Question

$KMn{O}_4$ reacts with $KI$ , in basic medium to form $I_2$ and $Mn{O}_2$ . When $250ml$ of $0.1MKI$ solution is mixed with $250ml$ of $0.02MKMn{O}_4$ 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 $=\frac{0.015}{2}$
$=0.0075$