Let's work through the step by step: --- ## **Step 1: Write All Four Reactions** Given reactions: 1. \( 2\text{Al}(s) + 2\text{OH}^-(aq) + 6\text{HOH}(l) \rightarrow 2\text{Al(OH)}_4^-(aq) + 3\text{H}_2(g) \) 2. \( 2\text{Al(OH)}_4^-(aq) + 2\text{H}^+(aq) \rightarrow 2\text{Al(OH)}_3(s) + 4\text{HOH}(l) \) 3. \( 2\text{Al(OH)}_3(s) + 6\text{H}^+(aq) \rightarrow 2\text{Al}^{3+}(aq) + 6\text{HOH}(l) \) 4. \( 2\text{K}^+(aq) + 2\text{Al}^{3+}(aq) + 4\text{SO}_4^{2-}(aq) + 24\text{HOH}(l) \rightarrow 2\text{KAl(SO}_4)_2 \cdot 12\text{H}_2\text{O}(s) \) ("alum") --- ## **Step 2: Add All Equations Together** Let’s write all reactants on one side and all products on the other: **Reactants:** - \( 2\text{Al}(s) \) - \( 2\text{OH}^-(aq) \) - \( 6\text{HOH}(l) \) - \( 2\text{H}^+(aq) \) - \( 6\text{H}^+(aq) \) - \( 2\text{K}^+(aq) \) - \( 4\text{SO}_4^{2-}(aq) \) - \( 24\text{HOH}(l) \) **Products:** - \( 2\text{Al(OH)}_4^-(aq) \) - \( 3\text{H}_2(g) \) - \( 2\text{Al(OH)}_3(s) \) - \( 4\text{HOH}(l) \) - \( 2\text{Al}^{3+}(aq) \) - \( 6\text{HOH}(l) \) - \( 2\text{KAl(SO}_4)_2 \cdot 12\text{H}_2\text{O}(s) \) --- ## **Step 3: Cancel Species That Appear on Both Sides** Let’s look for species that appear on both sides: - \( 2\text{Al(OH)}_4^-(aq) \): produced in step 1, consumed in step 2 → **cancel** - \( 2\text{Al(OH)}_3(s) \): produced in step 2, consumed in step 3 → **cancel** - \( 2\text{Al}^{3+}(aq) \): produced in step 3, consumed in step 4 → **cancel** - \( 6\text{HOH}(l) \): reactant in step 1, product in step 3 → **cancel** - \( 24\text{HOH}(l) \): reactant in step 4 only, keep - \( 4\text{HOH}(l) \): product in step 2 only, keep Now, let's be careful with water (\(\text{HOH}\)) since it appears in multiple steps: - Added up, total water on reactant side: \(6\) (step 1) + \(24\) (step 4) = \(30\) - On product side: \(4\) (step 2) + \(6\) (step 3) = \(10\) So, net water: \(30 - 10 = 20\) on the reactant side. --- ## **Step 4: Write the Net Ionic Equation** After cancellation: **Reactants:** - \( 2\text{Al}(s) \) - \( 2\text{OH}^-(aq) \) - \( 8\text{H}^+(aq) \) - \( 2\text{K}^+(aq) \) - \( 4\text{SO}_4^{2-}(aq) \) - \( 20\text{HOH}(l) \) **Products:** - \( 3\text{H}_2(g) \) - \( 2\text{KAl(SO}_4)_2 \cdot 12\text{H}_2\text{O}(s) \) --- ### **Final Net Ionic Equation** \[ 2\text{Al}(s) + 2\text{OH}^-(aq) + 8\text{H}^+(aq) + 2\text{K}^+(aq) + 4\text{SO}_4^{2-}(aq) + 20\text{H}_2\text{O}(l) \rightarrow 3\text{H}_2(g) + 2\text{KAl(SO}_4)_2 \cdot 12\text{H}_2\text{O}(s) \] --- ## **Summary Table** | Species | Reactant Side | Product Side | Net (Reactant - Product) | |----------------------------|:------------:|:------------:|:------------------------:| | \( \text{Al}(s) \) | 2 | | 2 | | \( \text{OH}^-(aq) \) | 2 | | 2 | | \( \text{H}^+(aq) \) | 8 | | 8 | | \( \text{K}^+(aq) \) | 2 | | 2 | | \( \text{SO}_4^{2-}(aq) \) | 4 | | 4 | | \( \text{HOH}(l) \) | 30 | 10 | 20 | | \( \text{H}_2(g) \) | | 3 | 3 | | Alum | | 2 | 2 | --- ## **Final Answer** \[ \boxed{ 2\text{Al}(s) + 2\text{OH}^-(aq) + 8\text{H}^+(aq) + 2\text{K}^+(aq) + 4\text{SO}_4^{2-}(aq) + 20\text{H}_2\text{O}(l) \rightarrow 3\text{H}_2(g) + 2\text{KAl(SO}_4)_2 \cdot 12\text{H}_2\text{O}(s) } \]

