How much benzene was consumed if 105 g of nitrobenzene were obtained at 80% yield?

To solve, we write:
X g -? m = 105 g; W = 80%
1. С6Н6 + HNO3 = C6H5NO2 + H2O – substitutions, nitrobenzene was formed;
1 mol 1 mol.
2. Let’s make calculations according to the formulas of substances:
M (C6H6) = 78 g / mol.
M (C6H5NO2) = 123 g / mol.
3. Calculations:
W = m (practical) / m (theoretical) * 100;
m (C6H6) = 105 / 0.80 = 131.25 g (theoretical weight).
Y (C6H5NO2) = m / M = 131.25 / 123 = 1 mol.
Y (C6H6) = 1 mol since the amount of substances according to the equation is 1 mol.
m (C6H6) = Y * M = 1 * 78 = 78 g.
Answer: to carry out nitration, benzene weighing 78 g is required.