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Redox titrations are used to determine the amounts of oxidizing and reducing agents in solution. For example, a solution of hydrogen peroxide, H2O2, can be titrated against a solution of potassium permanganate, KMnO4. The following equation represents the reaction: 2KMnO4(aq) H2O2(aq) 3H2SO4(aq)→3O2(g) 2MnSO4(aq) K2SO4(aq) 4H2O(l) A certain amount of hydrogen peroxide was dissolved in 100. mL of water and then titrated with 1.68 M KMnO4. What mass of H2O2 was dissolved if the titration required 20.8 mL of the KMnO4 solution

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Answer: 0.595 g of H₂O₂ were dissolved.

Explanation:

1) Balanced chemical equation of the titration (given):

  • 2KMnO₄(aq) + H₂O₂(aq) + 3H₂SO₄(aq) →

                                                → 3O₂(g) + 2MnSO₄(aq)+  K₂SO₄(aq) + 4H₂O(l)

2) Mole ratios of KMnO₄(aq) and H₂O₂(aq) (the reactants the question deals with):

  • 2 moles 2KMnO₄(aq) : 1 mol H₂O₂(aq)

3) Number of moles of KMnO₄ used:

  • V = 20.8 mL = 0.0208 liter
  • M = 1.68 M

  • M = n of solute / V of solution in liter
  • ⇒ n of KMnO₄ = M × V = 1.68 M × 0.0208 liter = 0.0349 moles

4) Number of moles of H₂O₂ used:

  • Set the proportion using the stoichimetric ratio and the actual number of moles of KMnO₄:

       2 moles KMnO₄ / 1 mol H₂O₂ = 0.0349 moles KMnO₄ / x

       ⇒ x = 0.0349 / 2 moles H₂O₂ = 0.0175 moles H₂O₂

5) Mass of H₂O₂ dissolved:

  • molar mass H₂O₂ = 2(1.008g/mol) + 2(15.999 g/mol) = 34.014 g/mol
  • molar mass = mass in grams / number of moles ⇒
  • ⇒ mass in grams = molar mass × number of moles = 34.01 g/mol × 0.0175 moles = 0.595 g.

So, the answer, using the appropiate number of signficant digits (3) is 0.595 g of H₂O₂.

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