Answer:
31.43 ml of sulfuric acid is needed to react with 23.1 g of barium peroxide.
Explanation:
For the reaction:
BaO₂(s) + H₂SO₄(aq) ⇒ BaSO₄(s) + H₂O₂(aq)
We need 1 mole of BaO₂ to react with 1 mole of H₂SO₄ and then forming 1 mole of BaSO₄ and 1 mole of H₂O₂.
We know that 4.5 M solution of H₂SO₄ means 4.5 moles of H₂SO₄ in 1000 ml of solution.
Then,
23.1 g BaO₂ × \(\frac{1 mole_{BaO_2}}{163.33g_{BaO_2}}\) × \(\frac{1 mole_{H_{2}SO_{4}}}{1mole_{BaO_2}}\) × \(\frac{1000ml}{4.5 mole_{H_{2}SO_{4}}}\) = 31.43 ml H₂SO₄
