How many mole of KI are oxidised by no. of mole of KIO₃ formed when 1 mole of I₂ is boiled with excess of KOH ??

KOH + I₂ → KI + KIO₃ + H₂O

KI + KIO₃ → H⁺→ I₂ + H₂O

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To determine how many moles of potassium iodide (KI) are oxidized by the number of moles of potassium iodate (KIO3) formed when 1 mole of iodine (I2) reacts with excess potassium hydroxide (KOH), we need to analyze the given chemical reactions step by step.

The Reactions Involved

We have two key reactions to consider:

• First Reaction: KOH + I2 → KI + KIO3 + H2O
• Second Reaction: KI + KIO3 → H+ → I2 + H2O

Analyzing the First Reaction

In the first reaction, 1 mole of I2 reacts with KOH to produce KI and KIO3. The stoichiometry of this reaction indicates that for every mole of I2, we produce 1 mole of KI and 1 mole of KIO3. Therefore, if we start with 1 mole of I2, we will generate:

• 1 mole of KI
• 1 mole of KIO3

Understanding the Second Reaction

In the second reaction, KI reacts with KIO3. Here, KI acts as a reducing agent, and KIO3 is reduced to I2. The stoichiometry of this reaction shows that 1 mole of KI is oxidized for every mole of KIO3 consumed. Since we produced 1 mole of KIO3 from the first reaction, it will require 1 mole of KI to be oxidized.

Final Calculation

From the analysis of both reactions, we can conclude that:

• 1 mole of I2 produces 1 mole of KIO3.
• 1 mole of KIO3 requires 1 mole of KI to be oxidized.

Thus, when 1 mole of I2 is boiled with excess KOH, it results in the oxidation of 1 mole of KI. Therefore, the answer to your question is that 1 mole of KI is oxidized by the number of moles of KIO3 formed.

Summary

In summary, the stoichiometric relationships in the reactions show a direct correlation between the moles of I2, KI, and KIO3. This clear relationship allows us to confidently state that 1 mole of KI is oxidized for every mole of KIO3 produced when 1 mole of I2 reacts with excess KOH.