Types of Chemical Reaction Worksheet | Lecture notes Chemistry ... - Free Printable

Let's solve each of the problems step by step. We'll balance the chemical equations and then identify the type of reaction for each.

---

Part 1: Balance the reactions and identify the type

---

#### 1.
Reaction:
\[
\_\_\_\_ \text{NaBr} + \_\_\_\_ \text{Ca(OH)}_2 \rightarrow \_\_\_\_ \text{CaBr}_2 + \_\_\_\_ \text{NaOH}
\]

- Balance:
- Ca: 1 on both sides → OK
- Br: 2 on right → need 2 NaBr on left
- Na: 2 on left → need 2 NaOH on right
- O and H: Check OH groups — 2 OH on left (from Ca(OH)₂), 2 on right (from 2 NaOH) → balanced

\[
\boxed{2} \text{NaBr} + \boxed{1} \text{Ca(OH)}_2 \rightarrow \boxed{1} \text{CaBr}_2 + \boxed{2} \text{NaOH}
\]

- Type: Double displacement (metathesis)

---

#### 2.
Reaction:
\[
\_\_\_\_ \text{NH}_3 + \_\_\_\_ \text{H}_2\text{SO}_4 \rightarrow \_\_\_\_ (\text{NH}_4)_2\text{SO}_4
\]

- Balance:
- NH₄⁺ needs 2 N atoms → 2 NH₃
- SO₄²⁻ is one → 1 H₂SO₄
- Products: 1 (NH₄)₂SO₄

\[
\boxed{2} \text{NH}_3 + \boxed{1} \text{H}_2\text{SO}_4 \rightarrow \boxed{1} (\text{NH}_4)_2\text{SO}_4
\]

- Type: Acid-base neutralization (also a synthesis)

---

#### 3.
Reaction:
\[
\_\_\_\_ \text{C}_5\text{H}_9\text{O} + \_\_\_\_ \text{O}_2 \rightarrow \_\_\_\_ \text{CO}_2 + \_\_\_\_ \text{H}_2\text{O}
\]

- This is a combustion reaction.
- C₅H₉O → 5 C, 9 H, 1 O
- CO₂ and H₂O are products

Let’s balance:
- C: 5 → 5 CO₂
- H: 9 → 9/2 H₂O → so 9 H₂O → need to double everything later
- O: Left: 1 from fuel + 2×O₂ → Right: 5×2 = 10 O in CO₂ + 9 O in H₂O → total 19 O

So:
- Right: 10 (CO₂) + 9 (H₂O) = 19 O atoms
- Left: 1 O from fuel → so 18 O from O₂ → 9 O₂ molecules

But we have fractional H₂O → let's use whole numbers.

Multiply entire equation by 2:

- Fuel: 2 C₅H₉O → 10 C, 18 H, 2 O
- CO₂: 10 CO₂
- H₂O: 9 H₂O → 18 H
- O atoms: Right: 10×2 = 20 (CO₂) + 9 = 29 O
- Left: 2 O from fuel → need 27 O from O₂ → 27/2 = 13.5 → not good

Try again with coefficients.

Let’s assume:
\[
a \text{C}_5\text{H}_9\text{O} + b \text{O}_2 \rightarrow c \text{CO}_2 + d \text{H}_2\text{O}
\]

Balancing:
- C: 5a = c
- H: 9a = 2d → d = 9a/2
- O: a + 2b = 2c + d

Substitute:
- c = 5a
- d = 9a/2
- O: a + 2b = 2(5a) + 9a/2 = 10a + 4.5a = 14.5a
→ 2b = 13.5a → b = 6.75a

Set a = 4 → smallest integer that eliminates decimals

Then:
- a = 4 → C₅H₉O: 4
- c = 20 → CO₂: 20
- d = 9×4 / 2 = 18 → H₂O: 18
- b = 6.75 × 4 = 27 → O₂: 27

Check O:
- Left: 4 (from fuel) + 2×27 = 54 → total 58
- Right: 20×2 = 40 (CO₂) + 18 = 58 → OK

\[
\boxed{4} \text{C}_5\text{H}_9\text{O} + \boxed{27} \text{O}_2 \rightarrow \boxed{20} \text{CO}_2 + \boxed{18} \text{H}_2\text{O}
\]

