What mass of bromobenzene is formed by the interaction of 7.8 g of benzene with an excess of bromine

What mass of bromobenzene is formed by the interaction of 7.8 g of benzene with an excess of bromine in the presence of FeBr3, the reaction product yield is 90%?

To solve, we will compose the equation of the process:
C6H6 + Br2 (FeBr3) = C6H5Br + HBr – substitutions, bromobenzene was formed;

Calculations according to the formulas of substances:
M (C6H6) = 78 g / mol; M (C6H5Br) = 156.9 g / mol.

Determine the amount of the original substance, if the mass is known:
Y (C6H6) = m / M = 7.8 / 78 = 0.1 mol.

Y (C6H5Br) = 0.1 mol since the amount of substances is 1 mol.

Find the mass of the product:
m (C6H5Br) = Y * M = 0.1 * 156.9 = 15.69 g (theoretical weight);

m (C6H5B r) = 0.90 * 15.69 = 14.12 g (practical weight).

Answer: bromobenzene was obtained, the mass of which, taking into account the product yield, is 14.12 g