What is the mass of the precipitate formed by mixing 180 g of a 15% solution of lithium hydroxide

What is the mass of the precipitate formed by mixing 180 g of a 15% solution of lithium hydroxide with a solution containing 0.3 mol of iron sulfate (3), if the practical yield of the precipitate is 78%?

Given:
m solution (LiOH) = 180 g
ω (LiOH) = 15%
n (Fe2 (SO4) 3) = 0.3 mol
η (sediment) = 78%

To find:
m (draft) -?

1) 6LiOH + Fe2 (SO4) 3 => 3Li2SO4 + 2Fe (OH) 3 ↓;
2) m (LiOH) = ω * m solution / 100% = 15% * 180/100% = 27 g;
3) n (LiOH) = m / M = 27/24 = 1.13 mol;
4) n theory. (Fe (OH) 3) = n (LiOH) * 2/6 = 1.13 * 2/6 = 0.38 mol;
5) n practical (Fe (OH) 3) = η * n theory. / 100% = 78% * 0.38 / 100% = 0.3 mol;
6) m practical. (Fe (OH) 3) = n practical. * M = 0.3 * 107 = 32.1 g.

Answer: The mass of Fe (OH) 3 is 32.1 g.