Question

Benzene,C6H6(ℓ), reacts with bromine, Br2 to form bromobenzene, C6H5Br(ℓ) as described below C6H6(ℓ)    +    Br2(ℓ)   →...

Benzene,C6H6(ℓ), reacts with bromine, Br2 to form bromobenzene, C6H5Br(ℓ) as described below

C6H6(ℓ)    +    Br2(ℓ)   →        C6H5Br(ℓ)       +         HBr(g)

Calculate the amount of bromobenzene, C6H5Br(ℓ), in grams that can be produced from reaction of 30.0 g of C6H6 and 50.0 g Br2?

Molar Mass C6H6 = 78.11 g/mol    Molar Mass Br2 = 159.8 g/mol     Molar Mass C6H5Br = 157.0 g/mol

Hint: This is a limiting reactant problem.

Homework Answers

Answer #1

For the given reaction,

moles of benzene = 30 g/78.11 g/mol

                             = 0.3841 mols

moles of Br2 = 50 g/159.8 g/mol

                     = 0.3130 mols

Here,

Br2 is the imiting reagent

Thus,

mass of bromobenzene formed = 0.3130 mol x 157 g/mol

                                                   = 49.141 g

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