Hydrogen peroxide can be prepared by the reaction of barium peroxide with sulfuric acid according to

BaO₂ + H₂SO₄ ⇒ BaSO₄ + H₂O₂ 169g : 98g 23.1g : mx
mx = [23.1*98g]/169g ≈ 13.4g Mr = 98g/mol
n = 13.4g/98g/mol ≈ 0.14mol
Cm = 4.5M n = 0.14mol
V = 0.14mol/4.5mol/dm³ ≈ 0.031L = 31mL

Answered by Yipes | 2024-06-10

**31.43 ml **of sulfuric acid is needed to react with 23.1 g of barium peroxide. ;

Answered by valenbraca | 2024-06-12

To prepare hydrogen peroxide from barium peroxide and sulfuric acid, you need to follow a stoichiometric approach. For 23.1 grams of barium peroxide, approximately 30.2 mL of sulfuric acid is required. The reaction follows the balanced equation BaO₂ + H₂SO₄ → BaSO₄ + H₂O₂.
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Answered by valenbraca | 2024-12-26