An excess of sodium hydroxide solution was added to a solution of ferric chloride 2 with a mass fraction of 5%

An excess of sodium hydroxide solution was added to a solution of ferric chloride 2 with a mass fraction of 5%, as a result of the reaction, a precipitate weighing 4.5 g was formed. Determine the mass of the initial salt solution.

Let us find the amount of the substance of the iron precipitate Fe (OH) 2.

n = m: M.

M (Fe (OH) 2) = 56 + 34 = 90 g / mol.

n = 4.5 g: 90 g / mol = 0.05 mol.

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

FeCl2 + 2NaOH = Fe (OH) 2 + 2NaCl.

For 1 mole of iron hydroxide, there is 1 mole of iron chloride. The substances are in quantitative ratios of 1: 1. The amount of the substance will be the same.

n (FeCl2) = n (Fe (OH) 2) = 0.05 mol.

Find the mass of FeCl2.

m = n M.

M (FeCl2) = 56 + 70 = 126 g / mol.

m = 0.05 mol × 126 g / mol = 6.3 g.

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

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

m (FeCl2 solution) = (6.3: 5%) × 100% = 126 g.