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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+O2In 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 1: Assign oxidation numbers to each of the atoms in the equation and write the numbers above the atom as:K+7MnO4+H2−1O2→+2Mn2++0O2Step 2: Identify the atoms that are oxidized and those that are reduced as:Reduction: K+7MnO4→+2Mn2+Oxidation: H2−1O2→0O2Step 3: oxidation-number change is:Reduction: : K+7MnO4→+2Mn2+: Gain of 5 electrons totalOxidation: H2−1O2→0O2: Loss of total 2 electronsStep 4: Balance the total change in oxidation number as:Reduction: : K+7MnO4→+2Mn2+×2: Gain of 10 electrons Oxidation: : H2−1O2→0O2×5: Loss of total 10 electronsORReduction: : 2K+7MnO4→2+2Mn2+Oxidation: : 5H2−1O2→50O2Step 5: Balance K, O atoms in reduction reaction by adding H2O and then balance H by H+ as:Reduction: : 2K+7MnO4+16H+→2+2Mn2++2K++8H2OOxidation: : 5H2−1O2→50O2+10H+Thus overall Balanced reaction is:2K+7MnO4+16H++5H2−1O2→2+2Mn2++2K++8H2O+50O2+10H+cancel H+ from both sides:2KMnO4+6H++5H2O2→2Mn2++2K++8H2O+5O2Thus 2 mol of KMnO4 require 5 mol of H2O2 for decolorization. Therefore for decolourisation of 1 mole of acidified KMnO4 the moles of H2O2 required are 5/2.

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