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The number of moles of $KMn{O_4}$reduced by one mole of $KI$in alkaline medium is
A.$\dfrac{1}{4}$
B.$2$
C.$\dfrac{3}{2}$
D.$4$

Answer
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Hint: At first think about the balanced reaction between $KMn{O_4}$ and $KI$ in the presence of an alkaline medium. In a balanced chemical reaction, the number of atoms on the right side should be equal to the number of atoms on the left side.


Complete answer:
According to the question, the reaction between $KMn{O_4}$ and $KI$ in the presence of alkaline medium without balancing is as follows,
$KMn{O_4} + KI \to Mn{O_2} + KI{O_3}$
In order to balance the above equation, we need to follow the steps required for balancing equations.
They are,
Step $1$:- Write the oxidation numbers of the atoms and separate the equation into two halves. The first half is for the reduction of oxidation number of atoms and the other half is for the oxidation of atoms.
 $KMn{O_4} + KI \to Mn{O_2} + KI{O_3}$
   $ + 7$ $ - 1$ $ + 4$ $ + 5$
Reduction half reaction:-$KMn{O_4} \to Mn{O_2}$
Oxidation half reaction:-$KI \to KI{O_3}$
Step $2$:-The change of oxidation number in, reduction half reaction is gain of three electrons and in oxidation half reaction is loss of six electrons.
Step $3$:- Balance the total change in oxidation number by multiplying the reduction reaction with two and oxidation reactions with one.
$2KMn{O_4} \to 2Mn{O_2}$
$KI \to KI{O_3}$


Step $4$:-Balance O atoms in reduction reaction by adding ${H_2}O$ and then balance H by ${H^ + }$
$
  2KMn{O_4} + 8{H^ + } \to 2Mn{O_2} + 4{H_2}O + 2{K^ + } \\
  KI + 3{H_2}O \to KI{O_3} + 6{H^ + } \\
 $
Step $5$:- Add $O{H^{^ - }}$ions on both sides to neutralize ${H^ + }$ ions.
The overall reaction after balancing ${H^ + }$ ions.
$2KMn{O_4} + 8{H^ + } + 8O{H^ - } \to 2Mn{O_2} + 8O{H^ - } + 2{K^ + } + 4{H_2}O$
The overall reaction is ,
$
  2KMn{O_4} + 4{H_2}O + KI + 6O{H^ - } \to 2Mn{O_2} + 8O{H^ - } + 2{K^ + } + KI{O_3} + 3{H_2}O \\
  2KMn{O_4} + {H_2}O + KI \to 2Mn{O_2} + 2KOH + KI{O_3} \\
 $
From the above reaction, we can say that two moles of $KMn{O_4}$ are reduced by one mole of $KI$ in an alkaline medium.

So, the answer is B.

Note:
-For balancing chemical equations you need to learn exactly how to calculate the oxidation numbers of atoms. Recheck while balancing oxygen atoms and water molecules because of one mistake your total answer will be incorrect.