One mole of acidified `K_(2)Cr_(2)O_(7)` on reaction with excess of KCl will liberate….., moles of `I_(2)`.

Options

A. 3
B. 1
C. 7
D. 2

Answer

To solve the problem of how many moles of \( I_2 \) are liberated when one mole of acidified \( K_2Cr_2O_7 \) reacts with excess \( KI \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
The reaction involves \( K_2Cr_2O_7 \) (potassium dichromate) and \( KI \) (potassium iodide) in an acidic medium. The products of this reaction will include \( I_2 \) (iodine), \( K_2SO_4 \) (potassium sulfate), and \( Cr_2(SO_4)_3 \) (chromium(III) sulfate).

2. **Write the Balanced Reaction**:
The balanced chemical reaction can be represented as:
\[ K_2Cr_2O_7 + 6 KI + 7 H_2SO_4 \rightarrow 2 Cr_2(SO_4)_3 + 3 I_2 + 7 K_2SO_4 + 7 H_2O \]
Here, \( K_2Cr_2O_7 \) oxidizes \( I^- \) to \( I_2 \).

3. **Determine Oxidation States**:
In \( K_2Cr_2O_7 \), chromium is in the +6 oxidation state, and it gets reduced to +3 in \( Cr_2(SO_4)_3 \). The iodide ion \( I^- \) is oxidized to \( I_2 \) (0 oxidation state).

4. **Calculate the n-factor for \( K_2Cr_2O_7 \)**:
The change in oxidation state for chromium is from +6 to +3:
\[ \text{Change} = 6 - 3 = 3 \]
Since there are 2 chromium atoms in \( K_2Cr_2O_7 \), the total change is:
\[ \text{n-factor} = 2 \times 3 = 6 \]

5. **Calculate the n-factor for \( I_2 \)**:
The change in oxidation state for iodine is from -1 (in \( I^- \)) to 0 (in \( I_2 \)):
\[ \text{Change} = 0 - (-1) = 1 \]
Since 2 moles of \( I^- \) produce 1 mole of \( I_2 \), the n-factor for \( I_2 \) is:
\[ \text{n-factor} = 2 \times 1 = 2 \]

6. **Set Up the Equivalent Relationship**:
The equivalents of \( K_2Cr_2O_7 \) must equal the equivalents of \( I_2 \):
\[ \text{Equivalents of } K_2Cr_2O_7 = \text{Equivalents of } I_2 \]
Using the formula:
\[ \text{Equivalents} = \text{moles} \times \text{n-factor} \]
For \( K_2Cr_2O_7 \):
\[ \text{Equivalents} = 1 \text{ mole} \times 6 = 6 \]

7. **Calculate Moles of \( I_2 \)**:
Let \( x \) be the moles of \( I_2 \):
\[ x \times 2 = 6 \]
Solving for \( x \):
\[ x = \frac{6}{2} = 3 \]

### Final Answer:
Thus, one mole of acidified \( K_2Cr_2O_7 \) will liberate **3 moles of \( I_2 \)**.

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