Balancing Chemical Equations - Double Replacement Reactions by ... - Free Printable

Let's go through each reaction one by one, predict the products, determine if a reaction occurs, and balance the equations where applicable. We'll classify the reactions as double replacement, combustion, single replacement, or acid-base as appropriate.

---

1. Ag₂SO₄ + NaNO₃ →

- Double replacement:
Ag₂SO₄(aq) + 2NaNO₃(aq) → 2AgNO₃(aq) + Na₂SO₄(aq)

- Check solubility:
- AgNO₃ is soluble.
- Na₂SO₄ is soluble.
- No precipitate forms.

✔ No reaction (all products are soluble).

> Answer: No reaction

---

2. AlCl₃ + (NH₄)₃PO₄ →

- Double replacement:
AlCl₃(aq) + (NH₄)₃PO₄(aq) → AlPO₄(s) + NH₄Cl(aq)

- AlPO₄ is insoluble (phosphates of Al³⁺ are insoluble).
- NH₄Cl is soluble.

So a precipitate forms.

Balance:
AlCl₃ + (NH₄)₃PO₄ → AlPO₄(s) + 3NH₄Cl

> Balanced equation:
> AlCl₃(aq) + (NH₄)₃PO₄(aq) → AlPO₄(s) + 3NH₄Cl(aq)

---

3. C₃H₆ + O₂ →

- Combustion of hydrocarbon (C₃H₆ is propene or cyclopropane).

General combustion:
Hydrocarbon + O₂ → CO₂ + H₂O

C₃H₆ + O₂ → CO₂ + H₂O

Balance:
C₃H₆ + 9/2 O₂ → 3CO₂ + 3H₂O
Multiply by 2 to eliminate fraction:

2C₃H₆ + 9O₂ → 6CO₂ + 6H₂O

> Balanced equation:
> 2C₃H₆ + 9O₂ → 6CO₂ + 6H₂O

---

4. Cu(NO₃)₂ + NH₄CN →

- Double replacement:
Cu(NO₃)₂(aq) + 2NH₄CN(aq) → Cu(CN)₂(s) + 2NH₄NO₃(aq)

- Cu(CN)₂ is insoluble (cyanides of most metals are insoluble, especially transition metals like Cu²⁺).

So a precipitate forms.

Balance:
Cu(NO₃)₂ + 2NH₄CN → Cu(CN)₂(s) + 2NH₄NO₃

> Balanced equation:
> Cu(NO₃)₂(aq) + 2NH₄CN(aq) → Cu(CN)₂(s) + 2NH₄NO₃(aq)

---

5. Pb(NO₃)₂ + Cu(OH)₂ →

- Both in aqueous state? But Cu(OH)₂ is insoluble — it's not aqueous.

But assuming both are dissolved (as per instruction: "assume both reactants are in aqueous state"), but Cu(OH)₂ is not soluble, so it won't dissolve.

Wait — this is a problem. If Cu(OH)₂ is solid, then it’s not reacting in solution.

But let's assume we're forcing it into aqueous form for the exercise.

Double replacement:
Pb(NO₃)₂(aq) + Cu(OH)₂(s) → Pb(OH)₂(s) + Cu(NO₃)₂(aq)

Now check solubility:
- Pb(OH)₂ is insoluble (hydroxides of Pb²⁺ are insoluble).
- Cu(NO₃)₂ is soluble.

So a precipitate may form, but since Cu(OH)₂ is already a solid, the reaction may not proceed well.

But from a chemical standpoint, a double displacement reaction can occur if ions exchange.

However, since both Pb(OH)₂ and Cu(OH)₂ are insoluble, and NO₃⁻ salts are soluble, the net effect might be no net change.

But actually, Pb(OH)₂ is less soluble than Cu(OH)₂? Not really — both are insoluble.

But Pb²⁺ has higher tendency to form precipitates with OH⁻.

But in reality, no significant reaction occurs because both hydroxides are solids.

Alternatively, if both are dissolved (hypothetically), then:

Pb(NO₃)₂(aq) + Cu(OH)₂(aq) → Pb(OH)₂(s) + Cu(NO₃)₂(aq)

But Cu(OH)₂ is not aqueous — it’s insoluble.

So unless we dissolve it, no reaction.

✔ No reaction — Cu(OH)₂ does not dissolve, so no ion exchange.

