C6H6(l) Br2(l) = C6H5Br(l) + HBr(g)

Nov 2013
C6H6(l) Br2(l) = C6H5Br(l) + HBr(g)

My question is if 18.3 grams of Benzene (C6H6) reacts with 3.21 grams of Bromine (Br2) how many grams of Bromobenzene (C6H5Br) will be produced?

I have tried this multiple ways and I am really lost when it comes to setting this up. I tried converting the benzene and bromine to moles but not sure if this is the correct way to go about this problem because after I convert it to moles separately I can't figure where to go from there???
Apr 2013
C6H6(l) + Br2(l) = C6H5BR +HBr

You were on the right track. The first step is to convert grams to moles

So you have 18.3 grams of Benzene and 3.21 grams of Bromine, so you would determine the molar mass of each compound.

18.3 g C6H6 = 1 mole C6H6 = 0.23427 moles
78.114 g/mol

3.21 g Br2 = 1 mole = 0.02 moles
159.808 g/mol

If you look at the molar ratio You see that the ratio on either C6H6 or Br2 to C6H5BR is 1:1. So 18.3 grams will product .23427 moles of Bromobenzine, and 3.21 grams of Bromine will produce .02 moles of Bromobenzine.

Before you solve, you need to determine if there is a limiting reactant. In this case the limiting reactant is Bromine, at .02 moles. Therefore the maximum amount of product from the reaction is .02 moles.

Now that we have the molar amount of the limiting reactant, we now need the molar mass of Bromobenzine. So, adding the molar mass of each element in the compound, the molar mass of Bromobenzene is 157.02 g/mol

The final step in the problem is to determine the mass of 0.02 mol of Bromobenzine

.02 mol Bromobenzine 157.02 grams = 3.14 g C6H5Br
1 mole

Just remember, these problems look very daunting at first, but more often than not it's easy to break a complex problem into several modular steps.
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