Sumedha mixed two solutions P and Q. She recorded the following observations and conclusions in her notebook
I. A yellow precipitate is formed.
II. It is a double displacement reaction. The solutions P, Q and the yellow precipitate formed are respectively.

Options

A. `Pb(NO)_(3), KI and PbI_(2)`
B. `AgNO_(3), NaCl and AgCl`
C. `Na_(2)SO_(4), BaCl_(2), and BaSO_(4)`
D. `FeCl_(3), NH_(4),OH and Fe(OH)_(2)`

Answer

To solve the question, we need to identify the two solutions (P and Q) that, when mixed, result in a yellow precipitate, and confirm that the reaction is a double displacement reaction.

### Step-by-Step Solution:

1. **Identify the Yellow Precipitate**:
- The problem states that a yellow precipitate is formed. Common yellow precipitates in chemistry include lead iodide (PbI2).

2. **Determine the Nature of the Reaction**:
- The question mentions that it is a double displacement reaction. In a double displacement reaction, the cations and anions of two different compounds exchange places to form two new compounds.

3. **Identify Possible Solutions**:
- To produce lead iodide (PbI2), we can use lead nitrate (Pb(NO3)2) and potassium iodide (KI).
- When these two solutions are mixed, the lead ions (Pb²⁺) from lead nitrate react with iodide ions (I⁻) from potassium iodide to form lead iodide (PbI2), which is the yellow precipitate.

4. **Write the Reaction**:
- The balanced chemical equation for the reaction is:
\[ \text{Pb(NO}_3\text{)}_2 (aq) + 2 \text{KI} (aq) \rightarrow 2 \text{KNO}_3 (aq) + \text{PbI}_2 (s) \]
- Here, Pb(NO3)2 is solution P, KI is solution Q, and PbI2 is the yellow precipitate formed.

5. **Conclusion**:
- Therefore, the solutions P and Q and the yellow precipitate formed are:
- Solution P: Lead Nitrate (Pb(NO3)2)
- Solution Q: Potassium Iodide (KI)
- Yellow Precipitate: Lead Iodide (PbI2)

### Final Answer:
- Solutions P and Q are Lead Nitrate (Pb(NO3)2) and Potassium Iodide (KI) respectively, and the yellow precipitate formed is Lead Iodide (PbI2).

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