50 Balancing Equations Worksheet Answers | Chemical equation ... - Free Printable

Let's solve the Balancing Chemical Equations Worksheet step by step and then classify each reaction type.

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🔧 Step 1: Balancing Each Equation

We'll balance each chemical equation by ensuring the number of atoms of each element is equal on both sides.

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#### 1.
Zn (s) + AgNO₃ (aq) → Zn(NO₃)₂ (aq) + Ag (s)

- Zn: 1 → 1 ✔
- Ag: 1 → 1 ✘ (Right has 1 Ag, left has 1 in AgNO₃)
- NO₃: 3 → 6 ✘

Balance:
- Use 2 AgNO₃ to get 2 Ag and 2 NO₃ groups
- Then Zn(NO₃)₂ has 2 NO₃ → so Zn = 1
- Ag on right must be 2

✔ Balanced:
Zn + 2AgNO₃ → Zn(NO₃)₂ + 2Ag

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#### 2.
N₂ (g) + H₂ (g) → NH₃ (g)

- N: 2 → 1 ✘
- H: 2 → 3 ✘

Use 2NH₃ → 2N, 6H
Then N₂ gives 2N → OK
H₂ needs 3 molecules to give 6H

✔ Balanced:
N₂ + 3H₂ → 2NH₃

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#### 3.
NaCl (aq) + AgC₂H₃O₂ (aq) → NaC₂H₃O₂ (aq) + AgCl (s)

This is a double displacement.

All elements are already balanced as written:

- Na: 1 → 1
- Cl: 1 → 1
- Ag: 1 → 1
- C₂H₃O₂: 1 → 1

✔ Already balanced:
NaCl + AgC₂H₃O₂ → NaC₂H₃O₂ + AgCl

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#### 4.
Mg(OH)₂ (aq) + H₃PO₄ (aq) → H₂O (l) + Mg₃(PO₄)₂ (aq)

- Mg: 1 → 3 → need 3 Mg(OH)₂
- PO₄: 1 → 2 → need 2 H₃PO₄
- H: from OH and H₃PO₄
- O: check later

Try:
- 3 Mg(OH)₂
- 2 H₃PO₄
- Products: Mg₃(PO₄)₂ (has 3 Mg, 2 PO₄)
- Water: H comes from 3×2 = 6 H from OH, and 2×3 = 6 H from H₃PO₄ → total 12 H → 6 H₂O

✔ Balanced:
3Mg(OH)₂ + 2H₃PO₄ → 6H₂O + Mg₃(PO₄)₂

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#### 5.
HNO₃ (aq) + Ni (s) → Ni(NO₃)₂ (aq) + H₂ (g)

- Ni: 1 → 1
- NO₃: 1 → 2 → need 2 HNO₃
- H: 2 → 2 → H₂ formed

So:
- 2HNO₃ → 2H⁺ and 2NO₃⁻
- Ni → Ni²⁺ → Ni(NO₃)₂
- 2H⁺ → H₂

✔ Balanced:
2HNO₃ + Ni → Ni(NO₃)₂ + H₂

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#### 6.
Ba(HCO₃)₂ (s) → BaCO₃ (s) + H₂O (g) + CO₂ (g)

Decomposition of bicarbonate.

- Ba: 1 → 1
- H: 2 → 2 (in H₂O)
- C: 2 → 1 in BaCO₃ + 1 in CO₂ → 2 ✔
- O: 6 → 3 (BaCO₃) + 1 (H₂O) + 2 (CO₂) = 6 ✔

But we have:
- Ba(HCO₃)₂ → BaCO₃ + H₂O + CO₂

Wait: HCO₃⁻ decomposes: 2HCO₃⁻ → CO₃²⁻ + H₂O + CO₂

So yes:
- 1 Ba(HCO₃)₂ → 1 BaCO₃ + 1 H₂O + 1 CO₂

✔ Balanced:
Ba(HCO₃)₂ → BaCO₃ + H₂O + CO₂

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#### 7.
BaCl₂ (aq) + Na₂SO₄ (aq) → NaCl (aq) + BaSO₄ (s)

Double displacement.

- Ba: 1 → 1
- Cl: 2 → 1 → need 2 NaCl
- Na: 2 → 2 → OK
- SO₄: 1 → 1

✔ Balanced:
BaCl₂ + Na₂SO₄ → 2NaCl + BaSO₄

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#### 8.
Al₂(CO₃)₃ (s) → Al₂O₃ (s) + CO₂ (g)

Decomposition of carbonate.

