Question

What is the theoretical yield of bromobenzene in this reaction when 32.0 g of benzene reacts with 69.3 g of bromine?

Answer 1

First calculate the number of moles of bromine and benzene.

Number of moles = $\frac{given mass in g}{molar mass}$

Mass of benzene = $32.0 g$ (given)

Molar mass of benzene=  $78.11 g/mol$

Substitute the above values in formula, we get

Number of moles of benzene= $\frac{32.0 g}{78.11 g/mol}$

= $0.409 mol$

Mass of bromine = 69.3 g

Molar mass of bromine = $2\times 79.9 g/mol$ = $159.8 g/mol$

Substitute the above values in formula, we get

Number of moles of benzene= $\frac{69.3 g}{159.8 g/mol}$

= $0.43 g/mol$

Now, ratio of benzene and bromine comes out be 1:1 respectively and limiting reagent is benzene because present less as compared to bromine.

Thus, number of moles of bromobenzene = $0.409 mol$

Theoretical yield = $number of moles \times molar mass$

= $0.409 mol \times 157.02 g/mol$   (molar mass of bromonenzene = 157.02 g/mol)

= $64.22118 g$

Hence, theoretical yield of bromobenzene is $64.22118 g$