HNO3 weighing 63 grams was added to benzene weighing 39 grams. Find the weight of the product?

1. The nitration reaction proceeds in the presence of sulfuric acid according to the scheme:

C6H6 + HNO3 → C6H5NO2 + H2O;

2.n (C6H6) = m (C6H6): M (C6H6);

M (C6H6) = 6 * 12 + 6 * 1 = 78 g / mol;

n (C6H6) = 39: 78 = 0.5 mol;

3.n (HNO3) = m (HNO3): M (HNO3);

M (HNO3) = 1 + 14 + 3 * 16 = 63 g / mol;

n (HNO3) = 63: 63 = 1 mol;

4.n (C6H6) is less than n (HNO3), therefore, the calculation is carried out according to the drawback:

M (C6H5NO2) = 6 * 12 + 5 * 1 + 14 + 2 * 16 = 103 g / mol;

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

m (C6H5NO2) = 0.5 * 103 = 51.5 g.

Answer: 51.5 g.