A suspension containing insoluble substances `ZnS, MnS, HgS, Ag_(2)S` and `FeS`, is treated with `2N HCl`.On filtering, the filtrate contains appreciable amounts of which one of the following ?

Options

A. Zinc and mercury
B. Silver and iron
C. Maganese and mercury
D. Zinc, manganese and iron

Answer

To solve the problem, we need to analyze the behavior of each of the insoluble substances when treated with 2N HCl. The substances in question are ZnS, MnS, HgS, Ag2S, and FeS.

### Step-by-Step Solution:

1. **Identify the Reaction with HCl:**
- When each of these sulfides is treated with hydrochloric acid (HCl), we need to determine whether they will dissolve or remain insoluble.

2. **Analyze Each Sulfide:**
- **ZnS (Zinc Sulfide):**
- ZnS reacts with HCl to form ZnCl2 (zinc chloride), which is soluble in water.
- Reaction: ZnS + 2HCl → ZnCl2 + H2S (gas)
- **MnS (Manganese Sulfide):**
- MnS also reacts with HCl to form MnCl2 (manganese chloride), which is soluble in water.
- Reaction: MnS + 2HCl → MnCl2 + H2S (gas)
- **HgS (Mercurous Sulfide):**
- HgS does not dissolve in HCl, as it is insoluble in dilute acids.
- **Ag2S (Silver Sulfide):**
- Ag2S does not dissolve in HCl; it remains insoluble.
- **FeS (Iron Sulfide):**
- FeS reacts with HCl to form FeCl2 (iron(II) chloride), which is soluble in water.
- Reaction: FeS + 2HCl → FeCl2 + H2S (gas)

3. **Filtration Process:**
- After treating the mixture with HCl, we filter the solution. The soluble salts (ZnCl2, MnCl2, and FeCl2) will pass through the filter paper into the filtrate, while the insoluble sulfides (HgS, Ag2S) will remain as residue on the filter paper.

4. **Conclusion:**
- The filtrate will contain appreciable amounts of ZnCl2, MnCl2, and FeCl2. Therefore, the answer to the question is that the filtrate contains appreciable amounts of **ZnCl2, MnCl2, and FeCl2.**

### Final Answer:
The filtrate contains appreciable amounts of **ZnCl2, MnCl2, and FeCl2.**

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