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How much \mathrm{KMnO_{4}} is required to react with equimolar solution of \mathrm{KI \; and\; KBr} in acidic medium?

Option: 1

1/5


Option: 2

2/5


Option: 3

3/5


Option: 4

7/5


Answers (1)

best_answer

Reaction of \mathrm{KMnO_{4}} with KI is : 2 \mathrm{KMnO}_4+10 \mathrm{KI}+8 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow 2 \mathrm{MnSO}_4+6 \mathrm{~K}_2 \mathrm{SO}_4+5 \mathrm{I}_2

Reaction of \mathrm{KMnO_{4}} with KBr is :2 \mathrm{KMnO}_4+10 \mathrm{KBr}+8 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow 2 \mathrm{MnSO}_4+6 \mathrm{~K}_2 \mathrm{SO}_4+5 \mathrm{Br}_2

So, 10 moles of KI require 2 moles of \mathrm{KMnO_{4}} for the reaction \Rightarrow 1 mole of KI requires \frac{1}{5} mole of \mathrm{KMnO_{4}}.

Similarly, 1 mole of KBr requires \frac{1}{5} mole of \mathrm{KMnO_{4}}

So, for a mixture of 1 mole of KI and 1 mole of KBr requires \frac{1}{5}+\frac{1}{5}=\frac{2}{5} moles of \mathrm{KMnO_{4}}

Posted by

vishal kumar

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