Question

If $x$ moles of $KI$ are oxidized by number of moles of \[KI{O_3}\]formed when 1 mole of \[{I_2}\]is boiled with excess of\[KOH\], then what is the value of\[\frac{1}{x}\]?
\[6KOH{\text{ }} + {\text{ }}3{I_2} \to {\text{ }}5KI{\text{ }} + {\text{ }}KI{O_3} + {\text{ }}3{H_2}O\]
\[KI{\text{ }} + {\text{ }}KI{O_3}-{\text{ }}{I_2} + {\text{ }}{H_2}O\]
A.1
B.2
C.3
D.None of these

Answer 1

Hint: In this reaction, the 3 moles of iodine is forming 1 mole of \[KI{O_3}\] and in the question we have taken 1 mole of iodine and after this calculation, the number of moles of $KI$ which will be oxidized will be equal to number of moles of KIO3 present and this will finally give the answer.

Complete step by step answer:
In the given question, two reactions are occurring and we need to find the number of moles of oxidized by KIO3 and which could be found by using these two of the reactions.
In first reaction, we will firstly examine whether the reaction is balanced or not, and after looking number of elements on the reactants and the product side we found that reaction is balanced which is represented as: -
\[6KOH{\text{ }} + {\text{ }}3{I_2} \to {\text{ }}5KI{\text{ }} + {\text{ }}KI{O_3} + {\text{ }}3{H_2}O\]
So, from this reaction, we can say that 3 moles of Iodine i.e. \[{I_2}\] are forming 1 mole of \[KI{O_3}\] and in the question we have taken 1 mole of iodine initially with excess of\[KOH\]. So, the equation becomes: -
3 moles \[{I_2}\] gives = 1 mole \[KI{O_3}\]
1 mole \[{I_2}\] will give = \[\frac{1}{3}\]moles \[KI{O_3}\]
So, \[\frac{1}{3}\] moles of \[KI{O_3}\] will be produced when 1 mole of Iodine is reacted with excess of \[KOH\].
Now if we look at the second reaction, we found that 1 mole of \[KI\]are oxidized by 1 mole of \[KI{O_3}\] only and we can say \[KI\] will be oxidized equal to amount of \[KI{O_3}\] present.
So, we can say \[\frac{1}{3}\] moles \[KI{O_3}\] = \[\frac{1}{3}\] moles KI
(although the amount of KI formed is more i.e. \[\frac{5}{3}\] moles but it will be oxidized as much amount the \[KI{O_3}\] is present)
So, number of moles of KI oxidized by \[KI{O_3}\] = \[\frac{1}{3}\] which is equal to x
So, the value of \[\frac{1}{x}\] = 3

Therefore, the correct answer is C i.e. 3

Note:
This reaction in which iodine is generally used in titrations which are called as iodometric titrations and iodimetric titrations. Both are different types of titrations and the difference is making of Iodine gas. In one it is prepared in-situ and in another it is added directly.