Practice worksheet for balancing chemical equations and writing net ionic equations.

Let's solve each of these net ionic equations step by step. The goal is to:

1. Balance the molecular equation.
2. Write the complete ionic equation, showing all strong electrolytes as dissociated ions.
3. Cancel out spectator ions (ions that appear unchanged on both sides).
4. Write the net ionic equation.

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1) Mg(NO₃)₂(aq) + Na₂CrO₄(aq) → MgCrO₄(s) + NaNO₃(aq)

Balanced Molecular Equation:
Mg(NO₃)₂(aq) + Na₂CrO₄(aq) → MgCrO₄(s) + 2NaNO₃(aq)

Complete Ionic Equation:
Mg²⁺(aq) + 2NO₃⁻(aq) + 2Na⁺(aq) + CrO₄²⁻(aq) → MgCrO₄(s) + 2Na⁺(aq) + 2NO₃⁻(aq)

Cancel spectator ions (Na⁺ and NO₃⁻):
Mg²⁺(aq) + CrO₄²⁻(aq) → MgCrO₄(s)

✔ Net Ionic Equation:
Mg²⁺(aq) + CrO₄²⁻(aq) → MgCrO₄(s)

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2) Na₃PO₄(aq) + Ni(ClO₄)₂(s) → Ni₃(PO₄)₂(s) + NaClO₄(aq)

Note: Ni(ClO₄)₂ is solid, so it does not dissociate.

Balance the equation:
2Na₃PO₄(aq) + 3Ni(ClO₄)₂(s) → Ni₃(PO₄)₂(s) + 6NaClO₄(aq)

Complete Ionic Equation:
6Na⁺(aq) + 2PO₄³⁻(aq) + 3Ni(ClO₄)₂(s) → Ni₃(PO₄)₂(s) + 6Na⁺(aq) + 6ClO₄⁻(aq)

But since Ni(ClO₄)₂ is solid, it stays intact.

So:
6Na⁺(aq) + 2PO₄³⁻(aq) + 3Ni(ClO₄)₂(s) → Ni₃(PO₄)₂(s) + 6Na⁺(aq) + 6ClO₄⁻(aq)

Spectator ion: Na⁺

Cancel 6Na⁺ from both sides:
2PO₄³⁻(aq) + 3Ni(ClO₄)₂(s) → Ni₃(PO₄)₂(s) + 6ClO₄⁻(aq)

But ClO₄⁻ comes from solid, so it's not free. We must be careful.

Actually, since Ni(ClO₄)₂ is solid, it doesn't dissociate. So we can’t write it as ions.

Thus, in the complete ionic equation, only aqueous compounds are broken down.

So:
2Na₃PO₄(aq) → 6Na⁺(aq) + 2PO₄³⁻(aq)
3Ni(ClO₄)₂(s) → remains as solid
→ Ni₃(PO₄)₂(s) + 6NaClO₄(aq) → 6Na⁺(aq) + 6ClO₄⁻(aq)

Now full ionic:
6Na⁺(aq) + 2PO₄³⁻(aq) + 3Ni(ClO₄)₂(s) → Ni₃(PO₄)₂(s) + 6Na⁺(aq) + 6ClO₄⁻(aq)

Cancel 6Na⁺:
2PO₄³⁻(aq) + 3Ni(ClO₄)₂(s) → Ni₃(PO₄)₂(s) + 6ClO₄⁻(aq)

But this is not ideal because ClO₄⁻ appears as product but not from dissociation.

Wait — the problem says Ni(ClO₄)₂(s), so it's a solid reactant. That means it's not dissolved, so no ions.

So in net ionic, we cannot break it up.

Therefore, the net ionic should include solid Ni(ClO₄)₂ as reactant.

But typically, net ionic equations involve aqueous ions reacting to form precipitates.

Here, a solid reacts with an aqueous compound to form another solid.

So, let’s re-express:

We have:
2Na₃PO₄(aq) + 3Ni(ClO₄)₂(s) → Ni₃(PO₄)₂(s) + 6NaClO₄(aq)

Ionic form:
6Na⁺(aq) + 2PO₄³⁻(aq) + 3Ni(ClO₄)₂(s) → Ni₃(PO₄)₂(s) + 6Na⁺(aq) + 6ClO₄⁻(aq)

Cancel 6Na⁺:
2PO₄³⁻(aq) + 3Ni(ClO₄)₂(s) → Ni₃(PO₄)₂(s) + 6ClO₄⁻(aq)

But ClO₄⁻ is now a product ion, but it came from the solid — so it's not accurate.

