Question
Write the total ionic and net ionic equatio, consider the balance equation as well as stage for; Ba(OH)2(aq) + Fe2(SO4)3(aq) produce fe(OH)3(s)+BaSO4(aq)
Answers
GPT-4o mini
To write the total ionic and net ionic equations for the reaction between barium hydroxide and iron(III) sulfate, we will start with the balanced molecular equation.
The balanced molecular equation is:
\[
3 \, \text{Ba(OH)}_2(aq) + \text{Fe}_2(\text{SO}_4)_3(aq) \rightarrow 2 \, \text{Fe(OH)}_3(s) + 3 \, \text{BaSO}_4(aq)
\]
### Step 1: Write the Total Ionic Equation
To write the total ionic equation, we need to separate the strong electrolytes into their ions while keeping insoluble compounds (such as Fe(OH)₃) in their molecular form.
The total ionic equation is:
\[
3 \, \text{Ba}^{2+}(aq) + 6 \, \text{OH}^-(aq) + 2 \, \text{Fe}^{3+}(aq) + 3 \, \text{SO}_4^{2-}(aq) \rightarrow 2 \, \text{Fe(OH)}_3(s) + 3 \, \text{Ba}^{2+}(aq) + 3 \, \text{SO}_4^{2-}(aq)
\]
### Step 2: Write the Net Ionic Equation
Now, we remove the spectator ions. The spectator ions in this reaction are \( \text{Ba}^{2+} \) and \( \text{SO}_4^{2-} \), which appear on both sides of the equation:
The net ionic equation is:
\[
6 \, \text{OH}^-(aq) + 2 \, \text{Fe}^{3+}(aq) \rightarrow 2 \, \text{Fe(OH)}_3(s)
\]
### Summary
**Total Ionic Equation:**
\[
3 \, \text{Ba}^{2+}(aq) + 6 \, \text{OH}^-(aq) + 2 \, \text{Fe}^{3+}(aq) + 3 \, \text{SO}_4^{2-}(aq) \rightarrow 2 \, \text{Fe(OH)}_3(s) + 3 \, \text{Ba}^{2+}(aq) + 3 \, \text{SO}_4^{2-}(aq)
\]
**Net Ionic Equation:**
\[
6 \, \text{OH}^-(aq) + 2 \, \text{Fe}^{3+}(aq) \rightarrow 2 \, \text{Fe(OH)}_3(s)
\]
The balanced molecular equation is:
\[
3 \, \text{Ba(OH)}_2(aq) + \text{Fe}_2(\text{SO}_4)_3(aq) \rightarrow 2 \, \text{Fe(OH)}_3(s) + 3 \, \text{BaSO}_4(aq)
\]
### Step 1: Write the Total Ionic Equation
To write the total ionic equation, we need to separate the strong electrolytes into their ions while keeping insoluble compounds (such as Fe(OH)₃) in their molecular form.
The total ionic equation is:
\[
3 \, \text{Ba}^{2+}(aq) + 6 \, \text{OH}^-(aq) + 2 \, \text{Fe}^{3+}(aq) + 3 \, \text{SO}_4^{2-}(aq) \rightarrow 2 \, \text{Fe(OH)}_3(s) + 3 \, \text{Ba}^{2+}(aq) + 3 \, \text{SO}_4^{2-}(aq)
\]
### Step 2: Write the Net Ionic Equation
Now, we remove the spectator ions. The spectator ions in this reaction are \( \text{Ba}^{2+} \) and \( \text{SO}_4^{2-} \), which appear on both sides of the equation:
The net ionic equation is:
\[
6 \, \text{OH}^-(aq) + 2 \, \text{Fe}^{3+}(aq) \rightarrow 2 \, \text{Fe(OH)}_3(s)
\]
### Summary
**Total Ionic Equation:**
\[
3 \, \text{Ba}^{2+}(aq) + 6 \, \text{OH}^-(aq) + 2 \, \text{Fe}^{3+}(aq) + 3 \, \text{SO}_4^{2-}(aq) \rightarrow 2 \, \text{Fe(OH)}_3(s) + 3 \, \text{Ba}^{2+}(aq) + 3 \, \text{SO}_4^{2-}(aq)
\]
**Net Ionic Equation:**
\[
6 \, \text{OH}^-(aq) + 2 \, \text{Fe}^{3+}(aq) \rightarrow 2 \, \text{Fe(OH)}_3(s)
\]