# Determine the mass of the sediment formed by the interaction of 200 g of 40% sodium hydroxide solution

**Determine the mass of the sediment formed by the interaction of 200 g of 40% sodium hydroxide solution with 50 g of iron susphate 2.**

Find the mass of sodium hydroxide in the solution.

The mass fraction of a substance is calculated by the formula:

W = m (substance): m (solution) × 100%,

m (substance) = (m (solution) × W): 100%,

m (substance) = (200 g × 40%): 100% = 80 g.

Find the amount of sodium hydroxide substance by the formula:

M (NaOH) = 40 g / mol.

n = m: M.

n = 80 g: 40 g / mol = 2 mol.

M (FeSO4) = 150 g / mol.

n (FeSO4) = 50 g: 150 g / mol = 0.33 mol.

NaOH is in excess, we solve the problem due to the lack of FeSO4.

Let’s compose the reaction equation, find the quantitative ratios of substances.

2NaOH + FeSO4 = Fe (OH) 2 ↓ + Na2SO4.

According to the reaction equation, there is 1 mole of iron (II) hydroxide per 1 mole of iron sulfate.

Substances are in quantitative ratios 1: 1.

The amount of substance will be the same.

n (FeSO4) = n (Fe (OH) 2) = 0.33 mol.

Let’s find the mass of the sediment by the formula:

m = n × M,

M (Fe (OH) 2) = 90 g / mol.

m = 0.33 mol × 90 g / mol = 29.7 g.

Answer: 29.7 g.