What mass of barium sulfate is formed when 76.5 grams of barium oxide reacts with a sufficient amount of sulfuric acid?

univerkov | January 17, 2021 | education

Given:
m (BaO) = 76.5 g
To find:
m (BaSO4)
Decision:
BaO + H2SO4 = BaSO4 + H2O
n (BaO) = m / M = 76.5 g / 153 g / mol = 0.5 mol
n (BaO): n (BaSO4) = 1: 1
n (BaSO4) = 0.5 mol
m (BaSO4) = n * M = 0.5 mol * 233 g / mol = 116.5 g
Answer: 116.5 g

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