How many moles of iodine are liberated when 1 mole potassium dichromate reacts with potassium iodide?

Options

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

Explanation

To determine how many moles of iodine are liberated when 1 mole of potassium dichromate reacts with potassium iodide, we need to follow these steps:

### Step 1: Write the unbalanced chemical equation
The unbalanced reaction between potassium dichromate (K₂Cr₂O₇) and potassium iodide (KI) in the presence of sulfuric acid (H₂SO₄) can be written as:
\[ K_2Cr_2O_7 + KI + H_2SO_4 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + I_2 + H_2O + KOH \]

### Step 2: Balance the chemical equation
To balance the equation, we need to ensure that the number of atoms for each element is the same on both sides of the equation. The balanced equation is:
\[ K_2Cr_2O_7 + 6KI + 12H_2SO_4 \rightarrow 2K_2SO_4 + Cr_2(SO_4)_3 + 3I_2 + 6KOH + 6H_2O \]

### Step 3: Analyze the balanced equation
From the balanced equation, we can see that:
- 1 mole of K₂Cr₂O₇ reacts with 6 moles of KI.
- This reaction produces 3 moles of iodine (I₂).

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

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

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