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

1 Answer

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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


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