Determine the mass of the precipitate released by the interaction of 20 g of potassium sulfate containing 25%

Determine the mass of the precipitate released by the interaction of 20 g of potassium sulfate containing 25% impurities with an excess of sodium barium solution.

given:
m solution (K2SO4) = 20 g
w impurities (K2SO4) = 25% = 0.25
to find:
m (BaSO4)
decision:
K2SO4 + Ba (NO3) 2 = 2KNO3 + BaSO4
w of the pure substance (K2SO4) = 1-0.25 = 0.75
m pure substance (K2SO4) = 20 g * 0.75 = 15 g
n (K2SO4) = 15 g / 174 = 0.09 mol
n (K2SO4): n (BaSO4) = 1: 1
n (BaSO4) = 0.09 mol
m (BaSO4) = 0.09 mol * 233 = 20.97 g
Answer: 20.97