I really need help on this part please :(

Question

5. A student prepared bromobenzene in the Organic lab by reacting 10.0 g of benzene, C6H6, with bromine in the presence of a ferric bromide catalyst according to the reaction of

C6H6 (l) + Br2 (l) ––––––> C6H5Br (l) + HBr (g) FeBr3

If the student obtained 12.9 g of bromobenzene, what was the percent yield?

Answer 1

To find the % yield, you first must know the theoretical yield. Percent yield = actual yield / theoretical yield x 100%.

Starting with the balanced equation:

C₆H₆(l) + Br₂(l) ==> C₆H₅Br(l) + HBr(g) (note the FeBr₃ is the catalyst and doesn't enter into the reaction)

Use the stoichiometry of the balanced equation along with dimensional analysis to find the theoretical yield of C₆H₅Br: ... (molar mass C₆H₆ = 78.1 g/mol; molar mass C₆H₅Br = 157 g/mol)

10.0 g C₆H₆ x 1 mol C₆H₆ / 78.1 g x 1 mol C₆H₅Br / mol C₆H₆ x 157 g C₆H₅Br/mol = 20.1 g C₆H₅Br

% yield = 12.9 g / 20.1 g (x100%) = 64.2% yield