- Al: 2 → 2
- C: 3 → 3 → need 3 CO₂
- O: 9 → 3 (in Al₂O₃) + 6 (in 3CO₂) = 9 ✔

✔ Balanced:
Al₂(CO₃)₃ → Al₂O₃ + 3CO₂

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#### 9.
Ca (s) + H₂O (l) → Ca(OH)₂ (aq) + H₂ (g)

- Ca: 1 → 1
- H: 2 → 2 in Ca(OH)₂ + 2 in H₂? Wait:
- Left: 2H
- Right: Ca(OH)₂ has 2H, H₂ has 2H → total 4H → too many

So need 2H₂O → 4H

Then:
- Ca + 2H₂O → Ca(OH)₂ + H₂

Now:
- Ca: 1 → 1
- H: 4 → 2 (in OH) + 2 (in H₂) = 4 ✔
- O: 2 → 2 (in OH) ✔

✔ Balanced:
Ca + 2H₂O → Ca(OH)₂ + H₂

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#### 10.
LiHCO₃ (s) → Li₂CO₃ (s) + H₂O (g) + CO₂ (g)

Decomposition of bicarbonate.

- Li: 1 → 2 → need 2 LiHCO₃
- H: 2 → 2 (in H₂O)
- C: 2 → 1 (in CO₃) + 1 (in CO₂) = 2 ✔
- O: 6 → 3 (Li₂CO₃) + 1 (H₂O) + 2 (CO₂) = 6 ✔

So:
2LiHCO₃ → Li₂CO₃ + H₂O + CO₂

✔ Balanced.

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#### 11.
N₂ (g) + O₂ (g) → N₂O₅ (g)

- N: 2 → 2
- O: 2 → 5 → need 5/2 O₂ → use fractions?

Better: multiply by 2

→ 2N₂ + 5O₂ → 2N₂O₅

Check:
- N: 4 → 4
- O: 10 → 10

✔ Balanced:
2N₂ + 5O₂ → 2N₂O₅

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#### 12.
MgBr₂ (aq) + KOH (aq) → KBr (aq) + Mg(OH)₂ (s)

- Mg: 1 → 1
- Br: 2 → 1 → need 2 KBr
- K: 2 → 2 → need 2 KOH
- OH: 2 → 2 → OK

✔ Balanced:
MgBr₂ + 2KOH → 2KBr + Mg(OH)₂

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#### 13.
Mn (s) + CuCl (aq) → Cu (s) + MnCl₂ (s)

Wait: CuCl is not common; usually CuCl₂. But let’s assume it's CuCl.

But CuCl would give Cu⁺, but Mn → Mn²⁺

So likely it should be CuCl₂.

But assuming given is correct: CuCl

Then:
- Mn → Mn²⁺ → MnCl₂ → needs 2 Cl
- So need 2 CuCl

Then:
- 2CuCl → 2Cu + 2Cl⁻
- Mn → Mn²⁺ + 2e⁻
- 2Cu⁺ + 2e⁻ → 2Cu

So:
Mn + 2CuCl → 2Cu + MnCl₂

✔ Balanced:
Mn + 2CuCl → 2Cu + MnCl₂

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#### 14.
Zn (s) + S₈ (s) → ZnS (s)

S₈ is elemental sulfur.

- Zn: 1 → 1
- S: 8 → 1 → need 8 ZnS

So:
8Zn + S₈ → 8ZnS

✔ Balanced.

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#### 15.
NaOH (aq) + H₂SO₄ (aq) → H₂O (l) + Na₂SO₄ (aq)

Acid-base neutralization.

- H₂SO₄ is diprotic → needs 2 NaOH
- Na: 2 → 2
- H: 2 (from NaOH) + 2 (from H₂SO₄) = 4H → 2 H₂O
- SO₄: 1 → 1

✔ Balanced:
2NaOH + H₂SO₄ → 2H₂O + Na₂SO₄

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#### 16.
K (s) + H₂O (l) → KOH (aq) + H₂ (g)

- K: 1 → 1
- H: 2 → 1 (in KOH) + 2 (in H₂)? → total 3H → no

Better: 2K + 2H₂O → 2KOH + H₂

Check:
- K: 2 → 2
- H: 4 → 2 (in 2KOH) + 2 (in H₂) = 4 ✔
- O: 2 → 2 ✔

✔ Balanced:
2K + 2H₂O → 2KOH + H₂

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#### 17.
C₅H₁₂ (l) + O₂ (g) → H₂O (g) + CO₂ (g)

Combustion of pentane.

C₅H₁₂ → 5CO₂ + 6H₂O

Balance O:

- Right: 5×2 + 6×1 = 10 + 6 = 16 O → need 8 O₂

So:
C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O

✔ Balanced.

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#### 18.
KOH (aq) + H₂CO₃ (aq) → H₂O (l) + K₂CO₃ (aq)

Acid-base.

H₂CO₃ is diprotic → needs 2 KOH

So:
2KOH + H₂CO₃ → 2H₂O + K₂CO₃

✔ Balanced.

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#### 19.
C₄H₈O₂ (l) + O₂ (g) → H₂O (g) + CO₂ (g)

Assume this is a combustion reaction.

