Potassium iodide reacts with acidified `K_(2)Cr_(2)O_(7)`. How many moles of KI are required for one mole of `K_(2)Cr_(2)O_(7)` ?

Explanation

To determine how many moles of potassium iodide (KI) are required for one mole of potassium dichromate (K₂Cr₂O₇) when they react in an acidified solution, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
- The reactants are potassium iodide (KI) and acidified potassium dichromate (K₂Cr₂O₇).
- The products of the reaction will involve iodine (I₂) and chromium ions (Cr³⁺).

2. **Write the Half-Reactions**:
- The dichromate ion (Cr₂O₇²⁻) is reduced to chromium ions (Cr³⁺).
- The iodide ions (I⁻) from potassium iodide are oxidized to iodine (I₂).

The half-reaction for the reduction of dichromate is:
\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]

The half-reaction for the oxidation of iodide is:
\[ 6\text{I}^- \rightarrow 3\text{I}_2 + 6\text{e}^- \]

3. **Balance the Electrons**:
- In the reduction half-reaction, 6 electrons are gained.
- In the oxidation half-reaction, 6 electrons are lost.
- This means the two half-reactions are balanced in terms of electrons.

4. **Combine the Half-Reactions**:
- By combining the balanced half-reactions, we can see the overall reaction:
\[ \text{Cr}_2\text{O}_7^{2-} + 6\text{I}^- + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 3\text{I}_2 + 7\text{H}_2\text{O} \]

5. **Determine the Moles of KI Required**:
- From the balanced equation, we see that 6 moles of iodide ions (I⁻) are required for every 1 mole of dichromate ion (Cr₂O₇²⁻).
- Since each mole of KI provides one mole of I⁻, this means that 6 moles of KI are required for 1 mole of K₂Cr₂O₇.

### Final Answer:
**6 moles of KI are required for 1 mole of K₂Cr₂O₇.**