Chemistry Question: How to do this?

When benzene (c6H6) reacts with bromine (br2), bromobenzene (C6H5Br) is obtained:
C6H6 + Br2 ---> c6h5Br + HBr

(a) what is the theoretical yield of bromobenzene in this reaction with 65.0g of bromine?
(b) if the actual yield of bromobenzene was 56.7g, what was the percentage yield?

PLease explain TYVM

Answer:
molecular weight of br2 = 159.808... so do moles of br2 = 65.0/159.808 = 0.406738.. but you're using 3 sig figs so 0.407 mol br2 available
that means if all 0.407mol of br2 react then you'll get 0.407 mol of c6h5br... use the molecular weights c=12, h=1, br = 79.9 so c6h5br = 156.9 g/mol...
0.407 mol*157g/mol = 63.8 g bromobenzene theoretical.
actual yield = 56.7/theoretical = 88.8% (which is pretty high)

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