Find the mass of the precipitate that is formed by the interaction of 2 mol FeCl of ferric chloride (3) with sodium hydroxide.

Let us find the amount of the substance of iron (III) chloride.

n = m: M.

M (FeCl3) = 56 + 35 × 3 = 161 g / mol.

n = 214 g: 90 g / mol = 2.38 mol.

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

FeCl3 + 3NaOH = Fe (OH) 3 ↓ + 3NaCl.

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

n (FeCl3) = n (Fe (OH) 3) = 2 mol.

Let’s find the mass of the sediment.

m = n M.

M (Fe (OH) 3) = 56 + 3 × (16 + 1) = 107 g / mol.

m = 2 mol × 107 g / mol = 214 g.

Answer:; m = 214 g.