Question

9) Consider the reaction of C6H6 Br2 C6H5Br HBr a. What is the theoretical yield of C6H5Br if 42.1 g of C6H6 react with 73.0 g of Br2

Answer 1

Answer:

71.7 g of C₆H₅Br are produced at the theoretical yield.

Explanation:

The reaction is:

C₆H₆ + B₂ → C₆H₅Br + HBr

This is an easy excersie, stoichiometry is 1:1

We determine moles of reactants:

42.1 g / 78g/mol = 0.540 moles of benzene

73 g / 159.80 g/mol = 0.457 moles of hydrogen bromide

Certainly, the HBr is the limiting reactant, because I need 0.540 moles of HBr for 0.540 moles of benzene and I only have, 0.457 moles.

0.457 moles of HBr will produce 0.457 of bromobenzene, at the 100 % yield reaction (theoretical yield)

We convert the moles to mass: 0.457 g . 157 g/mol = 71.7 g