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
