Dry hydrogen chloride was passed through a mixture of aniline, benzene and phenol with a total weight of 95.4 g.

Dry hydrogen chloride was passed through a mixture of aniline, benzene and phenol with a total weight of 95.4 g. This formed 51.8 g of a precipitate, which was filtered off. The filtrate was treated with an excess of bromine water to obtain 99.3 g of precipitate. Calculate the mass of benzene (in grams) in the original mixture.

Only aniline will react with hydrogen chloride from the entire mixture (aniline, benzene, phenol), forming phenylammonium chloride with a mass of 51.8 grams. The molecular weight of the resulting compound is 129.6 g / mol, which means that the amount of substance (according to the formula v = m (mass) / Mr (molecular weight)) of phenylammonium chloride will be v = 51.8 / 129.6 = 0.4 mol. Aniline will have the same amount of substance (0.4 mol), since its ratio with the final product is 1: 1. This means that the mass of aniline is m = v * Mr = 0.4 * 93 = 37.2 g.

The precipitate was removed, benzene and phenol remained in the filtrate. With an excess of bromine water, only phenol will react, this is a qualitative reaction to it, 2,4,6-tribromophenol is formed with a mass of 99.3 g. The amount of substance of this compound is v = 99.3 / 330.8 = 0.3 mol. The amount of phenol substance will also be 0.3 mol. Phenol mass m = 0.3 * 94 = 28.2 g.

To find the mass of benzene, it is necessary to subtract from the total mass of the mixture the masses of phenol and aniline that have entered into the reaction: m benzene = 95.4-37.2-28.2 = 30 g.