The only cations present in the slightly acidic solution are `Fe^(3+),Zn^(2+) and Cu^(2+)`. The reagent that when added in exess to this solution would identify and separate `Fe^(3+)` in one step is:

Options

A. 2M HCl
B. 6M `NH_(3)`
C. 6M NaOH
D. `H_(2)S` gas

Answer

To solve the problem of identifying and separating `Fe^(3+)` from a solution containing `Fe^(3+)`, `Zn^(2+)`, and `Cu^(2+)`, we need to find a reagent that will precipitate `Fe^(3+)` while keeping `Zn^(2+)` and `Cu^(2+)` in solution.

### Step-by-Step Solution:

1. **Identify the Cations**: The cations present in the solution are `Fe^(3+)`, `Zn^(2+)`, and `Cu^(2+)`. We need to separate `Fe^(3+)` from the others.

2. **Choose a Suitable Reagent**: We need a reagent that will selectively precipitate `Fe^(3+)` while keeping `Zn^(2+)` and `Cu^(2+)` soluble.

3. **Consider Common Reagents**:
- **HCl**: When hydrochloric acid (HCl) is added, `Cu^(2+)` forms a precipitate of `CuCl2`, which is not desired.
- **NH3**: Ammonia (NH3) is a better option. In the presence of excess NH3, `Zn^(2+)` remains soluble as `[Zn(NH3)4]^(2+)` complex, and `Cu^(2+)` can form a soluble complex as well. However, `Fe^(3+)` will precipitate as `Fe(OH)3`.

4. **Precipitation Reaction**: When excess NH3 is added to the solution, `Fe^(3+)` reacts to form `Fe(OH)3`, which is insoluble and precipitates out of the solution:
\[ Fe^{3+} + 3OH^- \rightarrow Fe(OH)_3 \downarrow \]

5. **Final Result**: Thus, the reagent that can be added in excess to identify and separate `Fe^(3+)` from the solution is **6 Molar Ammonia (NH3)**.

### Conclusion:
The answer to the question is that the reagent that can be added in excess to separate `Fe^(3+)` is **6 Molar NH3**.