The number of moles of $$KMn{O_4}$$  reduced by one mole of $$KI$$  in alkaline medium is

Solution :
In alkaline medium, $$KMn{O_4}$$  is reduced to $${K_2}Mn{O_4}$$
$$\eqalign{ & KI + {H_2}O \to KOH + HI \cr & \mathop {2KMn{O_4}}\limits^{ + 7} + 2KOH \to 2{K_2}\mathop {\mathop {Mn{O_4}}\limits^{ + 6} + {H_2}O + \left[ O \right]}\limits^{} \cr} $$
Hence, one mole of $$KMn{O_4}$$  is reduced by one mole of $$KI.$$