120.0 g of sulfuric acid are added to 230.0 g of barium peroxide. Which reactant is in excess and by how much? How much barium sulfate is precipitated in this reaction?

Answer 1

H₂SO₄(aq) + BaO₂ ==> BaSO₄ + H₂O₂

Find limiting reactant:

moles H₂SO₄ = 120.0 g x 1 mol/98 g = 1.22 moles

moles BaO₂ = 230.0 g x 1 mol/169 g = 1.36 moles

Because they react in a 1:1 mol ratio, the H₂SO₄ is limiting, leaving BaO₂ as the excess reactant

The amount of excess BaO₂ = 1.36 mol - 1.22 mol = 0.141 moles = 23.8 g excess BaO₂

Mass BaSO₄ formed = 1.22 mol H₂SO₄ x 1 mol BaSO₄/mol H₂SO₄ x 233 g/mol = 284.3 g BaSO₄