Find the mass of the precipitate formed by the interaction of 50 grams of sodium sulfate with an excess of barium chloride.

univerkov | January 17, 2021 | education

Given:
m (Na2SO4) = 50 g
To find:
m (BaSO4)
Decision:
Na2SO4 + BaCl2 = 2NaCl + BaSO4
n (Na2SO4) = m / M = 50 g / 142 g / mol = 0.35 mol
n (Na2SO4): n (BaSO4) = 1: 1
n (BaSO4) = 0.35 mol
m (BaSO4) = n * M = 0.35 mol * 233 g / mol = 81.55 g
Answer: 81.55 g

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