# Chemistry! Test tomorrow, pls help !

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we're given the equation:

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

you need to prepare 50.0g pf bromobenzene and you expect no more than a 75% yield. How much benzene should you begin with if the yield is 75%.

i know that % yield = (actual/theoretical) * 100 %

but i don't know how to get the actual/theoretical values. would 50.0 g be put as an actual or theoretical value? The way i understand it, is that it should be used as the actual value because that is the amount you want to prepare, and if its put as a theoretical value the actual amount would always be less than theoretical. Am i thinking right?

This question confuses me, maybe it confuses me because they say 75% twice and it throws me off lol. please explain steps as well, thank you in advance!

• Chemistry! Test tomorrow, pls help ! - ,

This is a stoichiometry problem with an extra step of % thrown into the mix.

C6H6 + Br2 ==> C6H5Br + HBr.
mols bromobenzene in 50 g is 50/molar mass bromobenzene.

Now you wan to convert mols bromobenzene to mols benzene. Since the equation shows you it is 1 mol benzene to 1 mol bromobenzene, then mols benzene = mols bromobenzene.

Now convert mols benzene to grams. That's g = mols x molar mass.

That gives you grams benzene you would need to start with to prepare 50.0g bromobenzene (that's what the problem wanted) BUT that is for the theoretical yield which is 100%. Let's call grams benzene you've calculated something like n.
I find it easy to say, "What number grams benzene x 0.75 = n grams benzene?" In equation form that is
? g benzene x 0.75 = n g. Solve for ?g benzene.

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