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 are liberated when one mole of potassium dichromate reacts with an excess of potassium iodide in the presence of concentrated sulfuric acid, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
- The reactants are potassium dichromate (K₂Cr₂O₇) and potassium iodide (KI).
- The products of the reaction include potassium sulfate (K₂SO₄), chromium(II) sulfate (Cr₂(SO₄)₃), water (H₂O), and iodine (I₂).

2. **Write the Balanced Chemical Equation**:
- The balanced chemical equation for the reaction is:
\[ K_2Cr_2O_7 + 6 KI + 7 H_2SO_4 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + 3 I_2 + 7 H_2O \]
- This equation shows that potassium dichromate reacts with potassium iodide and sulfuric acid to produce potassium sulfate, chromium(III) sulfate, iodine, and water.

3. **Determine the Moles of Iodine Produced**:
- From the balanced equation, we can see that 1 mole of potassium dichromate (K₂Cr₂O₇) produces 3 moles of iodine (I₂).
- Therefore, when 1 mole of K₂Cr₂O₇ reacts, it liberates 3 moles of iodine.

4. **Conclusion**:
- The number of moles of iodine liberated when one mole of potassium dichromate reacts with excess potassium iodide is **3 moles**.

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

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