Calculate the mass of the precipitate formed when draining a sodium hydroxide solution weighing 80 g

Calculate the mass of the precipitate formed when draining a sodium hydroxide solution weighing 80 g with a mass fraction of alkali of 5.2% and an excess of magnesium chloride solution.

Dissolved sodium hydroxide mass:
80 * 0.052 = 4.16 (g);
Molar mass of NaOH:
23 + 16 + 1 = 40 (g / mol);
Find the number of mol of NaOH:
4.16 / 40 = 0.104 (mol);
Let’s write the reaction equation:
MgCl2 + 2NaOH = Mg (OH) 2 + 2NaCl – magnesium hydroxide precipitates.
Find the amount of precipitated Mg (OH) 2:
2 mol NaOH -> 1 mol Mg (OH) 2;
0.104 mol HCl -> x mol Mg (OH) 2;
x = 0.104 / 2 = 0.052 (mol).
Molar mass of magnesium hydroxide:
24 + 1 * 2 + 16 * 2 = 58 (g / mol);
The mass of the sediment formed:
0.052 * 58 = 3.016 (g).
Answer: 3.016 g.