
1 mole of benzene (C6H6) reacts with 1 mole of bromine (Br2) to produce 1 mole of bromobenzene (C6H5Br).
The molar mass of benzene (C6H6) is 78.11 g/mol.
To calculate the theoretical yield, we follow these steps:
1. Convert the given mass of benzene (C6H6) to moles:
moles of C6H6 = mass of benzene (g) / molar mass of benzene (g/mol)
moles of C6H6 = 15.0 g / 78.11 g/mol
2. Using the stoichiometry of the reaction, we can see that the ratio of benzene to bromobenzene is 1:1. So, the moles of C6H5Br (bromobenzene) formed will be the same as the moles of benzene used.
3. Convert the moles of C6H5Br to grams:
mass of C6H5Br (g) = moles of C6H5Br × molar mass of C6H5Br
Now you can calculate the maximum amount of C6H5Br (bromobenzene) that can be obtained from 15.0 g of benzene using the steps above.
b) To find out how much C6H6 was not converted to C6H5Br, we need to compare the stoichiometric ratio of benzene (C6H6) to bromobenzene (C6H5Br). From the balanced equation:
1 mole of C6H6 (benzene) reacts with 1 mole of C6H5Br (bromobenzene).
So, if 1 mole of C6H5Br2+ (dibromobenzene) is obtained, this means that 1 mole of bromobenzene was formed. From the reaction equation, we can see that 1 mole of bromobenzene is formed from 1 mole of benzene. Therefore, all the benzene must have been converted to bromobenzene, and none is left unreacted.
c) The actual yield of C6H5Br (bromobenzene) obtained by the student is given as 2.50 g. This is the experimental yield.
d) To calculate the percentage yield, we use the formula:
Percentage yield = (actual yield / theoretical yield) × 100
Substitute the values for the actual yield and theoretical yield (calculated in part a) into the formula to calculate the percentage yield.