How many moles of iodin are liberated when 2 moles of potassium permanganate rect with potassium iodide?

Explanation

To determine how many moles of iodine (I2) are liberated when 2 moles of potassium permanganate (KMnO4) react with potassium iodide (KI), we need to analyze the redox reaction that occurs between these substances.

### Step-by-Step Solution:

1. **Identify the Reactants and Products:**
- The reactants are potassium permanganate (KMnO4) and potassium iodide (KI).
- The products of the reaction are manganese ions (Mn2+) and iodine (I2).

2. **Write the Half-Reactions:**
- The reduction half-reaction involves the permanganate ion (MnO4-) being reduced to Mn2+:
\[ \text{MnO}_4^- + 8H^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4H_2O \]
- The oxidation half-reaction involves iodide ions (I-) being oxidized to iodine (I2):
\[ 2I^- \rightarrow I_2 + 2e^- \]

3. **Balance the Half-Reactions:**
- The reduction half-reaction shows that 1 mole of MnO4- requires 5 electrons.
- The oxidation half-reaction shows that 2 moles of I- produce 1 mole of I2 by losing 2 electrons.

4. **Combine the Half-Reactions:**
- To balance the electrons, we need to multiply the oxidation half-reaction by 5:
\[ 5(2I^- \rightarrow I_2 + 2e^-) \rightarrow 10I^- \rightarrow 5I_2 + 10e^- \]
- Now, we can combine the balanced half-reactions:
\[ 2\text{MnO}_4^- + 16H^+ + 10I^- \rightarrow 2\text{Mn}^{2+} + 8H_2O + 5I_2 \]

5. **Determine the Moles of I2 Produced:**
- From the balanced equation, we see that 2 moles of KMnO4 produce 5 moles of I2.
- Therefore, when 2 moles of potassium permanganate react, they liberate 5 moles of iodine.

### Final Answer:
**5 moles of iodine (I2) are liberated when 2 moles of potassium permanganate react with potassium iodide.**