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Question

# For decolourisation of 1 mole of acidified KMnO4 the moles of H2O2 required are:

A
1/2
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B
3/2
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C
5/2
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D
7/2
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Solution

## The correct option is C 5/2Reaction involved is: KMnO4+H2O2+H+→K++Mn2++H2O+O2 In a balanced chemical reaction number of all atoms at right side should be equal to the left side of the reaction. Balancing a chemical reaction as: Step I: Assign oxidation numbers to each of the atoms in the equation and write the numbers above the atoms as: +7KMnO4+−1H2O2→+2Mn2++0O2 Step 2: Identify the atoms that are oxidized and those that are reduced as: Reduction: +7KMnO4→+2Mn2+ Oxidation: −1H2O2→0O2 Step 3: oxidation-number change is: Reduction: +7KMnO4→+2Mn2+: Gain of 5 electrons total Oxidation: −1H2O2→0O2: Loss of total 2 electrons Step 4: Balance the total change in oxidation number as: Reduction: +7KMnO4→+2Mn2+×2: Gain of 10 electrons Oxidation: :−1H2O2→0O2×5: Loss of total 10 electrons OR Reduction: :+72KMnO4→+22Mn2+ Oxidation: :−15H2O2→02O2 Thus overall Balanced reaction is: +72KMnO4+16H++−15H2O2→+22Mn2++2K++8H2O+05O2+10H+ cancel H+ from both sides: 2KMnO4+6H++5H2O2→2Mn2++2K++8H2O+5O2 Thus 2 mol of KMnO4 require 5 mol of H2O2 for decolorization. Therefore for decolourisaiton of 1 mole of acidified KMnO4 the moles of H2O2 required are 5/2.

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