`AlCl_(3)+3NaOH to Al(OH)_(3)darr+3NaCl`

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 chemical reaction \( \text{AlCl}_3 + 3\text{NaOH} \rightarrow \text{Al(OH)}_3 + 3\text{NaCl} \), we can analyze the reaction step by step.

### Step 1: Identify the Reactants
The reactants in the reaction are:
- Aluminum chloride (\( \text{AlCl}_3 \))
- Sodium hydroxide (\( \text{NaOH} \))

### Step 2: Write the Products
When aluminum chloride reacts with sodium hydroxide, it produces:
- Aluminum hydroxide (\( \text{Al(OH)}_3 \)), which is a precipitate
- Sodium chloride (\( \text{NaCl} \))

### Step 3: Determine the Type of Reaction
This reaction can be classified based on the following characteristics:
- **Precipitate Formation**: The formation of aluminum hydroxide (\( \text{Al(OH)}_3 \)) indicates that a solid precipitate is formed when the two reactants combine.
- **Double Displacement Reaction**: Since the anions and cations are exchanging partners, it can also be classified as a double displacement reaction.

### Step 4: Conclusion
Since the main characteristic of this reaction is the formation of a precipitate, we conclude that this is a **precipitate formation reaction**.

### Final Answer
The correct classification of the reaction \( \text{AlCl}_3 + 3\text{NaOH} \rightarrow \text{Al(OH)}_3 + 3\text{NaCl} \) is **precipitate formation reaction**.

---