The number of moles of `KMnO_(4)` reduced by `1 "mol of" KI` in alkaline medium is (a)1 / 5 (b)2 (c)3 / 2 (d)4

Options

A. `1//5`
B. 2
C. `3//2`
D. 4

Explanation

To solve the question regarding the number of moles of \( KMnO_4 \) reduced by \( 1 \) mole of \( KI \) in alkaline medium, we can follow these steps:

### Step 1: Write the balanced chemical equation
In alkaline medium, potassium permanganate (\( KMnO_4 \)) reacts with potassium iodide (\( KI \)). The reaction can be represented as follows:

\[ KMnO_4 + KI + KOH \rightarrow MnO_2 + KIO_3 + H_2O \]

### Step 2: Identify the oxidation states
In this reaction:
- Manganese in \( KMnO_4 \) is in the +7 oxidation state and is reduced to +4 in \( MnO_2 \).
- Iodine in \( KI \) is in the -1 oxidation state and is oxidized to +5 in \( KIO_3 \).

### Step 3: Balance the equation
To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides. The balanced equation is:

\[ 2 KMnO_4 + 10 KI + 8 KOH \rightarrow 2 MnO_2 + 5 KIO_3 + 8 H_2O \]

### Step 4: Determine the moles of \( KMnO_4 \) reduced
From the balanced equation, we can see that:
- 10 moles of \( KI \) reduce 2 moles of \( KMnO_4 \).

To find out how many moles of \( KMnO_4 \) are reduced by 1 mole of \( KI \):

\[ \text{Moles of } KMnO_4 \text{ reduced by } 1 \text{ mole of } KI = \frac{2 \text{ moles of } KMnO_4}{10 \text{ moles of } KI} = \frac{1}{5} \text{ moles of } KMnO_4 \]

### Conclusion
Thus, the number of moles of \( KMnO_4 \) reduced by \( 1 \) mole of \( KI \) in alkaline medium is \( \frac{1}{5} \).

### Final Answer
(a) \( \frac{1}{5} \)

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