Chemistry lab worksheet on reactions in aqueous solutions, featuring five problems to write balanced molecular, ionic, and net ionic equations.

Let's solve each of the five problems on your lab worksheet step by step. We will write balanced molecular, ionic, and net ionic equations for each reaction, or write NR if no reaction occurs.

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1.

Molecular Equation:
$$
2\text{NaCl}(aq) + \text{Pb(NO}_3)_2(aq) \rightarrow \text{PbCl}_2(s) + 2\text{NaNO}_3(aq)
$$

Explanation:
This is a double displacement (precipitation) reaction. Lead(II) chloride (PbCl₂) is insoluble in water and forms a precipitate.

Ionic Equation:
Break all soluble strong electrolytes into ions:
- NaCl(aq) → Na⁺(aq) + Cl⁻(aq)
- Pb(NO₃)₂(aq) → Pb²⁺(aq) + 2NO₃⁻(aq)
- NaNO₃(aq) → Na⁺(aq) + NO₃⁻(aq)
- PbCl₂(s) remains as solid (not dissociated)

So:
$$
2\text{Na}^+(aq) + 2\text{Cl}^-(aq) + \text{Pb}^{2+}(aq) + 2\text{NO}_3^-(aq) \rightarrow \text{PbCl}_2(s) + 2\text{Na}^+(aq) + 2\text{NO}_3^-(aq)
$$

Net Ionic Equation:
Cancel spectator ions (Na⁺ and NO₃⁻):
$$
\text{Pb}^{2+}(aq) + 2\text{Cl}^-(aq) \rightarrow \text{PbCl}_2(s)
$$

✔ Answer:
- Molecular: $2\text{NaCl}(aq) + \text{Pb(NO}_3)_2(aq) \rightarrow \text{PbCl}_2(s) + 2\text{NaNO}_3(aq)$
- Ionic: $2\text{Na}^+(aq) + 2\text{Cl}^-(aq) + \text{Pb}^{2+}(aq) + 2\text{NO}_3^-(aq) \rightarrow \text{PbCl}_2(s) + 2\text{Na}^+(aq) + 2\text{NO}_3^-(aq)$
- Net Ionic: $\text{Pb}^{2+}(aq) + 2\text{Cl}^-(aq) \rightarrow \text{PbCl}_2(s)$

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2.

Molecular Equation:
$$
\text{Mg(OH)}_2(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + 2\text{H}_2\text{O}(l)
$$

Explanation:
This is an acid-base neutralization. Mg(OH)₂ is a weak base (insoluble), and HCl is a strong acid. They react to form salt and water.

Ionic Equation:
- Mg(OH)₂(s) remains as solid (not dissolved)
- HCl(aq) → H⁺(aq) + Cl⁻(aq)
- MgCl₂(aq) → Mg²⁺(aq) + 2Cl⁻(aq)
- H₂O(l) remains as liquid

So:
$$
\text{Mg(OH)}_2(s) + 2\text{H}^+(aq) + 2\text{Cl}^-(aq) \rightarrow \text{Mg}^{2+}(aq) + 2\text{Cl}^-(aq) + 2\text{H}_2\text{O}(l)
$$

Net Ionic Equation:
Cancel spectator ions (Cl⁻):
$$
\text{Mg(OH)}_2(s) + 2\text{H}^+(aq) \rightarrow \text{Mg}^{2+}(aq) + 2\text{H}_2\text{O}(l)
$$

✔ Answer:
- Molecular: $\text{Mg(OH)}_2(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + 2\text{H}_2\text{O}(l)$
- Ionic: $\text{Mg(OH)}_2(s) + 2\text{H}^+(aq) + 2\text{Cl}^-(aq) \rightarrow \text{Mg}^{2+}(aq) + 2\text{Cl}^-(aq) + 2\text{H}_2\text{O}(l)$
- Net Ionic: $\text{Mg(OH)}_2(s) + 2\text{H}^+(aq) \rightarrow \text{Mg}^{2+}(aq) + 2\text{H}_2\text{O}(l)$

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3.

Molecular Equation:
$$
\text{Na}_2\text{CO}_3(aq) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l)
$$

Explanation:
Carbonate reacts with acid to produce CO₂ gas (gas evolution reaction).

Ionic Equation:
- Na₂CO₃(aq) → 2Na⁺(aq) + CO₃²⁻(aq)
- HCl(aq) → H⁺(aq) + Cl⁻(aq)
- NaCl(aq) → Na⁺(aq) + Cl⁻(aq)
- CO₂(g) and H₂O(l) remain as is

So:
$$
2\text{Na}^+(aq) + \text{CO}_3^{2-}(aq) + 2\text{H}^+(aq) + 2\text{Cl}^-(aq) \rightarrow 2\text{Na}^+(aq) + 2\text{Cl}^-(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l)
$$

Net Ionic Equation:
Cancel spectator ions (Na⁺ and Cl⁻):
$$
\text{CO}_3^{2-}(aq) + 2\text{H}^+(aq) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l)
$$

✔ Answer:
- Molecular: $\text{Na}_2\text{CO}_3(aq) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l)$
- Ionic: $2\text{Na}^+(aq) + \text{CO}_3^{2-}(aq) + 2\text{H}^+(aq) + 2\text{Cl}^-(aq) \rightarrow 2\text{Na}^+(aq) + 2\text{Cl}^-(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l)$
- Net Ionic: $\text{CO}_3^{2-}(aq) + 2\text{H}^+(aq) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l)$

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4.

