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

Hydrogen Peroxide Can Be Prepared By The Reaction Of Barium Peroxide With Sulfuric Acid According To class=

Sagot :

Yipesid
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

Valenbracaid

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

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