Determine the mass of 2,4,6-trinitrophenol formed by the interaction of 9.2 g of phenol with 21 g of nitric acid.

Let’s execute the solution:

In accordance with the condition of the problem, we compose the equation of the process:
m = 9.2 g m = 21 g Xg -?

С6Н5ОН + 3HNO3 = C6H2 (NO2) 3OH + 3H2O – nitration, formed 2, 4, 6 trinitrophenol;

We make calculations:
M (C6H5OH) = 94 g / mol;

M (HNO3) = 63 g / mol;

M (trinitrophenol) = 229 g / mol;

Y (C6H5OH) = m / M = 9.2 / 94 = 0.09 mol;

Y (trinitrophenol) = 0.09 mol since the amount of substances is 1 mol.

Find the mass of the product:
m (trinitrophenol) = Y * M = 0.09 * 229 = 20.61 g

Answer: received trinitrophenol (picric acid) weighing 20.61 g