Give a reaction where Lead iodide is formed as one of the product.

Explanation

To form lead iodide (PbI2) as one of the products, we can carry out a double displacement reaction between lead(II) nitrate and potassium iodide. Here’s the step-by-step solution:

### Step 1: Identify the Reactants
We will use lead(II) nitrate and potassium iodide as our reactants. The chemical formulas are:
- Lead(II) nitrate: Pb(NO3)2
- Potassium iodide: KI

### Step 2: Write the Chemical Reaction
In a double displacement reaction, the cations and anions of the reactants exchange partners. The reaction can be written as:
\[ \text{Pb(NO}_3\text{)}_2 (aq) + 2 \text{KI} (aq) \rightarrow \text{PbI}_2 (s) + 2 \text{KNO}_3 (aq) \]

### Step 3: Identify the Products
From the reaction, we can see that lead iodide (PbI2) is formed as a solid precipitate, and potassium nitrate (KNO3) remains in the aqueous solution.

### Step 4: Balance the Equation
The equation is balanced as written:
- 1 Pb on both sides
- 2 I on both sides
- 2 K on both sides
- 2 NO3 on both sides

### Final Balanced Equation
The final balanced equation for the reaction is:
\[ \text{Pb(NO}_3\text{)}_2 (aq) + 2 \text{KI} (aq) \rightarrow \text{PbI}_2 (s) + 2 \text{KNO}_3 (aq) \]

### Conclusion
In this reaction, lead iodide (PbI2), which is a yellow solid precipitate, is formed along with potassium nitrate in the aqueous solution.

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