Question

Part A What is the theoretical yield of bromobenzene in this reaction when 35.0 g of benzene reacts with 75.8 g of bromine? m = 70.3 g

Part B If the actual yield of bromobenzene was 65.4 g , what was the percentage yield?

Answer 1

Th reaction is

Benzene + Br2 ---> Bromobenzene

Benzene moles = mass of benzene / Molar mass of benzene

               = (35 g / 78.11 g/mol ) = 0.448

Br2 moles = mass / molar mass of Br2   = (75.8 g / 159.8 g/mol ) = 0.47434

since Benzene and Br2 reacts in 1:1 ratio , benzene being relatively less is limiting reagent

Bromobenzene moles produced =0.448

Mass of bromobenzene = 0.448 mol x 157 g/mol                ( mass = moles x molar mass)

          = 70.35 g                            

Theoretical yield = 70.35 g

actual yield = 65.4 g

percent yield = ( 100 x actual yield ) / ( theoretical yield)

           = ( 100 x 65.4 g / 70.35 g)

       = 93 %