# Calculate the mass of 2,4,6-trinitrophenol formed by the interaction of 250 g of a 15%

**Calculate the mass of 2,4,6-trinitrophenol formed by the interaction of 250 g of a 15% phenol solution with 300 g of a 25% nitric acid solution.**

Let’s write down given:

m solution (С6Н5ОН) = 250 g

w (C6H5OH) = 15% or 0.15

m solution (HNO3) = 300 g

w (HNO3) = 25% or 0.25

m (C6H2OH (NO2) 3) -?

Decision.

1) Let us write the equation for the reaction of the interaction of phenol and nitric acid:

С6Н5ОН + 3 HNO3 = C6H2OH (NO2) 3 + 3 Н2О

1 mol 3 mol

2) Calculate the masses of the substances participating in the reaction:

m (C6H5OH) = 250 * 0.15 = 37.5 g

m (HNO3) = 300 g * 0.25 = 75 g

3) Let’s calculate the amount of substance of each of the reacting substances:

M (C6H5OH) = 12 * 6 + 1 * 6 + 16 = 94 g / mol

n (С6Н5ОН) = 37.5 g / 94 g / mol = 0.399 mol

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

n (HNO3) = 75 g / 63 g / mol = 1.19 mol

4) Find a substance that is in excess:

According to the reaction equation, 1 mol (С6Н5ОН) interacts with 3 mol (HNO3),

so for 0.4 mol (C6H5OH) 1.2 mol (HNO3) is needed, and we have 1.19 mol of acid, therefore, phenol is given in excess and we make the calculation for the lack of nitric acid.

5) Calculate the mass of the formed 2,4,6 – trinitrophenol:

M (C6H2OH (NO2) 3) = 12 * 6 + 3 + 16 + 3 * (14 + 16 * 3) = 277 g / mol

3 * 63 g (HNO3) forms 277 g (C6H2OH (NO2) 3)

75 g (HNO3) forms x g (C6H2OH (NO2) 3)

X = 75g * 277g / 189g

X = 109.92 g (C6H2OH (NO2) 3)

Answer: 109.92 g (C6H2OH (NO2) 3)