Reduction of 123 g of nitrobenzene gave 88 g of aniline. calculate the yield of the product as a percentage

Reduction of 123 g of nitrobenzene gave 88 g of aniline. calculate the yield of the product as a percentage of the theoretically possible.

1. Let us draw up a scheme for the reduction of nitrobenzene to aniline:

C6H5NO2 + 3H2 → C6H5NH2 + 2H2O.

2. Let’s calculate the chemical amount of nitrobenzene:

n (C6H5NO2) = m (C6H5NO2): M (C6H5NO2);

M (C6H5NO2) = 6 * 12 + 5 + 14 + 2 * 16 = 123 g / mol;

n (C6H5NO2) = 123: 123 = 1 mol.

3. Determine the theoretical amount of aniline:

ntheor (С6H5NH2) = n (C6H5NO2) = 1 mol.

4. Let us calculate the theoretical mass of the obtained aniline:

mtheor (C6H5NH2) = ntheor (C6H5NH2) * M (C6H5NH2);

M (C6H5NH2) = 12 * 6 + 5 + 14 + 2 = 93 g / mol;

mtheor (C6H5NH2) = 1 * 93 = 93 g.

5. Let’s find the output of aniline:

ν = mpract (С6H5NH2): mtheor (С6H5NH2) = 88: 93 = 0.9462 or 94.62%.

Answer: 94.62%.