How many moles of iodine are liberated when one mol of potassium dichromate reacts with excess of potassium iodide in the presence of concentrated sulphuric acid?

Options

A. 1
B. 2
C. 3
D. 4

Answer

To determine how many moles of iodine (I2) are liberated when one mole of potassium dichromate (K2Cr2O7) reacts with excess potassium iodide (KI) in the presence of concentrated sulfuric acid (H2SO4), we can follow these steps:

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The reaction between potassium dichromate, potassium iodide, and sulfuric acid can be represented as follows:
\[ K_2Cr_2O_7 + 7 H_2SO_4 + 6 KI \rightarrow Cr_2(SO_4)_3 + 3 I_2 + 4 K_2SO_4 + 7 H_2O \]

2. **Identify the Stoichiometric Coefficients:**
From the balanced equation, we can see the stoichiometric coefficients:
- For potassium dichromate (K2Cr2O7): 1
- For iodine (I2): 3

3. **Determine the Moles of Iodine Produced:**
According to the balanced equation, 1 mole of K2Cr2O7 produces 3 moles of I2. Since the question states that we have 1 mole of K2Cr2O7, we can directly use the stoichiometric ratio:
\[ \text{Moles of I2} = 1 \text{ mole of K2Cr2O7} \times \frac{3 \text{ moles of I2}}{1 \text{ mole of K2Cr2O7}} = 3 \text{ moles of I2} \]

4. **Conclusion:**
Therefore, when 1 mole of potassium dichromate reacts with excess potassium iodide, 3 moles of iodine are liberated.

### Final Answer:
**3 moles of iodine (I2) are liberated.**