`Pb(AcO)_(2)+H_(2)S to PbS darr+2AcOH`

Options

A. For precipitate formation formation reaction
B. For precipitate dissolution reaction
C. For precipitate exchange reaction
D. For no reaction

Answer

To solve the given reaction \( \text{Pb(AcO)}_{2} + \text{H}_{2}\text{S} \rightarrow \text{PbS} \downarrow + 2\text{AcOH} \), we will analyze the reaction step by step.

### Step 1: Identify the Reactants
The reactants in the equation are:
- Lead(II) acetate, \( \text{Pb(AcO)}_{2} \)
- Hydrogen sulfide, \( \text{H}_{2}\text{S} \)

### Step 2: Identify the Products
The products formed from the reaction are:
- Lead(II) sulfide, \( \text{PbS} \) (which is a precipitate)
- Acetic acid, \( \text{AcOH} \)

### Step 3: Determine the Type of Reaction
In this reaction, we observe that:
- On the reactant side, there are no precipitates present.
- Upon the reaction, lead(II) sulfide \( \text{PbS} \) is formed, which is insoluble in water and precipitates out of the solution.

This indicates that the reaction is a **precipitate formation reaction** since a solid precipitate is formed from the reactants.

### Step 4: Conclusion
Based on the analysis above, the correct classification of the reaction is:
- **Precipitate Formation Reaction**

### Final Answer
The reaction \( \text{Pb(AcO)}_{2} + \text{H}_{2}\text{S} \rightarrow \text{PbS} \downarrow + 2\text{AcOH} \) is classified as a precipitate formation reaction.

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