- Type: Combustion

---

#### 4.
Reaction:
\[
\_\_\_\_ \text{Pb} + \_\_\_\_ \text{H}_3\text{PO}_4 \rightarrow \_\_\_\_ \text{H}_2 + \_\_\_\_ \text{Pb}_3(\text{PO}_4)_2
\]

- Pb₃(PO₄)₂ has 3 Pb and 2 PO₄
- So need 3 Pb and 2 H₃PO₄
- H₃PO₄ provides H → H₂ gas
- Each H₃PO₄ has 3 H → 2 H₃PO₄ → 6 H → 3 H₂

So:
\[
\boxed{3} \text{Pb} + \boxed{2} \text{H}_3\text{PO}_4 \rightarrow \boxed{3} \text{H}_2 + \boxed{1} \text{Pb}_3(\text{PO}_4)_2
\]

- Type: Single displacement

---

#### 5.
Reaction:
\[
\_\_\_\_ \text{Li}_3\text{N} + \_\_\_\_ \text{NH}_4\text{NO}_3 \rightarrow \_\_\_\_ \text{LiNO}_3 + \_\_\_\_ (\text{NH}_4)_3\text{N}
\]

This looks like a double displacement, but note: Li₃N and NH₄NO₃ exchange ions.

Products: LiNO₃ and (NH₄)₃N

Balance:
- Li: 3 on left → 3 LiNO₃ on right
- N: Let’s count carefully

Left:
- Li₃N: 1 N
- NH₄NO₃: 2 N (one in NH₄⁺, one in NO₃⁻)
→ Total N: depends on stoichiometry

Right:
- LiNO₃: 1 N per molecule
- (NH₄)₃N: 4 N (3 from NH₄⁺, 1 from N³⁻)

Let’s suppose:
- 1 Li₃N → gives 3 Li⁺ and 1 N³⁻
- To make (NH₄)₃N → need 3 NH₄⁺ and 1 N³⁻ → so 3 NH₄⁺ needed

Each NH₄NO₃ gives 1 NH₄⁺ and 1 NO₃⁻

So need 3 NH₄NO₃ → gives 3 NH₄⁺ and 3 NO₃⁻

Now:
- Li₃N: 3 Li⁺, 1 N³⁻
- 3 NH₄NO₃: 3 NH₄⁺, 3 NO₃⁻

Products:
- (NH₄)₃N: 3 NH₄⁺ and 1 N³⁻ → uses up 3 NH₄⁺ and 1 N³⁻
- Remaining: 3 Li⁺ and 3 NO₃⁻ → 3 LiNO₃

Perfect!

So:
\[
\boxed{1} \text{Li}_3\text{N} + \boxed{3} \text{NH}_4\text{NO}_3 \rightarrow \boxed{3} \text{LiNO}_3 + \boxed{1} (\text{NH}_4)_3\text{N}
\]

- Type: Double displacement

---

#### 6.
Reaction:
\[
\_\_\_\_ \text{HBr} + \_\_\_\_ \text{Al(OH)}_3 \rightarrow \_\_\_\_ \text{H}_2\text{O} + \_\_\_\_ \text{AlBr}_3
\]

- Al(OH)₃ is base, HBr is acid → acid-base reaction
- Al³⁺ combines with 3 Br⁻ → AlBr₃
- 3 H⁺ from HBr combine with 3 OH⁻ → 3 H₂O

So:
- Al(OH)₃ → 1 Al, 3 OH
- Need 3 HBr → 3 H⁺ and 3 Br⁻
- Products: AlBr₃ and 3 H₂O

\[
\boxed{3} \text{HBr} + \boxed{1} \text{Al(OH)}_3 \rightarrow \boxed{3} \text{H}_2\text{O} + \boxed{1} \text{AlBr}_3
\]

