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

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valenbraca valenbraca · Answer 2 · 2019-10-21T19:47:43-04:00

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 × [tex]\frac{1 mole_{BaO_2}}{163.33g_{BaO_2}}[/tex] × [tex]\frac{1 mole_{H_{2}SO_{4}}}{1mole_{BaO_2}}[/tex] × [tex]\frac{1000ml}{4.5 mole_{H_{2}SO_{4}}}[/tex] = 31.43 ml H₂SO₄