Potassium sulfate was added to 200 grams of a 5% barium hydroxide solution. Calculate the mass of the precipitate formed.

Given:
m (Ba (OH) 2) = 200 g
w (Ba (OH) 2) = 5% = 0.05
To find:
m (BaSO4)
Decision:
Ba (OH) 2 + K2SO4 = BaSO4 + 2KOH
m in-islands (Ba (OH) 2) = m solution * w = 200 g * 0.05 = 10 g
n (Ba (OH) 2) = m / M = 10 g / 171 g / mol = 0.06 mol
n (Ba (OH) 2): n (BaSO4) = 1: 1
n (BaSO4) = 0.06 mol
m (BaSO4) = n * M = 0.06 mol * 233 g / mol = 13.98 g
Answer: 13.98 g