Question

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.

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