195 g of benzene containing 20% of impurities was treated with a solution of nitric acid weighing 200 g

195 g of benzene containing 20% of impurities was treated with a solution of nitric acid weighing 200 g with a mass fraction of acid of 70%, received 196.8 g of nitrobenzene. Determine the proportion of the output of nitrobenzene?

Let’s compose the reaction equation.

C6H6 + HNO3 = C6H5NO2 + H2O.

Let us determine the mass of pure benzene and the molar mass.

m (C6H6) = 195 * 0.8 = 156 g.

M (C6H6) = 78 g / mol.

Determine the mass and molar mass of nitric acid.

m (HNO3) = 200 * 0.7 = 140 g.

M = 63 g / mol.

Determine which substance is in excess.

156 g – x g.

78 g / mol – 63 g / mol.

X = 126 g.

Calculated by nitric acid.

M (C6H5NO2) = 123 g / mol.

140 g – at g.

63 g / mol – 123 g / mol.

Y = 273 g.

W = 196.8 / 273 = 72%.

Answer: the yield of nitrobenzene is 72 percent.