`KMnO_(4)` reacts with KI,in basic medium to form `I_(2)` and `MnO_(2)`When 250 mL of 0.1M KI solution is mixed with 250 mL of 0.02 M `KMnO_(4)`, in basic medium, what is the number of moles of `I_(2)` formed?

Options

A. 0.015
B. 0.0075
C. 0.005
D. 0.01

Answer

(b)

Explanation

`2KMnO_(4)+ 6KI + 4H_(2)0 rarr 2MnO_(2) + 3I_(2) + 8KOH`
In the above reaction, `KMnO_(4)` acts as oxidising agent and oxidises KI to `I_(2)`. Therefore, number of moles of `I_(2)`formed depends upon number of moles of`KMnO_(4)` used.
`Molarity = W/Mxx 1000/V`(in ml)No. of moles of `KMnO_(4)` = `Molarity xx V/1000 =0.005`
For 2 moles of KMNO,----3 moles of KI are formed
0.005 moles of `KMnO_(4)`-----`0.005 xx 3/2 = 0.0075`
`:.`0.0075 moles of KI are formed