- Type: Acid-base neutralization

---

Part 2: Identify the type of reaction

---

#### 7.
\[
\text{Na}_3\text{PO}_4 + 3 \text{KOH} \rightarrow 3 \text{NaOH} + \text{K}_3\text{PO}_4
\]

- Ions swap: Na⁺ ↔ K⁺
- Double displacement

---

#### 8.
\[
\text{MgCl}_2 + \text{Li}_2\text{CO}_3 \rightarrow \text{MgCO}_3 + 2 \text{LiCl}
\]

- Exchange of ions → Mg²⁺ with Li⁺, Cl⁻ with CO₃²⁻
- Double displacement

---

#### 9.
\[
\text{C}_6\text{H}_{12} + 9 \text{O}_2 \rightarrow 6 \text{CO}_2 + 6 \text{H}_2\text{O}
\]

- Hydrocarbon + O₂ → CO₂ + H₂O
- Combustion

---

#### 10.
\[
\text{Pb} + \text{Fe(SO}_4) \rightarrow \text{Pb(SO}_4) + \text{Fe}
\]

- Pb displaces Fe from compound → Pb more reactive?
- Actually, Pb is less active than Fe → but assuming it works
- Pb replaces Fe → Single displacement

---

#### 11.
\[
\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2
\]

- One compound breaks into two → Decomposition

---

#### 12.
\[
\text{P}_4 + 3 \text{O}_2 \rightarrow 2 \text{P}_2\text{O}_3
\]

- Two substances form one → Synthesis

---

#### 13.
\[
2 \text{RbNO}_3 + \text{BeF}_2 \rightarrow \text{Be(NO}_3)_2 + 2 \text{RbF}
\]

- Rb⁺ swaps with Be²⁺ → Double displacement

---

#### 14.
\[
2 \text{AgNO}_3 + \text{Cu} \rightarrow \text{Cu(NO}_3)_2 + 2 \text{Ag}
\]

- Cu displaces Ag → Single displacement

---

#### 15.
\[
\text{C}_3\text{H}_6\text{O} + 4 \text{O}_2 \rightarrow 3 \text{CO}_2 + 3 \text{H}_2\text{O}
\]

- Organic compound burns → Combustion

---

#### 16.
\[
2 (\text{C}_5\text{H}_5) + \text{Fe} \rightarrow \text{Fe}(\text{C}_5\text{H}_5)_2
\]

- Two cyclopentadienyl radicals combine with Fe → Synthesis

---

#### 17.
\[
\text{SeCl}_6 + \text{O}_2 \rightarrow \text{SeO}_2 + 3 \text{Cl}_2
\]

- SeCl₆ decomposes? But O₂ is reactant → oxidation?

Actually:
- SeCl₆ → SeO₂ + Cl₂ → Se is oxidized? Wait: SeCl₆ has Se⁶⁺, SeO₂ has Se⁴⁺ → reduced!
- O₂ is reduced → so redox

But overall: one compound reacts with O₂ → products → Redox, but also could be seen as decomposition with oxygen.

But better classification: Redox reaction (specifically, oxidation-reduction)

Alternatively, since it involves formation of new compounds, maybe combination? No — multiple products.

It's not simple decomposition or synthesis.

Better: Redox reaction (oxidation of Cl⁻ and reduction of O₂)

But actually: SeCl₆ → SeO₂ + 3 Cl₂ → Se goes from +6 to +4 → reduced, Cl from -1 to 0 → oxidized → so Se is reduced, Cl oxidized → O₂ must be involved?

Wait: O₂ is reactant → likely O₂ oxidizes Cl⁻?

But Se is being reduced → contradiction.

Wait: SeCl₆ + O₂ → SeO₂ + 3 Cl₂

- Se: +6 in SeCl₆ → +4 in SeO₂ → reduced
- Cl: -1 → 0 → oxidized
- O: 0 → -2 → reduced

So both O₂ and Se are reduced? Impossible.

Wait: O₂ is reduced (0 → -2), Se is reduced (+6 → +4)? That can't happen unless something else is oxidized.

But Cl⁻ is oxidized (-1 → 0)

So:
- O₂: reduced (0 → -2)
- Cl⁻: oxidized (-1 → 0)
- Se: reduced (+6 → +4)

But only one species can be reduced if another is oxidized.

But here, both O₂ and Se are reduced? Contradiction.

Wait: In SeCl₆, Se is +6, Cl is -1

In SeO₂, Se is +4, O is -2

In Cl₂, Cl is 0

So:
- Se: +6 → +4 → reduced
- Cl: -1 → 0 → oxidized
- O: 0 → -2 → reduced

So two reductions and one oxidation → impossible.

