Net Ionic Equation Worksheet Answers showing balanced molecular, ionic, and net ionic equations for six reactions in aqueous solution.

Problem Overview:

The task involves writing balanced molecular equations, ionic equations, and net ionic equations (NIE) for a series of chemical reactions. The reactions are assumed to occur in aqueous solution. Here's a step-by-step explanation of how to solve each part:

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General Approach:

1. Molecular Equation: Write the balanced equation using the formulas of the reactants and products.
2. Ionic Equation: Break down all soluble ionic compounds into their respective ions. Include spectator ions (ions that do not participate in the reaction).
3. Net Ionic Equation (NIE): Eliminate spectator ions from the ionic equation to focus only on the ions that participate in the reaction.

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Solutions for Each Reaction:

#### 1. \( 2\text{NaCl}(aq) + \text{Pb(NO}_3\text{)}_2(aq) \rightarrow \text{PbCl}_2(s) + 2\text{NaNO}_3(aq) \)

- Molecular Equation: Already given.
\[
2\text{NaCl}(aq) + \text{Pb(NO}_3\text{)}_2(aq) \rightarrow \text{PbCl}_2(s) + 2\text{NaNO}_3(aq)
\]

- Ionic Equation: Dissociate all soluble ionic compounds into their ions.
\[
2\text{Na}^+(aq) + 2\text{Cl}^-(aq) + \text{Pb}^{2+}(aq) + 2\text{NO}_3^-(aq) \rightarrow \text{PbCl}_2(s) + 2\text{Na}^+(aq) + 2\text{NO}_3^-(aq)
\]

- Net Ionic Equation (NIE): Eliminate spectator ions (\(\text{Na}^+\) and \(\text{NO}_3^-\)).
\[
2\text{Cl}^-(aq) + \text{Pb}^{2+}(aq) \rightarrow \text{PbCl}_2(s)
\]

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#### 2. \( \text{Na}_2\text{CO}_3(aq) + \text{FeCl}_2(aq) \rightarrow \text{FeCO}_3(s) + 2\text{NaCl}(aq) \)

- Molecular Equation: Already given.
\[
\text{Na}_2\text{CO}_3(aq) + \text{FeCl}_2(aq) \rightarrow \text{FeCO}_3(s) + 2\text{NaCl}(aq)
\]

- Ionic Equation: Dissociate all soluble ionic compounds into their ions.
\[
2\text{Na}^+(aq) + \text{CO}_3^{2-}(aq) + \text{Fe}^{2+}(aq) + 2\text{Cl}^-(aq) \rightarrow \text{FeCO}_3(s) + 2\text{Na}^+(aq) + 2\text{Cl}^-(aq)
\]

- Net Ionic Equation (NIE): Eliminate spectator ions (\(\text{Na}^+\) and \(\text{Cl}^-\)).
\[
\text{CO}_3^{2-}(aq) + \text{Fe}^{2+}(aq) \rightarrow \text{FeCO}_3(s)
\]

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#### 3. \( \text{Mg(OH)}_2(aq) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + 2\text{H}_2\text{O}(l) \)

- Molecular Equation: Already given.
\[
\text{Mg(OH)}_2(aq) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + 2\text{H}_2\text{O}(l)
\]

- Ionic Equation: Dissociate all soluble ionic compounds into their ions.
\[
\text{Mg}^{2+}(aq) + 2\text{OH}^-(aq) + 2\text{H}^+(aq) + 2\text{Cl}^-(aq) \rightarrow \text{Mg}^{2+}(aq) + 2\text{Cl}^-(aq) + 2\text{H}_2\text{O}(l)
\]

- Net Ionic Equation (NIE): Eliminate spectator ions (\(\text{Mg}^{2+}\) and \(\text{Cl}^-\)).
\[
2\text{OH}^-(aq) + 2\text{H}^+(aq) \rightarrow 2\text{H}_2\text{O}(l)
\]
Simplify by dividing through by 2:
\[
\text{OH}^-(aq) + \text{H}^+(aq) \rightarrow \text{H}_2\text{O}(l)
\]

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#### 4. \( \text{K}_2\text{C}_2\text{O}_4(aq) + \text{CaCl}_2(aq) \rightarrow 2\text{KCl}(aq) + \text{CaC}_2\text{O}_4(s) \)

- Molecular Equation: Already given.
\[
\text{K}_2\text{C}_2\text{O}_4(aq) + \text{CaCl}_2(aq) \rightarrow 2\text{KCl}(aq) + \text{CaC}_2\text{O}_4(s)
\]

- Ionic Equation: Dissociate all soluble ionic compounds into their ions.
\[
2\text{K}^+(aq) + \text{C}_2\text{O}_4^{2-}(aq) + \text{Ca}^{2+}(aq) + 2\text{Cl}^-(aq) \rightarrow 2\text{K}^+(aq) + 2\text{Cl}^-(aq) + \text{CaC}_2\text{O}_4(s)
\]

