A) one fifth
B) five
C) one
D) two
Correct Answer: B
Solution :
In alkaline solution, KMn04 is reduced to\[\text{Mn}{{\text{O}}_{\text{2}}}\] (colourless). \[\begin{align} & \underline{\begin{align} & 2KMn{{O}_{4}}+2{{H}_{2}}O\xrightarrow{{}}Mn{{O}_{2}}+2KOH+3[O] \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,KI+3[O]\xrightarrow{{}}KI{{O}_{3}} \\ \end{align}} \\ & 2KMn{{O}_{4}}+2{{H}_{2}}O+KI\xrightarrow{{}}2Mn{{O}_{2}}+2KOH \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+KI{{O}_{3}} \\ \end{align}\] Hence, two moles of \[KMn{{O}_{4}}\]are reduced by one mole of KI.
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