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 =
`(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=
`(32.0 g)/(78.11 g/mol)`

=
`0.409 mol`

Mass of bromine = 69.3 g

Molar mass of bromine =
`2* 79.9 g/mol` =
`159.8 g/mol`

Substitute the above values in formula, we get

Number of moles of benzene=
`(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 * molar mass`

=
`0.409 mol * 157.02 g/mol` (molar mass of bromonenzene = 157.02 g/mol)

=
`64.22118 g`

Hence, theoretical yield of bromobenzene is
`64.22118 g`