- Net Ionic Equation (NIE): Eliminate spectator ions (\(\text{K}^+\) and \(\text{Cl}^-\)).
\[
\text{C}_2\text{O}_4^{2-}(aq) + \text{Ca}^{2+}(aq) \rightarrow \text{CaC}_2\text{O}_4(s)
\]

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#### 5. \( 2\text{(NH}_4\text{)}_3\text{PO}_4(aq) + 3\text{Zn(NO}_3\text{)}_2(aq) \rightarrow 6\text{NH}_4\text{NO}_3(aq) + \text{Zn}_3(\text{PO}_4)_2(s) \)

- Molecular Equation: Already given.
\[
2\text{(NH}_4\text{)}_3\text{PO}_4(aq) + 3\text{Zn(NO}_3\text{)}_2(aq) \rightarrow 6\text{NH}_4\text{NO}_3(aq) + \text{Zn}_3(\text{PO}_4)_2(s)
\]

- Ionic Equation: Dissociate all soluble ionic compounds into their ions.
\[
6\text{NH}_4^+(aq) + 2\text{PO}_4^{3-}(aq) + 3\text{Zn}^{2+}(aq) + 6\text{NO}_3^-(aq) \rightarrow 6\text{NH}_4^+(aq) + 6\text{NO}_3^-(aq) + \text{Zn}_3(\text{PO}_4)_2(s)
\]

- Net Ionic Equation (NIE): Eliminate spectator ions (\(\text{NH}_4^+\) and \(\text{NO}_3^-\)).
\[
2\text{PO}_4^{3-}(aq) + 3\text{Zn}^{2+}(aq) \rightarrow \text{Zn}_3(\text{PO}_4)_2(s)
\]

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#### 6. \( 3\text{LiOH}(aq) + \text{VCl}_3(aq) \rightarrow 3\text{LiCl}(aq) + \text{V(OH)}_3(s) \)

- Molecular Equation: Already given.
\[
3\text{LiOH}(aq) + \text{VCl}_3(aq) \rightarrow 3\text{LiCl}(aq) + \text{V(OH)}_3(s)
\]

- Ionic Equation: Dissociate all soluble ionic compounds into their ions.
\[
3\text{Li}^+(aq) + 3\text{OH}^-(aq) + \text{V}^{3+}(aq) + 3\text{Cl}^-(aq) \rightarrow 3\text{Li}^+(aq) + 3\text{Cl}^-(aq) + \text{V(OH)}_3(s)
\]

- Net Ionic Equation (NIE): Eliminate spectator ions (\(\text{Li}^+\) and \(\text{Cl}^-\)).
\[
3\text{OH}^-(aq) + \text{V}^{3+}(aq) \rightarrow \text{V(OH)}_3(s)
\]

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Final Answers:

1. NIE: \( 2\text{Cl}^-(aq) + \text{Pb}^{2+}(aq) \rightarrow \text{PbCl}_2(s) \)
2. NIE: \( \text{CO}_3^{2-}(aq) + \text{Fe}^{2+}(aq) \rightarrow \text{FeCO}_3(s) \)
3. NIE: \( \text{OH}^-(aq) + \text{H}^+(aq) \rightarrow \text{H}_2\text{O}(l) \)
4. NIE: \( \text{C}_2\text{O}_4^{2-}(aq) + \text{Ca}^{2+}(aq) \rightarrow \text{CaC}_2\text{O}_4(s) \)
5. NIE: \( 2\text{PO}_4^{3-}(aq) + 3\text{Zn}^{2+}(aq) \rightarrow \text{Zn}_3(\text{PO}_4)_2(s) \)
6. NIE: \( 3\text{OH}^-(aq) + \text{V}^{3+}(aq) \rightarrow \text{V(OH)}_3(s) \)

Boxed Final Answer:

\[
\boxed{
\begin{aligned}
1. & \quad 2\text{Cl}^-(aq) + \text{Pb}^{2+}(aq) \rightarrow \text{PbCl}_2(s) \\
2. & \quad \text{CO}_3^{2-}(aq) + \text{Fe}^{2+}(aq) \rightarrow \text{FeCO}_3(s) \\
3. & \quad \text{OH}^-(aq) + \text{H}^+(aq) \rightarrow \text{H}_2\text{O}(l) \\
4. & \quad \text{C}_2\text{O}_4^{2-}(aq) + \text{Ca}^{2+}(aq) \rightarrow \text{CaC}_2\text{O}_4(s) \\
5. & \quad 2\text{PO}_4^{3-}(aq) + 3\text{Zn}^{2+}(aq) \rightarrow \text{Zn}_3(\text{PO}_4)_2(s) \\
6. & \quad 3\text{OH}^-(aq) + \text{V}^{3+}(aq) \rightarrow \text{V(OH)}_3(s)
\end{aligned}
}
\]