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