KMn O_ 4 reacts with KI , in basic medium to form I_ 2 and Mn O_ 2...

Number of mtextlli equivalents of \text{M}\text{n}{\text{O}_{4}^{-} = 0.02 \times 3 \times 250 = 15

Number of mtextlli equivalents of {\text{l}_{2} = 0.1 \times 1 \times 250 = 25

Thus, here \text{M}\text{n}{\text{O}_{4}^{-} is limtextting reagent.

$∴$ Number of mtextlli equivalents of {\text{l}_{2} formed = Number of mtextlli equivalent of \text{M}\text{n}{\text{O}_{4}^{-} = 15

Or number of equivalent of {\text{l}_{2} formed $= \dfrac{15}{1000} = 0.015$

$∴$ Number of moles of {\text{l}_{2} formed $= \dfrac{0.015}{2}$

$= 0.0075$