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1 March, 03:25

How many milliliters of 0.021 moles of oxalic acid is needed to react with 80 mL of 0.11 moles of KMnO4

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  1. 1 March, 04:36
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    6.11 mL of H₂C₂O₄ are needed

    Explanation:

    We determine the reaction which is a redox one, in acidic medium

    C₂O₄⁻² + MnO₄⁻ → CO₂ + Mn²⁺

    C₂O₄⁻² → 2CO₂ + 2e⁻ Oxidation

    Carbon changes the oxidation state, from + 3 to + 4

    5e⁻ + MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O Reduction

    We add 4 water to the product side, in order to balance the oxygen and, we have 8H + in the reactant side, in order to balance the H

    Mn changes the oxidation state from + 7 to + 2

    (C₂O₄⁻² → 2CO₂ + 2e⁻).5

    5C₂O₄⁻² → 10CO₂ + 10e⁻

    (5e⁻ + MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O).2

    10e⁻ + 2MnO₄⁻ + 16H⁺ → 2Mn²⁺ + 8H₂O

    10e⁻ + 2MnO₄⁻ + 16H⁺ + 5C₂O₄⁻² → 2Mn²⁺ + 8H₂O + 10CO₂ + 10e⁻

    The electrons are cancelled, so the balanced reaction is:

    2KMnO₄ + 6HCl + 5H₂C₂O₄ → 2MnCl₂ + 8H₂O + 10CO₂ + 2KCl

    Concentration of KMnO₄ = 0.11 mol / 0.080mL = 1.375M

    Imagine that the reactants are in molar concentration (mol/L)

    Ratio in stoichiometry is 2:5

    2 moles of KMnO₄ react to 5 moles of oxalic acid

    Then, 1.375 moles of KMnO₄ will react to (1.375 moles. 5) / 2 = 3.437 M

    Concentration of H₂C₂O₄ = 3.437 M (mol/L)

    3.437 mol/L = 0.021 moles / Volume (L)

    0.021 moles / 3.437 mol/L = Volume (L) → 0.00611 L

    0.00611 L. 1000 mL / 1L = 6.11 mL
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