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13 July, 16:28

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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  1. 13 July, 16:56
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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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