Fe(Finely powdered)++HCl(dil.)`toP+Quarr`
Compound 'P' does not precipitate with:

Options

A. `AgnO_(3)`
B. `K_(3)[Fe(CN)_(6)]`
C. `(NH_(4))_(2)S`
D. `NH_(4)Cl+NH_(4)OH`

Answer

To solve the question, we need to analyze the reaction of finely powdered iron (Fe) with dilute hydrochloric acid (HCl) and determine which compound does not precipitate with the resulting product.

### Step-by-Step Solution:

1. **Identify the Reaction**:
When finely powdered iron (Fe) reacts with dilute hydrochloric acid (HCl), it forms iron(II) chloride (FeCl2) and hydrogen gas (H2).
\[ \text{Fe (s)} + 2 \text{HCl (aq)} \rightarrow \text{FeCl}_2 \text{ (aq)} + \text{H}_2 \text{ (g)} \]

2. **Identify Compound P**:
In this reaction, the compound P is identified as iron(II) chloride (FeCl2).

3. **Analyze the Precipitation Reactions**:
Now we will analyze how FeCl2 reacts with the given options to see if they form a precipitate or not:

- **Option 1: AgNO3**:
\[ \text{FeCl}_2 + 2 \text{AgNO}_3 \rightarrow 2 \text{AgCl (s)} + \text{Fe(NO}_3)_2 \]
This reaction produces a white precipitate of silver chloride (AgCl).

- **Option 2: K3[Fe(CN)6]**:
\[ \text{FeCl}_2 + K_3[\text{Fe(CN)}_6] \rightarrow \text{Fe}_3[\text{Fe(CN)}_6] + 6 \text{KCl} \]
This reaction forms a blue precipitate known as Turnbull's Blue.

- **Option 3: (NH4)2S**:
\[ \text{FeCl}_2 + (NH_4)_2S \rightarrow \text{FeS (s)} + 2 \text{NH}_4\text{Cl} \]
This reaction produces a black precipitate of iron(II) sulfide (FeS).

- **Option 4: NH4Cl + NH4OH**:
\[ \text{FeCl}_2 + \text{NH}_4\text{Cl} + \text{NH}_4\text{OH} \rightarrow \text{Fe(OH)}_2 + 2 \text{NH}_4\text{Cl} \]
This reaction does not produce a precipitate; it remains in solution.

4. **Conclusion**:
The compound P (FeCl2) does not precipitate with option 4 (NH4Cl + NH4OH). Therefore, the correct answer is that the compound P does not precipitate with NH4Cl + NH4OH.

### Final Answer:
The compound P does not precipitate with **NH4Cl + NH4OH**.

---