Question
Write the total ionic and net ionic equation, consider the balance equation as well as stage for HCl(aq)+ Na2CO3(aq)produce H2O(l)+CO2(g)+Nacl(aq)
Answers
GPT-4o mini
To generate the total ionic and net ionic equations for the reaction between hydrochloric acid (HCl) and sodium carbonate (Na2CO3), let's start with the balanced molecular equation:
\[
2 \, \text{HCl(aq)} + \text{Na}_2\text{CO}_3\text{(aq)} \rightarrow 2 \, \text{NaCl(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2(g)
\]
### Total Ionic Equation
Next, we'll write the total ionic equation by dissociating all the strong electrolytes (the soluble ionic compounds) in the reaction.
1. **HCl** dissociates into ions:
\[
2 \, \text{HCl(aq)} \rightarrow 2 \, \text{H}^+(aq) + 2 \, \text{Cl}^-(aq)
\]
2. **Na2CO3** dissociates into ions:
\[
\text{Na}_2\text{CO}_3\text{(aq)} \rightarrow 2 \, \text{Na}^+(aq) + \text{CO}_3^{2-}(aq)
\]
3. **NaCl** dissociates into ions:
\[
2 \, \text{NaCl(aq)} \rightarrow 2 \, \text{Na}^+(aq) + 2 \, \text{Cl}^-(aq)
\]
Putting it all together, the total ionic equation is:
\[
2 \, \text{H}^+(aq) + 2 \, \text{Cl}^-(aq) + 2 \, \text{Na}^+(aq) + \text{CO}_3^{2-}(aq) \rightarrow 2 \, \text{Na}^+(aq) + 2 \, \text{Cl}^-(aq) + \text{H}_2\text{O(l)} + \text{CO}_2(g)
\]
### Net Ionic Equation
To derive the net ionic equation, we eliminate the spectator ions from the total ionic equation. The spectator ions here are Na\(^+\) and Cl\(^-\) because they appear on both sides of the equation unchanged.
After removing the spectator ions, the net ionic equation becomes:
\[
2 \, \text{H}^+(aq) + \text{CO}_3^{2-}(aq) \rightarrow \text{H}_2\text{O(l)} + \text{CO}_2(g)
\]
### Summary
**Balanced Molecular Equation:**
\[
2 \, \text{HCl(aq)} + \text{Na}_2\text{CO}_3\text{(aq)} \rightarrow 2 \, \text{NaCl(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2(g)
\]
**Total Ionic Equation:**
\[
2 \, \text{H}^+(aq) + 2 \, \text{Cl}^-(aq) + 2 \, \text{Na}^+(aq) + \text{CO}_3^{2-}(aq) \rightarrow 2 \, \text{Na}^+(aq) + 2 \, \text{Cl}^-(aq) + \text{H}_2\text{O(l)} + \text{CO}_2(g)
\]
**Net Ionic Equation:**
\[
2 \, \text{H}^+(aq) + \text{CO}_3^{2-}(aq) \rightarrow \text{H}_2\text{O(l)} + \text{CO}_2(g)
\]
\[
2 \, \text{HCl(aq)} + \text{Na}_2\text{CO}_3\text{(aq)} \rightarrow 2 \, \text{NaCl(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2(g)
\]
### Total Ionic Equation
Next, we'll write the total ionic equation by dissociating all the strong electrolytes (the soluble ionic compounds) in the reaction.
1. **HCl** dissociates into ions:
\[
2 \, \text{HCl(aq)} \rightarrow 2 \, \text{H}^+(aq) + 2 \, \text{Cl}^-(aq)
\]
2. **Na2CO3** dissociates into ions:
\[
\text{Na}_2\text{CO}_3\text{(aq)} \rightarrow 2 \, \text{Na}^+(aq) + \text{CO}_3^{2-}(aq)
\]
3. **NaCl** dissociates into ions:
\[
2 \, \text{NaCl(aq)} \rightarrow 2 \, \text{Na}^+(aq) + 2 \, \text{Cl}^-(aq)
\]
Putting it all together, the total ionic equation is:
\[
2 \, \text{H}^+(aq) + 2 \, \text{Cl}^-(aq) + 2 \, \text{Na}^+(aq) + \text{CO}_3^{2-}(aq) \rightarrow 2 \, \text{Na}^+(aq) + 2 \, \text{Cl}^-(aq) + \text{H}_2\text{O(l)} + \text{CO}_2(g)
\]
### Net Ionic Equation
To derive the net ionic equation, we eliminate the spectator ions from the total ionic equation. The spectator ions here are Na\(^+\) and Cl\(^-\) because they appear on both sides of the equation unchanged.
After removing the spectator ions, the net ionic equation becomes:
\[
2 \, \text{H}^+(aq) + \text{CO}_3^{2-}(aq) \rightarrow \text{H}_2\text{O(l)} + \text{CO}_2(g)
\]
### Summary
**Balanced Molecular Equation:**
\[
2 \, \text{HCl(aq)} + \text{Na}_2\text{CO}_3\text{(aq)} \rightarrow 2 \, \text{NaCl(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2(g)
\]
**Total Ionic Equation:**
\[
2 \, \text{H}^+(aq) + 2 \, \text{Cl}^-(aq) + 2 \, \text{Na}^+(aq) + \text{CO}_3^{2-}(aq) \rightarrow 2 \, \text{Na}^+(aq) + 2 \, \text{Cl}^-(aq) + \text{H}_2\text{O(l)} + \text{CO}_2(g)
\]
**Net Ionic Equation:**
\[
2 \, \text{H}^+(aq) + \text{CO}_3^{2-}(aq) \rightarrow \text{H}_2\text{O(l)} + \text{CO}_2(g)
\]