Find the mass of the precipitate that forms when 95 g of 40% NAOH interacts with a Fe2 (So4) 3 solution.

Given:
m s (NaOH) = 95 g
w (NaOH) = 40% = 0.4
To find:
m (Fe (OH) 3)
Decision:
6NaOH + Fe2 (SO4) 3 = 2Fe (OH) 3 + 3Na2SO4
m in (NaOH) = w * m c = 0.4 * 95 = 38 g
n (NaOH) = m / M = 38 g / 40 g / mol = 0.95 mol
n (NaOH): n (Fe2 (SO4) 3 = 6: 1
n (Fe2 (SO4) 3) = 0.95 / 6 = 0.16 mol
m (Fe2 (SO4) 3) = n * M = 0.16 mol * 400 g / mol = 64 g
Answer: 64 g