The number of moles of `KMnO_(4)` reduced by `1 "mol of" KI` in alkaline medium is

Options

A. one
B. two
C. five
D. one-fifthe

Answer

B

Explanation

In alkaline medium, ` KMnO_4` oxidixes iodide to iodate and gets reduced to manganese dioxide :
`overset(+7)(M)nO_(4)^(-) rarr overset(+4)(M)n O_(2)`
`overset(-1)(I^(-)) rarr overset(+5)(I)O_(3)^(-)`
Thus, 1 mol of `KMnO_4` contains `3` equivalents (because its oxidation number decreases nu `3` unts : ` +7 rarr +4`). On the other hand, 1 mol of `KI` contains `6` equivalents (because oxidation number of iodine increases ny `6` units, ` -1 rarr +5`) According to the law of equivalents, we need `6` equivalents (2 mol) of `KMnO_4` to react with `6` requivalents (1 mol) of ` KI`. Alternatively, we can get the answer by means fo chemical equations :
`{:(2KMnO(4)+H_(2)O rarr 2KOH+2MnO_(2)+3[O]),(KI+3[O] rarr KIO_(3)),(bar(2KMnO_(4) +KI +H_(2)O rarr 2KOH +2MnO_(2) +KIO_(3))):}`