Molecular Equation:
$$
2\text{FeCl}_3(aq) + 3\text{Mg}(s) \rightarrow 3\text{MgCl}_2(aq) + 2\text{Fe}(s)
$$

Explanation:
This is a single replacement (redox) reaction. Magnesium is more active than iron and displaces it from solution.

Ionic Equation:
- FeCl₃(aq) → Fe³⁺(aq) + 3Cl⁻(aq)
- Mg(s) → Mg(s) (solid)
- MgCl₂(aq) → Mg²⁺(aq) + 2Cl⁻(aq)
- Fe(s) → Fe(s) (solid)

So:
$$
2\text{Fe}^{3+}(aq) + 6\text{Cl}^-(aq) + 3\text{Mg}(s) \rightarrow 3\text{Mg}^{2+}(aq) + 6\text{Cl}^-(aq) + 2\text{Fe}(s)
$$

Net Ionic Equation:
Cancel spectator ions (Cl⁻):
$$
2\text{Fe}^{3+}(aq) + 3\text{Mg}(s) \rightarrow 3\text{Mg}^{2+}(aq) + 2\text{Fe}(s)
$$

✔ Answer:
- Molecular: $2\text{FeCl}_3(aq) + 3\text{Mg}(s) \rightarrow 3\text{MgCl}_2(aq) + 2\text{Fe}(s)$
- Ionic: $2\text{Fe}^{3+}(aq) + 6\text{Cl}^-(aq) + 3\text{Mg}(s) \rightarrow 3\text{Mg}^{2+}(aq) + 6\text{Cl}^-(aq) + 2\text{Fe}(s)$
- Net Ionic: $2\text{Fe}^{3+}(aq) + 3\text{Mg}(s) \rightarrow 3\text{Mg}^{2+}(aq) + 2\text{Fe}(s)$

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5.

Molecular Equation:
$$
3(\text{NH}_4)_2\text{C}_2\text{O}_4(aq) + 2\text{Al(ClO}_4)_3(aq) \rightarrow 6\text{NH}_4\text{ClO}_4(aq) + \text{Al}_2(\text{C}_2\text{O}_4)_3(s)
$$

Explanation:
Double displacement reaction. Check solubility:
- Al₂(C₂O₄)₃ (aluminum oxalate) is insoluble → precipitate forms.

Ionic Equation:
Break all soluble compounds into ions:
- (NH₄)₂C₂O₄(aq) → 2NH₄⁺(aq) + C₂O₄²⁻(aq)
- Al(ClO₄)₃(aq) → Al³⁺(aq) + 3ClO₄⁻(aq)
- NH₄ClO₄(aq) → NH₄⁺(aq) + ClO₄⁻(aq)
- Al₂(C₂O₄)₃(s) remains solid

So:
$$
6\text{NH}_4^+(aq) + 3\text{C}_2\text{O}_4^{2-}(aq) + 2\text{Al}^{3+}(aq) + 6\text{ClO}_4^-(aq) \rightarrow 6\text{NH}_4^+(aq) + 6\text{ClO}_4^-(aq) + \text{Al}_2(\text{C}_2\text{O}_4)_3(s)
$$

Net Ionic Equation:
Cancel spectator ions (NH₄⁺ and ClO₄⁻):
$$
3\text{C}_2\text{O}_4^{2-}(aq) + 2\text{Al}^{3+}(aq) \rightarrow \text{Al}_2(\text{C}_2\text{O}_4)_3(s)
$$

✔ Answer:
- Molecular: $3(\text{NH}_4)_2\text{C}_2\text{O}_4(aq) + 2\text{Al(ClO}_4)_3(aq) \rightarrow 6\text{NH}_4\text{ClO}_4(aq) + \text{Al}_2(\text{C}_2\text{O}_4)_3(s)$
- Ionic: $6\text{NH}_4^+(aq) + 3\text{C}_2\text{O}_4^{2-}(aq) + 2\text{Al}^{3+}(aq) + 6\text{ClO}_4^-(aq) \rightarrow 6\text{NH}_4^+(aq) + 6\text{ClO}_4^-(aq) + \text{Al}_2(\text{C}_2\text{O}_4)_3(s)$
- Net Ionic: $3\text{C}_2\text{O}_4^{2-}(aq) + 2\text{Al}^{3+}(aq) \rightarrow \text{Al}_2(\text{C}_2\text{O}_4)_3(s)$

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✔ Summary Table:

| Problem | Type of Reaction | Net Ionic Equation |
|--------|------------------|--------------------|
| 1 | Precipitation | $\text{Pb}^{2+}(aq) + 2\text{Cl}^-(aq) \rightarrow \text{PbCl}_2(s)$ |
| 2 | Acid-Base | $\text{Mg(OH)}_2(s) + 2\text{H}^+(aq) \rightarrow \text{Mg}^{2+}(aq) + 2\text{H}_2\text{O}(l)$ |
| 3 | Gas Evolution | $\text{CO}_3^{2-}(aq) + 2\text{H}^+(aq) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l)$ |
| 4 | Redox (Single Replacement) | $2\text{Fe}^{3+}(aq) + 3\text{Mg}(s) \rightarrow 3\text{Mg}^{2+}(aq) + 2\text{Fe}(s)$ |
| 5 | Precipitation | $3\text{C}_2\text{O}_4^{2-}(aq) + 2\text{Al}^{3+}(aq) \rightarrow \text{Al}_2(\text{C}_2\text{O}_4)_3(s)$ |

All reactions occur — no "NR" cases.

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