Actually, ClO₄⁻ is released into solution when the reaction occurs, but since Ni(ClO₄)₂ is solid, it’s not fully dissociated.

This suggests that this reaction may not proceed easily, or we're assuming dissolution.

But per standard practice, if a compound is labeled (s), it's not dissociated.

So, no net ionic equation can be written with ions unless it dissolves.

But perhaps the problem intends for us to assume Ni(ClO₄)₂ is soluble? But it says (s).

Wait — nickel perchlorate is actually soluble, so likely a typo.

Check solubility rules:
- Most perchlorates are soluble.
- Ni(ClO₄)₂ is soluble.

So probably (s) is a mistake; it should be (aq).

Assuming it's (aq) instead of (s):

Corrected:
Na₃PO₄(aq) + Ni(ClO₄)₂(aq) → Ni₃(PO₄)₂(s) + NaClO₄(aq)

Balance:
2Na₃PO₄(aq) + 3Ni(ClO₄)₂(aq) → Ni₃(PO₄)₂(s) + 6NaClO₄(aq)

Ionic:
6Na⁺(aq) + 2PO₄³⁻(aq) + 3Ni²⁺(aq) + 6ClO₄⁻(aq) → Ni₃(PO₄)₂(s) + 6Na⁺(aq) + 6ClO₄⁻(aq)

Cancel spectators: Na⁺ and ClO₄⁻

Net ionic:
2PO₄³⁻(aq) + 3Ni²⁺(aq) → Ni₃(PO₄)₂(s)

✔ Net Ionic Equation:
2PO₄³⁻(aq) + 3Ni²⁺(aq) → Ni₃(PO₄)₂(s)

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3) AgNO₃(aq) + MgI₂(aq) → AgI(s) + Mg(NO₃)₂(aq)

Balance:
2AgNO₃(aq) + MgI₂(aq) → 2AgI(s) + Mg(NO₃)₂(aq)

Ionic:
2Ag⁺(aq) + 2NO₃⁻(aq) + Mg²⁺(aq) + 2I⁻(aq) → 2AgI(s) + Mg²⁺(aq) + 2NO₃⁻(aq)

Cancel spectators: Mg²⁺ and NO₃⁻

Net ionic:
Ag⁺(aq) + I⁻(aq) → AgI(s)

✔ Net Ionic Equation:
Ag⁺(aq) + I⁻(aq) → AgI(s)

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4) Ca(OH)₂(aq) + FeCl₃(aq) → Fe(OH)₃(s) + CaCl₂(aq)

Balance:
3Ca(OH)₂(aq) + 2FeCl₃(aq) → 2Fe(OH)₃(s) + 3CaCl₂(aq)

Ionic:
3Ca²⁺(aq) + 6OH⁻(aq) + 2Fe³⁺(aq) + 6Cl⁻(aq) → 2Fe(OH)₃(s) + 3Ca²⁺(aq) + 6Cl⁻(aq)

Cancel spectators: Ca²⁺ and Cl⁻

Net ionic:
Fe³⁺(aq) + 3OH⁻(aq) → Fe(OH)₃(s)

✔ Net Ionic Equation:
Fe³⁺(aq) + 3OH⁻(aq) → Fe(OH)₃(s)

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5) RbF(aq) + CuSO₄(aq) → Rb₂SO₄(aq) + CuF₂(aq)

Balance:
2RbF(aq) + CuSO₄(aq) → Rb₂SO₄(aq) + CuF₂(aq)

All compounds are soluble (except check CuF₂).

But CuF₂ is slightly soluble, but generally considered soluble enough to stay in solution.

So all are aqueous.

Ionic:
2Rb⁺(aq) + 2F⁻(aq) + Cu²⁺(aq) + SO₄²⁻(aq) → 2Rb⁺(aq) + SO₄²⁻(aq) + Cu²⁺(aq) + 2F⁻(aq)

Everything cancels — no reaction?

But wait: is there a precipitate?

Check solubility:
- All salts here are soluble.
- No precipitate, no gas, no weak electrolyte.

So no net reaction.

But the equation shows products as aqueous — so it's just a mixture.

✔ Net ionic: No reaction (all ions remain in solution)

But if we write it formally:

Net ionic:
No net reaction (or NR)

But let's see: is CuF₂ insoluble?

Actually, CuF₂ is soluble — yes, it's soluble.

So no precipitate.

