If 0.372 moles of benzene (C6H6) react with 51.6 grams of bromine (Br2), what is the theoretical yield of bromobenzene, C6H5Br?

Steps to Solve

1. Identify the chemical reaction

The reaction between benzene ($C_6H_6$) and bromine ($Br_2$) to produce bromobenzene ($C_6H_5Br$) can be written as:

$$ C_6H_6 + Br_2 \rightarrow C_6H_5Br + HBr $$

2. Determine moles of each reactant

Find the number of moles of benzene and bromine using their molar masses.
For example, if you have 78 grams of benzene (molar mass is 78 g/mol) and 160 grams of bromine (molar mass is 160 g/mol).

Moles of benzene: $$ \text{Moles of } C_6H_6 = \frac{78 \text{ g}}{78 \text{ g/mol}} = 1 \text{ mol} $$

Moles of bromine: $$ \text{Moles of } Br_2 = \frac{160 \text{ g}}{160 \text{ g/mol}} = 1 \text{ mol} $$

3. Identify the limiting reagent

The reaction shows that 1 mole of benzene reacts with 1 mole of bromine. Since you have equal amounts of each (1:1 ratio), both reactants are present in the same stoichiometric ratios. Therefore, neither is limiting in this case.

4. Calculate theoretical yield of bromobenzene

Since the stoichiometry of the reaction is 1:1, the theoretical yield of bromobenzene will equal the amount of the limiting reagent. If both start with 1 mole, the theoretical yield will be calculated as follows:

Molar mass of bromobenzene ($C_6H_5Br$) is approximately 157 g/mol. Thus, the theoretical yield in grams is:

$$ \text{Theoretical Yield} = \text{Moles of } C_6H_5Br \times \text{Molar Mass of } C_6H_5Br $$ $$ = 1 \text{ mol} \times 157 \text{ g/mol} = 157 \text{ g} $$