When lead nitrate reacts with potassium iodide, yellow precipitate of

Options

A. `PbI_(2)` is formed
B. `KNO_(3)` is formed
C. `Pb(NO_(3))_(2)` is formed
D. `PbIO_(3)` is formed.

Explanation

**Step-by-Step Solution:**

1. **Identify the Reactants:**
The reactants in this reaction are lead nitrate (Pb(NO₃)₂) and potassium iodide (KI).

2. **Write the Chemical Equation:**
The reaction between lead nitrate and potassium iodide can be represented as:
\[ \text{Pb(NO}_3\text{)}_2 + 2\text{KI} \rightarrow \text{PbI}_2 + 2\text{KNO}_3 \]

3. **Identify the Products:**
The products of this reaction are lead iodide (PbI₂) and potassium nitrate (KNO₃).

4. **Determine the Precipitate:**
Lead iodide (PbI₂) is an insoluble compound that forms a yellow precipitate when the two reactants are mixed.

5. **Balance the Chemical Equation:**
The equation is balanced as follows:
- On the left side, we have 1 lead (Pb), 2 nitrate (NO₃), 2 potassium (K), and 2 iodide (I).
- On the right side, we also have 1 lead (Pb), 2 iodide (I) in PbI₂, and 2 potassium (K) in 2KNO₃.
Thus, the balanced equation is:
\[ \text{Pb(NO}_3\text{)}_2 + 2\text{KI} \rightarrow \text{PbI}_2 + 2\text{KNO}_3 \]

6. **Conclusion:**
The yellow precipitate formed in this reaction is lead iodide (PbI₂).

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