> Answer: No reaction

---

6. AgNO₃ + MgCl₂ →

- Double replacement:
AgNO₃(aq) + MgCl₂(aq) → AgCl(s) + Mg(NO₃)₂(aq)

- AgCl is insoluble (chlorides of Ag⁺ are insoluble).
- Mg(NO₃)₂ is soluble.

So precipitate forms.

Balance:
2AgNO₃ + MgCl₂ → 2AgCl(s) + Mg(NO₃)₂

> Balanced equation:
> 2AgNO₃(aq) + MgCl₂(aq) → 2AgCl(s) + Mg(NO₃)₂(aq)

---

7. C₄H₈ + O₂ →

- Combustion of hydrocarbon (butene or cyclobutane).

C₄H₈ + O₂ → CO₂ + H₂O

Balance:
C₄H₈ + 6O₂ → 4CO₂ + 4H₂O

Check atoms:
- C: 4 = 4
- H: 8 = 8
- O: 12 = 8 + 4 = 12 ✔

> Balanced equation:
> C₄H₈ + 6O₂ → 4CO₂ + 4H₂O

---

8. NaF + CaBr₂ →

- Double replacement:
2NaF(aq) + CaBr₂(aq) → CaF₂(s) + 2NaBr(aq)

- CaF₂ is insoluble (fluorides of Ca²⁺ are insoluble).
- NaBr is soluble.

So precipitate forms.

> Balanced equation:
> 2NaF(aq) + CaBr₂(aq) → CaF₂(s) + 2NaBr(aq)

---

9. AlCl₃ + Ag₂SO₄ →

- Double replacement:
2AlCl₃(aq) + 3Ag₂SO₄(aq) → Al₂(SO₄)₃(aq) + 6AgCl(s)

- AgCl is insoluble.
- Al₂(SO₄)₃ is soluble.

So AgCl precipitate forms.

Balance:
2AlCl₃ + 3Ag₂SO₄ → Al₂(SO₄)₃ + 6AgCl

> Balanced equation:
> 2AlCl₃(aq) + 3Ag₂SO₄(aq) → Al₂(SO₄)₃(aq) + 6AgCl(s)

---

10. LiOH + Fe(NO₃)₃ →

- Double replacement:
3LiOH(aq) + Fe(NO₃)₃(aq) → Fe(OH)₃(s) + 3LiNO₃(aq)

- Fe(OH)₃ is insoluble (hydroxides of Fe³⁺ are insoluble).
- LiNO₃ is soluble.

Precipitate forms.

> Balanced equation:
> 3LiOH(aq) + Fe(NO₃)₃(aq) → Fe(OH)₃(s) + 3LiNO₃(aq)

---

11. NaOH + H₃PO₄ →

- Acid-base neutralization (H₃PO₄ is triprotic acid).

Can form different salts depending on ratio.

But typically, we write the complete neutralization:

3NaOH + H₃PO₄ → Na₃PO₄ + 3H₂O

> Balanced equation:
> 3NaOH(aq) + H₃PO₄(aq) → Na₃PO₄(aq) + 3H₂O(l)

(Other products like NaH₂PO₄ or Na₂HPO₄ possible, but full neutralization assumed.)

---

12. KOH + MgCl₂ →

- Double replacement:
2KOH(aq) + MgCl₂(aq) → Mg(OH)₂(s) + 2KCl(aq)

- Mg(OH)₂ is insoluble.
- KCl is soluble.

Precipitate forms.

> Balanced equation:
> 2KOH(aq) + MgCl₂(aq) → Mg(OH)₂(s) + 2KCl(aq)

---

13. Li₂O + FeCl₃ →

- This is a combination or double replacement?

Li₂O is basic oxide, reacts with water, but here with FeCl₃.

In aqueous solution:
Li₂O + H₂O → 2LiOH

Then: 3LiOH + FeCl₃ → Fe(OH)₃(s) + 3LiCl

So overall:
Li₂O + FeCl₃ → ?

But better to write stepwise.

But directly:
3Li₂O + 2FeCl₃ + 3H₂O → 2Fe(OH)₃(s) + 6LiCl

But that includes water.

Alternatively, assume both in aqueous solution.

Li₂O dissolves to give LiOH.

So: 3LiOH + FeCl₃ → Fe(OH)₃(s) + 3LiCl

So the net reaction is:

Li₂O + FeCl₃ → ?

To balance:
3Li₂O + 2FeCl₃ + 3H₂O → 2Fe(OH)₃(s) + 6LiCl

But that's messy.

