What is the mass of the precipitate that was formed when an excess of magnesium chloride interacts with 300 g

What is the mass of the precipitate that was formed when an excess of magnesium chloride interacts with 300 g of sodium hydroxide solution with a mass fraction of a dissolved substance of 30%?

Given:
m solution (NaOH) = 300 g
w (NaOH) = 30% = 0.3
To find:
m (Mg (OH) 2)
Decision:
2NaOH + MgCl2 = 2NaCl + Mg (OH) 2
m in-islands (NaOH) = w * m solution = 0.3 * 300 g = 90 g
n (NaOH) = m / M = 90 g / 40 g / mol = 2.25 mol
n (NaOH): n (Mg (OH) 2) = 2: 1
n (Mg (OH) 2) = 2.25 mol / 2 = 1.125 mol
m (Mg (OH) 2) = n * M = 1.125 mol * 58 g / mol = 65.25 g
Answer: 65.25 g