What mass of barium sulfate is formed when 30.6 grams of barium oxide reacts with a sufficient amount of sulfuric acid?

1. We make a chemical equation, not forgetting to put down the coefficients:

BaO + H2SO4 = BaSO4 + H2O.

2. Find the molar mass of BaO:

M (BaO) = 137 + 16 = 153g \ mol.

3. Find the amount of substance BaO:

n (BaO) = m (BaO) / M (BaO);

n (BaO) = 30.6 / 153 = 0.2 mol.

4. We make up the proportion:

n (BaО) / n (BaSO4) = 1/1.

5. Find the molar mass of BaSO4:

M (BaSO4) = 137 + 32 + 16 * 4 = 233g / mol.

6. Find the amount of substance BaSO4:

m (baSO4) = M (baSO4) * n (baSO4) = 233 * 0.2 = 46.6g.

Answer: m (BaSO4) = 46.6g.



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