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

Answer

To solve the problem, we need to find the number of moles of iodine (I₂) formed when 250 mL of 0.1 M KI is mixed with 250 mL of 0.02 M KMnO₄ in a basic medium.

### Step-by-Step Solution:

1. **Write the balanced chemical equation**:
The reaction between KMnO₄ and KI in basic medium can be represented as:
\[ \text{KMnO}_4 + 6 \text{KI} \rightarrow 3 \text{I}_2 + \text{MnO}_2 + 6 \text{KOH} \]
From this equation, we can see that 1 mole of KMnO₄ reacts with 6 moles of KI to produce 3 moles of I₂.

2. **Calculate the moles of KMnO₄**:
The molarity of KMnO₄ is given as 0.02 M and the volume is 250 mL (which is 0.250 L).
\[ \text{Moles of KMnO}_4 = \text{Molarity} \times \text{Volume} = 0.02 \, \text{mol/L} \times 0.250 \, \text{L} = 0.005 \, \text{mol} \]

3. **Calculate the moles of KI**:
The molarity of KI is given as 0.1 M and the volume is also 250 mL (0.250 L).
\[ \text{Moles of KI} = \text{Molarity} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.250 \, \text{L} = 0.025 \, \text{mol} \]

4. **Determine the limiting reactant**:
From the balanced equation, we see that 1 mole of KMnO₄ requires 6 moles of KI. Therefore, for 0.005 moles of KMnO₄, the required moles of KI would be:
\[ \text{Required moles of KI} = 0.005 \, \text{mol KMnO}_4 \times 6 \, \text{mol KI/mol KMnO}_4 = 0.030 \, \text{mol KI} \]
Since we only have 0.025 moles of KI available, KI is the limiting reactant.

5. **Calculate the moles of I₂ produced**:
According to the balanced equation, 6 moles of KI produce 3 moles of I₂. Therefore, the moles of I₂ produced from 0.025 moles of KI can be calculated as follows:
\[ \text{Moles of I}_2 = \left(\frac{3 \, \text{mol I}_2}{6 \, \text{mol KI}}\right) \times 0.025 \, \text{mol KI} = 0.0125 \, \text{mol I}_2 \]

### Final Answer:
The number of moles of I₂ formed is **0.0125 moles**.