Zinc pieces are added to acidified solution of `SO_(3)^(2-)`.Gas liberated can:

Options

A. turn lead acetane paper black
B. turn lime water milky
C. give white precipitate with `AgNO_(3)` solution
D. decolourize acidified `KMnO_(4)`

Explanation

To solve the question, we need to analyze the reaction between zinc (Zn) and the acidified solution of sulfite ions (SO3^2-). Here’s a step-by-step breakdown of the solution:

### Step 1: Write the reaction
When zinc is added to an acidified solution containing sulfite ions (SO3^2-), a reaction occurs. The zinc will displace the hydrogen ions (H+) from the acid, leading to the formation of hydrogen sulfide (H2S), zinc ions (Zn^2+), and water (H2O).

**Reaction:**
\[ \text{Zn} + \text{SO}_3^{2-} + 2\text{H}^+ \rightarrow \text{H}_2\text{S} + \text{Zn}^{2+} + \text{H}_2\text{O} \]

### Step 2: Identify the gas liberated
The gas liberated in this reaction is hydrogen sulfide (H2S).

### Step 3: Analyze the properties of H2S
Now, we need to evaluate the properties of the gas (H2S) in relation to the options provided:

1. **Turns lead acetate paper black**: This is correct. H2S reacts with lead acetate (Pb(C2H3O2)2) to form lead sulfide (PbS), which is black in color.

2. **Turns lime water milky**: This is incorrect. Lime water (Ca(OH)2) turns milky due to the presence of carbon dioxide (CO2), not H2S.

3. **Gives a white precipitate with AgNO3 solution**: This is incorrect. H2S reacts with silver nitrate (AgNO3) to form silver sulfide (Ag2S), which is black, not white.

4. **Decolorizes acidified KMnO4**: This is also incorrect. H2S does not decolorize acidified potassium permanganate (KMnO4); this reaction is typically associated with reducing agents like sulfite ions or other substances.

### Conclusion
From the analysis, the only correct statement regarding the gas liberated (H2S) is that it turns lead acetate paper black.

### Final Answer
The gas liberated can turn lead acetate paper black.

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