`KMnO_(4)` reacts with KI in basic medium to from `I_(2)` and `MnO_(2)`. When 250 mL of 0.1 M KI solution is mixed with 250 mL of 0.02 M `KMnO_(4)` in basic medium, what is the number of moles `I_(2)` formed ?

Options

A. 0.0075
B. 0.005
C. 0.01
D. 0.015

Explanation

To solve the problem, we need to determine the number of moles of \( I_2 \) formed when \( KMnO_4 \) reacts with \( KI \) in a basic medium. We will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between \( KMnO_4 \) and \( KI \) in basic medium can be written as:
\[ 2 KMnO_4 + 6 KI \rightarrow 2 MnO_2 + 3 I_2 + 4 K_2O \]

### Step 2: Calculate the moles of \( KI \)
We are given:
- Volume of \( KI \) solution = 250 mL
- Molarity of \( KI \) = 0.1 M

To find the moles of \( KI \):
\[ \text{Moles of } KI = \text{Molarity} \times \text{Volume (in L)} \]
Convert 250 mL to liters:
\[ 250 \, \text{mL} = 0.250 \, \text{L} \]
Now calculate the moles:
\[ \text{Moles of } KI = 0.1 \, \text{M} \times 0.250 \, \text{L} = 0.025 \, \text{moles} \]

### Step 3: Calculate the moles of \( KMnO_4 \)
We are given:
- Volume of \( KMnO_4 \) solution = 250 mL
- Molarity of \( KMnO_4 \) = 0.02 M

To find the moles of \( KMnO_4 \):
\[ \text{Moles of } KMnO_4 = \text{Molarity} \times \text{Volume (in L)} \]
Convert 250 mL to liters:
\[ 250 \, \text{mL} = 0.250 \, \text{L} \]
Now calculate the moles:
\[ \text{Moles of } KMnO_4 = 0.02 \, \text{M} \times 0.250 \, \text{L} = 0.005 \, \text{moles} \]

### Step 4: Determine the limiting reagent
From the balanced equation:
- 2 moles of \( KMnO_4 \) react with 6 moles of \( KI \).

To find the amount of \( KI \) required for 0.005 moles of \( KMnO_4 \):
\[ \text{Moles of } KI \text{ required} = 0.005 \, \text{moles } KMnO_4 \times \frac{6 \, \text{moles } KI}{2 \, \text{moles } KMnO_4} = 0.015 \, \text{moles} \]

Now compare the moles of \( KI \) available (0.025 moles) with the moles required (0.015 moles):
- Since \( KI \) is present in excess, \( KMnO_4 \) is the limiting reagent.

### Step 5: Calculate the moles of \( I_2 \) produced
From the balanced equation:
- 2 moles of \( KMnO_4 \) produce 3 moles of \( I_2 \).

Using the moles of \( KMnO_4 \):
\[ \text{Moles of } I_2 = 0.005 \, \text{moles } KMnO_4 \times \frac{3 \, \text{moles } I_2}{2 \, \text{moles } KMnO_4} = 0.0075 \, \text{moles} \]

### Final Answer
The number of moles of \( I_2 \) formed is \( 0.0075 \, \text{moles} \).

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