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The number of moles of KMnO4 reduced by one mole of KI in alkaline medium is :

  1. A
    one fifth
  2. B
    five
  3. C
    one
  4. D
    two

Solution & Step-by-step Explanation

The reaction between KMnO4 and KI in an alkaline medium can be represented by the equation: 2KMnO4 + KI + H2O → 2MnO2 + KIO3 + 2KOH. However, to balance the equation and determine the stoichiometry, we consider the oxidation states. KMnO4 is reduced to MnO2, and KI is oxidized to KIO3. The balanced equation considering the oxidation and reduction is: 2KMnO4 + KI + H2O → 2MnO2 + KIO3 + 2KOH. For every mole of KI oxidized to KIO3, 2 moles of KMnO4 are reduced. However, the exact stoichiometry can depend on the specific reaction conditions and the products formed. In general, for the reduction of KMnO4 to MnO2, the number of moles of KMnO4 reduced per mole of KI can vary based on the reaction. The correct answer depends on the balanced chemical equation for the specific reaction conditions.

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The number of moles of KMnO4 reduced by one mole of KI in alkaline medium is :
A
one fifth
B
five
C
one
D
two

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Inorganic Chemistry

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