Question

$KMnO_{4}$ reacts with $KI$ , in basic medium to form $I_{2}$ and $MnO_{2}$ . When $250\,ml$ of $0.1MKI$ solution is mixed with $250\,ml$ of $0.02MKMnO_{4}$ in basic medium, what are the number of moles of $I_{2}$ formed?

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

Answer 1

Answer: (B)

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.
$\therefore $ 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 $=\frac{15}{1000}=0.015$
$\therefore $ Number of moles of $\text{l}_{2}$ formed $=\frac{0.015}{2}$
$=0.0075$