During the catalytic chlorination of benzene, by-products, ortho- and para-dichlorobenzenes

During the catalytic chlorination of benzene, by-products, ortho- and para-dichlorobenzenes, can be formed. calculate the mass fraction of dichlorobenzenes in the mixture after the reaction of benzene weighing 39 g with chlorine with a volume of 12.32 liters (NU). The total yield of the reaction products is conventionally assumed to be 100%.

m is the mass of the substance
v – amount of substance
Mr – molecular weight
V is the volume of the substance
22.4 – molar volume of gases

C6H6 + Cl2 = C6H5Cl + HCl

Mr (C6H6) = 78 g / mol

Using the formula v = m / Mr, we find the amount of benzene substance:

v (C6H6) = 39/78 = 0.5 mol.

Using the formula v = V / 22.4, we find the amount of chlorine:

v (Cl2) = 12.32 / 22.4 = 0.55 mol.

According to the reaction coefficients, the ratio of benzene and chlorine is 1: 1, which means that 0.5 mol of chlorobenzene will be formed and 0.05 mol of chlorine will remain. It is these 0.05 mol that will interact with the already formed chlorobenzene:

C6H5Cl + Cl2 = C6H4Cl2 + HCl

0.05 mol of chlorobenzene will enter into the reaction, forming 0.05 mol of dichlorobenzene.

Mr (C6H4Cl2) = 146 g / mol

m (C6H4Cl2) = Mr * v = 0.05 * 146 = 7.3 g

In the reaction mixture of chlorobenzene, 0.05 mol less will remain, since this part will react with chlorine: v (C6H5Cl) = 0.5 – 0.05 = 0.45 mol

Mr (C6H5Cl) = 112 g / mol

Total, m (C6H5Cl) = Mr * v = 0.45 * 112 = 50.4 g

Total mass of the final mixture m (C6H5Cl + C6H4Cl2) = 50.4 + 7.3 = 57.7 g

Mass fraction w is equal to the ratio of the mass of the substance to the total mass of the mixture.

w (C6H4Cl2) = 7.3 / 57.7 = 0.126 or 12.6%.