Find the mass of the precipitate that is formed by the interaction of 2 mol FeCl of ferric chloride (3) with sodium hydroxide.
September 18, 2021 | education
| 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.
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