Must be error.

Actually, this reaction may not be balanced.

Check atoms:

Left: Se, 6 Cl, 2 O

Right: Se, 2 O, 6 Cl → OK

But oxidation states:
- Se: +6 → +4 → gain 2e⁻
- Cl: 6 × (-1) → 0 → lose 6e⁻
- O₂: 0 → 2 × (-2) = -4 → gain 4e⁻

Total e⁻ gained: 2 (by Se) + 4 (by O) = 6e⁻
Lost: 6e⁻ by Cl → OK

So:
- Se gains 2e⁻ (reduced)
- O₂ gains 4e⁻ (reduced)
- Cl loses 6e⁻ (oxidized)

So both Se and O₂ are reduced, Cl is oxidized → possible if O₂ is reducing agent? No, O₂ is usually oxidizing.

But in this case, O₂ is acting as a reducing agent? Unlikely.

Perhaps the reaction is written incorrectly.

But chemically, SeCl₆ is unstable and can hydrolyze, but this might be a hypothetical.

Regardless, this is a redox reaction where Cl⁻ is oxidized, and both Se and O₂ are reduced? That doesn't make sense.

Wait: Se is reduced, O₂ is reduced, Cl oxidized → net: electrons go from Cl to Se and O₂ → possible if Se and O₂ act as oxidizing agents.

But Se in SeCl₆ is already highly oxidized → can’t accept more electrons.

Wait: Se(+6) → Se(+4): accepts 2e⁻ → yes, possible.

O₂(0) → O(-2): accepts 4e⁻ → yes.

Cl(-1) → Cl(0): loses 1e⁻ per atom → 6 Cl lose 6e⁻

So total: 6e⁻ lost, 6e⁻ gained → balanced.

So yes: Redox reaction with multiple oxidizing agents.

But classification: Redox (oxidation-reduction)

Also, since it forms new compounds, perhaps displacement? Not really.

Best: Redox reaction

---

#### 18.
\[
2 \text{MgI}_2 + \text{Mn(SO}_3)_2 \rightarrow 2 \text{Mg(SO}_3) + \text{MnI}_4
\]

- Mg²⁺ swaps with Mn²⁺
- I⁻ and SO₃²⁻ exchange
- Double displacement

---

#### 19.
\[
\text{O}_3 \rightarrow \text{O}^\bullet + \text{O}_2
\]

- Ozone breaks down → Decomposition

---

#### 20.
\[
2 \text{NO}_2 \rightarrow 2 \text{O}_2 + \text{N}_2
\]

- NO₂ decomposes to N₂ and O₂ → Decomposition

---

✔ Final Answers

---

#### Balanced Equations & Types:

1. \( 2 \text{NaBr} + \text{Ca(OH)}_2 \rightarrow \text{CaBr}_2 + 2 \text{NaOH} \) → Double displacement
2. \( 2 \text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4 \) → Acid-base neutralization
3. \( 4 \text{C}_5\text{H}_9\text{O} + 27 \text{O}_2 \rightarrow 20 \text{CO}_2 + 18 \text{H}_2\text{O} \) → Combustion
4. \( 3 \text{Pb} + 2 \text{H}_3\text{PO}_4 \rightarrow 3 \text{H}_2 + \text{Pb}_3(\text{PO}_4)_2 \) → Single displacement
5. \( \text{Li}_3\text{N} + 3 \text{NH}_4\text{NO}_3 \rightarrow 3 \text{LiNO}_3 + (\text{NH}_4)_3\text{N} \) → Double displacement
6. \( 3 \text{HBr} + \text{Al(OH)}_3 \rightarrow 3 \text{H}_2\text{O} + \text{AlBr}_3 \) → Acid-base neutralization

---

#### Identify Reaction Type:

7. Double displacement
8. Double displacement
9. Combustion
10. Single displacement
11. Decomposition
12. Synthesis
13. Double displacement
14. Single displacement
15. Combustion
16. Synthesis
17. Redox reaction
18. Double displacement
19. Decomposition
20. Decomposition

---

✔ All solved! Let me know if you'd like explanations for any specific ones.