# 117 g of benzene are reacted with nitric acid to obtain nitrobenzene m (166) g.

**117 g of benzene are reacted with nitric acid to obtain nitrobenzene m (166) g. Calculate the yield of nitrobenzene from the theoretically possible.**

Given:

m (C6H6) = 117 g

m pract. (C6H5NO2) = 166 g

Find:

η (C6H5NO2) -?

Solution:

1) C6H6 + HNO3 => C6H5NO2 + H2O;

2) M (C6H6) = Mr (C6H6) = Ar (C) * N (C) + Ar (H) * N (H) = 12 * 6 + 1 * 6 = 78 g / mol;

3) n (C6H6) = m (C6H6) / M (C6H6) = 117/78 = 1.5 mol;

4) n theory. (C6H5NO2) = n (C6H6) = 1.5 mol;

5) M (C6H5NO2) = Mr (C6H5NO2) = Ar (C) * N (C) + Ar (H) * N (H) + Ar (N) * N (N) + Ar (O) * N (O ) = 12 * 6 + 1 * 5 + 14 * 1 + 16 * 2 = 123 g / mol;

6) m theor. (C6H5NO2) = n theory. (C6H5NO2) * M (C6H5NO2) = 1.5 * 123 = 184.5 g;

7) η (C6H5NO2) = m practical. (C6H5NO2) * 100% / m theor. (C6H5NO2) = 166 * 100% / 184.5 = 90%.

Answer: The yield of C6H5NO2 is 90%.