# Find the mass of the precipitate if 267 g of aluminum chloride and 20 g of sodium hydroxide were taken for the reaction.

First, write down the reaction equation. We get the following.
AlCl3 + 3NaOH = 3NaCl + Al (OH) 3.
First, we find the amount of the substance of aluminum chloride and sodium hydroxide. For this we use the following formula.
n = m / M.
M (AlCl3) = 27 + 35.5 × 3 = 27 + 106.5 = 133.5 g / mol.
M (NaOH) = 23 + 16 + 1 = 40 g / mol.
n (AlCl3) = 267 g / 133.5 g / mol = 2 mol.
n (NaOH) = 20 g / 40 g / mol = 0.5 mol.
Since there is less sodium hydroxide, it means that we calculate the sediment mass by sodium hydroxide.
0.5 mol NaOH – x mol Al (OH) 3.
3 mol NaOH – 1 mol Al (OH) 3.
Find the unknown value of x.
X = 0.5 mol × 1 mol ÷ 3 mol = 0.167 mol.
Next, we find the mass of the sediment.
M (Al (OH) 3) = 27 + 3 + 3 × 16 = 78 g / mol.
m = n × M = 0.167 mol × 78 g / mol = 13 g.
This means that the mass is 13 g.
Answer: m (Al (OH) 3) = 13 g.

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