How many grams of nitrobenzene can be obtained from 312 g of benzene if the mass fraction of the output is 75%?

In accordance with the condition of the problem, we write the equation:
С6Н6 + HNO3 = C6H5NO2 + H2O – the reaction of benzene nitration, nitrobenzene was obtained;
Let’s make the calculations:
M (C6H6) = 78 g / mol;
M (C6H5NO2) = 123 g / mol.
Let’s calculate the amount of benzene mol by the formula:
Y (C6H6) = m / M = 312/78 = 4 mol.
According to the equation, the amount of moles of benzene and the reaction product of nitrobenzene are equal to 1 mole, which means that Y (C6H5NO2) = 4 moles.
Find the theoretical mass С6Н5NO2:
m (C6H5NO2) = Y * M = 4 * 123 = 492 g.
practical weight:
W = m (practical) / m (theoretical) * 100;
m (practical) = 0.75 * 492 = 369 g.
Answer: the mass of С6Н5NO2 is 369 g.