Calculate the mass of the precipitate that forms when 190 g of a 20% magnesium

Calculate the mass of the precipitate that forms when 190 g of a 20% magnesium chloride solution reacts with an excess of potassium hydroxide solution.

Given:
m solution (MgCl2) = 190 g
ω (MgCl2) = 20%

To find:
m (draft) -?

1) MgCl2 + 2KOH => Mg (OH) 2 ↓ + 2KCl;
2) m (MgCl2) = ω * m solution / 100% = 20% * 190/100% = 38 g;
3) M (MgCl2) = Mr (MgCl2) = Ar (Mg) * N (Mg) + Ar (Cl) * N (Cl) = 24 * 1 + 35.5 * 2 = 95 g / mol;
4) n (MgCl2) = m / M = 38/95 = 0.4 mol;
5) n (Mg (OH) 2) = n (MgCl2) = 0.4 mol;
6) M (Mg (OH) 2) = Mr (Mg (OH) 2) = Ar (Mg) * N (Mg) + Ar (O) * N (O) + Ar (H) * N (H) = 24 * 1 + 16 * 2 + 1 * 2 = 58 g / mol;
7) m (Mg (OH) 2) = n * M = 0.4 * 58 = 23.2 g.

Answer: The mass of Mg (OH) 2 is 23.2 g.