## Step-by-Step Solution: Chemical Reactions To solve the problem, we will analyze the given reactions, add them together, cancel common species, and derive the net ionic equation. --- ## Step 1: Write All Four Reactions The reactions provided are as follows: 1. **Aluminum Reaction:** \[ 2\text{Al}(s) + 2\text{OH}^-(aq) + 6\text{H}_2\text{O}(l) \rightarrow 2\text{Al(OH)}_4^-(aq) + 3\text{H}_2(g) \] 2. **Formation of Aluminum Hydroxide:** \[ 2\text{Al(OH)}_4^-(aq) + 2\text{H}^+(aq) \rightarrow 2\text{Al(OH)}_3(s) + 4\text{H}_2\text{O}(l) \] 3. **Dissociation of Aluminum Hydroxide:** \[ 2\text{Al(OH)}_3(s) + 6\text{H}^+(aq) \rightarrow 2\text{Al}^{3+}(aq) + 6\text{H}_2\text{O}(l) \] 4. **Formation of Alum:** \[ 2\text{K}^+(aq) + 2\text{Al}^{3+}(aq) + 4\text{SO}_4^{2-}(aq) + 24\text{H}_2\text{O}(l) \rightarrow 2\text{KAl(SO}_4)_2 \cdot 12\text{H}_2\text{O}(s) \] --- ## Step 2: Combine All Equations ### Reactants & Products **Reactants:** - From reaction 1: \(2\text{Al}(s)\), \(2\text{OH}^-(aq)\), \(6\text{H}_2\text{O}(l)\) - From reaction 2: \(2\text{H}^+(aq)\) - From reaction 3: \(6\text{H}^+(aq)\) - From reaction 4: \(2\text{K}^+(aq)\), \(4\text{SO}_4^{2-}(aq)\), \(24\text{H}_2\text{O}(l)\) **Products:** - From reaction 1: \(2\text{Al(OH)}_4^-(aq)\), \(3\text{H}_2(g)\) - From reaction 2: \(2\text{Al(OH)}_3(s)\), \(4\text{H}_2\text{O}(l)\) - From reaction 3: \(2\text{Al}^{3+}(aq)\), \(6\text{H}_2\text{O}(l)\) - From reaction 4: \(2\text{KAl(SO}_4)_2 \cdot 12\text{H}_2\text{O}(s)\) --- ## Step 3: Cancel Common Species ### Identify Common Species - **Cancel \(2\text{Al(OH)}_4^-(aq)\)**: produced in step 1, consumed in step 2 - **Cancel \(2\text{Al(OH)}_3(s)\)**: produced in step 2, consumed in step 3 - **Cancel \(2\text{Al}^{3+}(aq)\)**: produced in step 3, consumed in step 4 - **Water (\(\text{H}_2\text{O}\))**: - Reactant side: \(6 + 24 = 30\) - Product side: \(4 + 6 = 10\) - Net water: \(30 - 10 = 20\) on the reactant side ### Remaining Components After canceling out, we are left with: **Reactants:** - \(2\text{Al}(s)\) - \(2\text{OH}^-(aq)\) - \(8\text{H}^+(aq)\) (from \(2 + 6\)) - \(2\text{K}^+(aq)\) - \(4\text{SO}_4^{2-}(aq)\) - \(20\text{H}_2\text{O}(l)\) **Products:** - \(3\text{H}_2(g)\) - \(2\text{KAl(SO}_4)_2 \cdot 12\text{H}_2\text{O}(s)\) --- ## Step 4: Write the Net Ionic Equation ### Final Form The net ionic equation is: \[ 2\text{Al}(s) + 2\text{OH}^-(aq) + 8\text{H}^+(aq) + 2\text{K}^+(aq) + 4\text{SO}_4^{2-}(aq) + 20\text{H}_2\text{O}(l) \rightarrow 3\text{H}_2(g) + 2\text{KAl(SO}_4)_2 \cdot 12\text{H}_2\text{O}(s) \] --- ## Summary Table | Species | Reactant Side | Product Side | Net (Reactant - Product) | |----------------------------|:-------------:|:------------:|:------------------------:| | \( \text{Al}(s) \) | 2 | | 2 | | \( \text{OH}^-(aq) \) | 2 | | 2 | | \( \text{H}^+(aq) \) | 8 | | 8 | | \( \text{K}^+(aq) \) | 2 | | 2 | | \( \text{SO}_4^{2-}(aq) \) | 4 | | 4 | | \( \text{H}_2\text{O}(l) \) | 30 | 10 | 20 | | \( \text{H}_2(g) \) | | 3 | 3 | | **Alum** | | 2 | 2 | --- ## Final Answer \[ \boxed{ 2\text{Al}(s) + 2\text{OH}^-(aq) + 8\text{H}^+(aq) + 2\text{K}^+(aq) + 4\text{SO}_4^{2-}(aq) + 20\text{H}_2\text{O}(l) \rightarrow 3\text{H}_2(g) + 2\text{KAl(SO}_4)_2 \cdot 12\text{H}_2\text{O}(s) } \]