✔ Net Ionic Equation:
No reaction (NR)

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6) (NH₄)₂S(aq) + Co(NO₃)₂(aq) → CoS(s) + NH₄NO₃(aq)

Balance:
(NH₄)₂S(aq) + Co(NO₃)₂(aq) → CoS(s) + 2NH₄NO₃(aq)

Ionic:
2NH₄⁺(aq) + S²⁻(aq) + Co²⁺(aq) + 2NO₃⁻(aq) → CoS(s) + 2NH₄⁺(aq) + 2NO₃⁻(aq)

Cancel spectators: NH₄⁺ and NO₃⁻

Net ionic:
Co²⁺(aq) + S²⁻(aq) → CoS(s)

✔ Net Ionic Equation:
Co²⁺(aq) + S²⁻(aq) → CoS(s)

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7) Na₃PO₄(aq) + FeCl₃(aq) → FePO₄(s) + NaCl(aq)

Balance:
Na₃PO₄(aq) + FeCl₃(aq) → FePO₄(s) + 3NaCl(aq)

Ionic:
3Na⁺(aq) + PO₄³⁻(aq) + Fe³⁺(aq) + 3Cl⁻(aq) → FePO₄(s) + 3Na⁺(aq) + 3Cl⁻(aq)

Cancel spectators: Na⁺ and Cl⁻

Net ionic:
Fe³⁺(aq) + PO₄³⁻(aq) → FePO₄(s)

✔ Net Ionic Equation:
Fe³⁺(aq) + PO₄³⁻(aq) → FePO₄(s)

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8) AgNO₃(aq) + HCl(aq) → AgCl(s) + HNO₃(aq)

Balance:
AgNO₃(aq) + HCl(aq) → AgCl(s) + HNO₃(aq)

Ionic:
Ag⁺(aq) + NO₃⁻(aq) + H⁺(aq) + Cl⁻(aq) → AgCl(s) + H⁺(aq) + NO₃⁻(aq)

Cancel spectators: H⁺ and NO₃⁻

Net ionic:
Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

✔ Net Ionic Equation:
Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

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9) MgBr₂(aq) + KOH(aq) → Mg(OH)₂(s) + KBr(aq)

Balance:
MgBr₂(aq) + 2KOH(aq) → Mg(OH)₂(s) + 2KBr(aq)

Ionic:
Mg²⁺(aq) + 2Br⁻(aq) + 2K⁺(aq) + 2OH⁻(aq) → Mg(OH)₂(s) + 2K⁺(aq) + 2Br⁻(aq)

Cancel spectators: K⁺ and Br⁻

Net ionic:
Mg²⁺(aq) + 2OH⁻(aq) → Mg(OH)₂(s)

✔ Net Ionic Equation:
Mg²⁺(aq) + 2OH⁻(aq) → Mg(OH)₂(s)

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10) AlCl₃(aq) + K₃PO₄(aq) → AlPO₄(s) + KCl(aq)

Balance:
AlCl₃(aq) + K₃PO₄(aq) → AlPO₄(s) + 3KCl(aq)

Ionic:
Al³⁺(aq) + 3Cl⁻(aq) + 3K⁺(aq) + PO₄³⁻(aq) → AlPO₄(s) + 3K⁺(aq) + 3Cl⁻(aq)

Cancel spectators: K⁺ and Cl⁻

Net ionic:
Al³⁺(aq) + PO₄³⁻(aq) → AlPO₄(s)

✔ Net Ionic Equation:
Al³⁺(aq) + PO₄³⁻(aq) → AlPO₄(s)

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## ✔ Final Answers:

| Problem | Net Ionic Equation |
|--------|---------------------|
| 1 | Mg²⁺(aq) + CrO₄²⁻(aq) → MgCrO₄(s) |
| 2 | 2PO₄³⁻(aq) + 3Ni²⁺(aq) → Ni₃(PO₄)₂(s) |
| 3 | Ag⁺(aq) + I⁻(aq) → AgI(s) |
| 4 | Fe³⁺(aq) + 3OH⁻(aq) → Fe(OH)₃(s) |
| 5 | No reaction (NR) |
| 6 | Co²⁺(aq) + S²⁻(aq) → CoS(s) |
| 7 | Fe³⁺(aq) + PO₄³⁻(aq) → FePO₄(s) |
| 8 | Ag⁺(aq) + Cl⁻(aq) → AgCl(s) |
| 9 | Mg²⁺(aq) + 2OH⁻(aq) → Mg(OH)₂(s) |
| 10 | Al³⁺(aq) + PO₄³⁻(aq) → AlPO₄(s) |

> 💡 Note: For problem #5, although the reaction is written, no precipitate forms, so no net reaction occurs. All ions remain in solution.

Let me know if you want the complete ionic equations written out too!