`AlCl_(2)` is soluble is axcess of `NaOH` forming sodium metaaluminate `Na[Al(OH)_(4)]`.

Explanation

To determine whether the statement regarding the solubility of AlCl3 in excess NaOH and the formation of sodium metaaluminate Na[Al(OH)4] is correct, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants:**
The reactants in this case are aluminum chloride (AlCl3) and sodium hydroxide (NaOH).

2. **Initial Reaction:**
When aluminum chloride is added to sodium hydroxide, it reacts to form aluminum hydroxide (Al(OH)3) and sodium chloride (NaCl). The balanced chemical equation for this reaction is:
\[ AlCl_3 + 3NaOH \rightarrow Al(OH)_3 \downarrow + 3NaCl \]
Here, Al(OH)3 precipitates out as a white solid.

3. **Formation of Sodium Metaaluminate:**
When excess sodium hydroxide is added to the reaction mixture, the precipitated aluminum hydroxide can dissolve. This occurs because aluminum hydroxide reacts with sodium hydroxide to form sodium metaaluminate. The reaction can be represented as:
\[ Al(OH)_3 + NaOH \rightarrow Na[Al(OH)_4] \]
This indicates that the aluminum hydroxide dissolves in excess NaOH to form sodium metaaluminate.

4. **Conclusion:**
Since aluminum chloride does indeed dissolve in excess sodium hydroxide to form sodium metaaluminate, the statement provided is correct.

### Final Statement:
The statement that AlCl3 is soluble in excess NaOH forming sodium metaaluminate Na[Al(OH)4] is correct.

---