If a student needs to prepare 50g of bromobenzene and expects no more than a 75% yieldl, how much benzene should the student begin with if the yield is 75%?

Question

C6H6(l)+ Br2(l) --> C6H5Br(l)+ HBr(g)

I calculated the moles of C6H5Br as .3185. But ion the end my answer is 18.7g

However, the correct answer is 33.2g. Which you get from using the mm of Br2 instead of C6H6.

Since i'm being asked about benzene would you use the mm of benzene?

Answer 1

You multiplied moles bromobenzene by 0.75. You should have divided by 0.75 to get 33.2%