Better: since Li₂O → 2LiOH, then:

3LiOH + FeCl₃ → Fe(OH)₃(s) + 3LiCl

So for Li₂O: need 3/2 Li₂O → 3LiOH

So:
(3/2)Li₂O + FeCl₃ → Fe(OH)₃(s) + 3LiCl

Multiply by 2:

3Li₂O + 2FeCl₃ → 2Fe(OH)₃(s) + 6LiCl

But wait — we need water! Oxide needs water to form hydroxide.

So unless water is present, no reaction.

But in aqueous solution, yes.

So assume water is present.

Thus, reaction occurs.

> Balanced equation:
> 3Li₂O(s) + 2FeCl₃(aq) + 3H₂O(l) → 2Fe(OH)₃(s) + 6LiCl(aq)

But if we just write as given:

It's implied aqueous, so:

Li₂O(s) + FeCl₃(aq) → ?

But Li₂O will react with water first.

So best to say: reaction occurs.

But perhaps simplify.

Standard approach: treat as double replacement after dissolution.

But since Li₂O isn’t ionic like salt, it's better to consider:

Li₂O + H₂O → 2LiOH

Then 3LiOH + FeCl₃ → Fe(OH)₃↓ + 3LiCl

So overall:
3Li₂O + 2FeCl₃ + 3H₂O → 2Fe(OH)₃(s) + 6LiCl

But since water is not listed, maybe we ignore it.

Alternatively, accept that a reaction occurs, forming Fe(OH)₃ precipitate.

So:

> Balanced equation (assuming water present):
> 3Li₂O(s) + 2FeCl₃(aq) + 3H₂O(l) → 2Fe(OH)₃(s) + 6LiCl(aq)

But if you must write without water, it's incomplete.

For simplicity, many textbooks write:

3Li₂O + 2FeCl₃ → 2Fe(OH)₃ + 6LiCl — but this is incorrect without water.

So better to say:

> Reaction occurs, but requires water.
> Final answer: 3Li₂O(s) + 2FeCl₃(aq) + 3H₂O(l) → 2Fe(OH)₃(s) + 6LiCl(aq)

---

14. C₂H₂ + O₂ →

- Combustion of acetylene.

C₂H₂ + O₂ → CO₂ + H₂O

Balance:
C₂H₂ + 5/2 O₂ → 2CO₂ + H₂O

Multiply by 2:

2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O

> Balanced equation:
> 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O

---

15. BaS + NH₄Cl →

- Double replacement:
BaS(aq) + 2NH₄Cl(aq) → BaCl₂(aq) + (NH₄)₂S(aq)

Now check solubility:
- BaCl₂ is soluble.
- (NH₄)₂S is soluble.
- All products are soluble.

No precipitate, no gas, no weak electrolyte.

So no reaction.

> Answer: No reaction

---

## ✔ Final Answers:

1. No reaction
2. AlCl₃(aq) + (NH₄)₃PO₄(aq) → AlPO₄(s) + 3NH₄Cl(aq)
3. 2C₃H₆ + 9O₂ → 6CO₂ + 6H₂O
4. Cu(NO₃)₂(aq) + 2NH₄CN(aq) → Cu(CN)₂(s) + 2NH₄NO₃(aq)
5. No reaction
6. 2AgNO₃(aq) + MgCl₂(aq) → 2AgCl(s) + Mg(NO₃)₂(aq)
7. C₄H₈ + 6O₂ → 4CO₂ + 4H₂O
8. 2NaF(aq) + CaBr₂(aq) → CaF₂(s) + 2NaBr(aq)
9. 2AlCl₃(aq) + 3Ag₂SO₄(aq) → Al₂(SO₄)₃(aq) + 6AgCl(s)
10. 3LiOH(aq) + Fe(NO₃)₃(aq) → Fe(OH)₃(s) + 3LiNO₃(aq)
11. 3NaOH(aq) + H₃PO₄(aq) → Na₃PO₄(aq) + 3H₂O(l)
12. 2KOH(aq) + MgCl₂(aq) → Mg(OH)₂(s) + 2KCl(aq)
13. 3Li₂O(s) + 2FeCl₃(aq) + 3H₂O(l) → 2Fe(OH)₃(s) + 6LiCl(aq)
14. 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
15. No reaction

---

Let me know if you want the answers formatted neatly for printing!