How many moles of iodin are liberated when 2 moles of potassium permanganate rect with potassium iodide?

How many moles of iodin are liberated when 2 moles of potassium perman

1.	How many moles of iodin are liberated when 2 moles of potassium permanganate rect with potassium iodide?
Answer» Solution :Balanced redox RECTION is `2MnO_(4)^(-)+10I^(-)+16H^(+)+5E^(2)+8H_(2)O` Thus 2 moles of `KMnO_(4)` react with KI to libreate 5 Moles of `I_(2)`