Choose the correct balanced chemical equations to show the reactions of the following:
Action of KOH on `Fe_(2)(SO_(4))_(3)`.

Options

A. `Fe_(2)(SO_(4))_(3) + 6KOH rarr 2Fe(OH)_(2) + K_(2)SO_(4)`
B. `Fe_(2)(SO_(4))_(3) + 6KOH rarr 2Fe(OH)_(3) + 3K_(2)SO_(4)`
C. `Fe_(2)(SO_(4))_(3) + 6KOH rarr 2Fe(OH)_(2) + 3KSO_(4)`
D. `Fe_(2)(SO_(4))_(3) + 6KOH rarr 2Fe(OH)_(2) + 3K_(2)SO`

Explanation

To solve the problem of balancing the chemical equation for the reaction between potassium hydroxide (KOH) and ferric sulfate (Fe₂(SO₄)₃), we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
- The reactants are ferric sulfate (Fe₂(SO₄)₃) and potassium hydroxide (KOH).
- The products will be ferric hydroxide (Fe(OH)₃) and potassium sulfate (K₂SO₄).

2. **Write the Unbalanced Equation**:
- The unbalanced equation can be written as:
\[ \text{Fe}_2(\text{SO}_4)_3 + \text{KOH} \rightarrow \text{Fe(OH)}_3 + \text{K}_2\text{SO}_4 \]

3. **Break Down the Compounds into Ions**:
- Ferric sulfate dissociates into:
- 2 Fe³⁺ ions
- 3 SO₄²⁻ ions
- Potassium hydroxide dissociates into:
- K⁺ ions
- OH⁻ ions

4. **Combine the Ions to Form Products**:
- The Fe³⁺ ions will combine with OH⁻ ions to form Fe(OH)₃.
- The K⁺ ions will combine with SO₄²⁻ ions to form K₂SO₄.

5. **Determine the Number of Each Ion Needed**:
- For every 2 Fe³⁺ ions, we need 6 OH⁻ ions to form 2 Fe(OH)₃.
- For every 3 SO₄²⁻ ions, we need 6 K⁺ ions to form 3 K₂SO₄.

6. **Write the Balanced Equation**:
- The balanced equation will be:
\[ \text{Fe}_2(\text{SO}_4)_3 + 6 \text{KOH} \rightarrow 2 \text{Fe(OH)}_3 + 3 \text{K}_2\text{SO}_4 \]

7. **Final Check for Balance**:
- Count the number of each atom on both sides:
- Fe: 2 on both sides
- SO₄: 3 on both sides
- K: 6 on both sides
- OH: 6 on both sides
- The equation is balanced.

### Final Balanced Equation:
\[ \text{Fe}_2(\text{SO}_4)_3 + 6 \text{KOH} \rightarrow 2 \text{Fe(OH)}_3 + 3 \text{K}_2\text{SO}_4 \]