C₄H₈O₂ → 4CO₂ + 4H₂O (since H: 8 → 4H₂O)

Now balance O:

Left: O from fuel = 2, plus O₂
Right: 4×2 = 8 O in CO₂ + 4×1 = 4 O in H₂O → total 12 O

So O₂ provides 10 O → 5 O₂

So:
C₄H₈O₂ + 5O₂ → 4CO₂ + 4H₂O

Check:
- C: 4 → 4
- H: 8 → 8
- O: 2 + 10 = 12 → 8 + 4 = 12 ✔

✔ Balanced.

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#### 20.
Al (s) + S₈ (s) → Al₂S₃ (s)

- Al: 1 → 2 → need 2 Al
- S: 8 → 3 → need LCM of 8 and 3 = 24

So:
- 24 Al → 12 Al₂S₃ (since 12×2=24 Al)
- S₈: 24 S → 3 S₈

So:
24Al + 3S₈ → 12Al₂S₃

Simplify: divide by 3 →
8Al + S₈ → 4Al₂S₃

✔ Balanced:
8Al + S₈ → 4Al₂S₃

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## ✔ Final Balanced Equations:

1. Zn + 2AgNO₃ → Zn(NO₃)₂ + 2Ag
2. N₂ + 3H₂ → 2NH₃
3. NaCl + AgC₂H₃O₂ → NaC₂H₃O₂ + AgCl
4. 3Mg(OH)₂ + 2H₃PO₄ → 6H₂O + Mg₃(PO₄)₂
5. 2HNO₃ + Ni → Ni(NO₃)₂ + H₂
6. Ba(HCO₃)₂ → BaCO₃ + H₂O + CO₂
7. BaCl₂ + Na₂SO₄ → 2NaCl + BaSO₄
8. Al₂(CO₃)₃ → Al₂O₃ + 3CO₂
9. Ca + 2H₂O → Ca(OH)₂ + H₂
10. 2LiHCO₃ → Li₂CO₃ + H₂O + CO₂
11. 2N₂ + 5O₂ → 2N₂O₅
12. MgBr₂ + 2KOH → 2KBr + Mg(OH)₂
13. Mn + 2CuCl → 2Cu + MnCl₂
14. 8Zn + S₈ → 8ZnS
15. 2NaOH + H₂SO₄ → 2H₂O + Na₂SO₄
16. 2K + 2H₂O → 2KOH + H₂
17. C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O
18. 2KOH + H₂CO₃ → 2H₂O + K₂CO₃
19. C₄H₈O₂ + 5O₂ → 4CO₂ + 4H₂O
20. 8Al + S₈ → 4Al₂S₃

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## 🧪 Types of Reactions

The five main types of chemical reactions:

1. Synthesis (Combination)
A + B → AB
Example: 2H₂ + O₂ → 2H₂O

2. Decomposition
AB → A + B
Example: 2H₂O → 2H₂ + O₂

3. Single Replacement (Displacement)
A + BC → AC + B
Example: Zn + 2HCl → ZnCl₂ + H₂

4. Double Replacement (Metathesis)
AB + CD → AD + CB
Example: AgNO₃ + NaCl → AgCl + NaNO₃

5. Combustion
Fuel + O₂ → CO₂ + H₂O
Example: CH₄ + 2O₂ → CO₂ + 2H₂O

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## 🔍 Classify Each Reaction:

| # | Reaction Type |
|---|---------------|
| 1 | Single Replacement |
| 2 | Synthesis |
| 3 | Double Replacement |
| 4 | Double Replacement |
| 5 | Single Replacement |
| 6 | Decomposition |
| 7 | Double Replacement |
| 8 | Decomposition |
| 9 | Single Replacement |
|10 | Decomposition |
|11 | Synthesis |
|12 | Double Replacement |
|13 | Single Replacement |
|14 | Synthesis |
|15 | Double Replacement |
|16 | Single Replacement |
|17 | Combustion |
|18 | Double Replacement |
|19 | Combustion |
|20 | Synthesis |

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## ✔ Final Answers:

TYPES OF REACTION (Fill in the blanks):

1. Synthesis – e.g., 2H₂ + O₂ → 2H₂O
2. Decomposition – e.g., 2H₂O → 2H₂ + O₂
3. Single Replacement – e.g., Zn + 2HCl → ZnCl₂ + H₂
4. Double Replacement – e.g., AgNO₃ + NaCl → AgCl + NaNO₃
5. Combustion – e.g., CH₄ + 2O₂ → CO₂ + 2H₂O

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Reaction Types for Each Equation:

1. Single Replacement
2. Synthesis
3. Double Replacement
4. Double Replacement
5. Single Replacement
6. Decomposition
7. Double Replacement
8. Decomposition
9. Single Replacement
10. Decomposition
11. Synthesis
12. Double Replacement
13. Single Replacement
14. Synthesis
15. Double Replacement
16. Single Replacement
17. Combustion
18. Double Replacement
19. Combustion
20. Synthesis

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✔ All equations balanced and classified!

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