Balancing Equations Worksheet featuring 25 chemical reactions to balance, including synthesis, decomposition, single replacement, double replacement, and combustion types.

Let's go through each of the chemical equations and balance them step by step. The goal is to ensure that the number of atoms of each element is equal on both sides of the equation (conservation of mass). We'll place small whole-number coefficients in front of the formulas.

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Synthesis Reactions

1. Ba + O₂ → BaO
- Left: 1 Ba, 2 O
- Right: 1 Ba, 1 O
- Balance O: put 2 BaO → now 2 O on right
- Now need 2 Ba on left
- Answer:
2 Ba + 1 O₂ → 2 BaO

2. C + S → CS₂
- Left: 1 C, 1 S
- Right: 1 C, 2 S
- Need 2 S on left → 2 S
- Answer:
1 C + 2 S → 1 CS₂

3. Li + O₂ → Li₂O
- Left: 1 Li, 2 O
- Right: 2 Li, 1 O
- Balance O: 2 Li₂O → 2 O, 4 Li
- So need 4 Li on left
- O₂ already has 2 O → OK
- Answer:
4 Li + 1 O₂ → 2 Li₂O

4. Mg + N₂ → Mg₃N₂
- Left: 1 Mg, 2 N
- Right: 3 Mg, 2 N
- Need 3 Mg on left
- Answer:
3 Mg + 1 N₂ → 1 Mg₃N₂

5. FeCl₂ + Cl₂ → FeCl₃
- Left: 1 Fe, 2 Cl from FeCl₂ + 2 Cl from Cl₂ = 4 Cl
- Right: 1 Fe, 3 Cl
- Need to balance Fe and Cl
- Try 2 FeCl₂ → 2 Fe, 4 Cl
- Add 1 Cl₂ → total 6 Cl
- Then 2 FeCl₃ → 2 Fe, 6 Cl
- Works!
- Answer:
2 FeCl₂ + 1 Cl₂ → 2 FeCl₃

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Decomposition or Analysis

6. KClO₃ → KCl + O₂
- Left: 1 K, 1 Cl, 3 O
- Right: 1 K, 1 Cl, 2 O
- O needs balancing → 2 KClO₃ → 6 O
- Then 3 O₂ → 6 O
- So 2 KClO₃ → 2 KCl + 3 O₂
- Answer:
2 KClO₃ → 2 KCl + 3 O₂

7. Ag₂O → Ag + O₂
- Left: 2 Ag, 1 O
- Right: 1 Ag, 2 O
- Balance O: 2 Ag₂O → 2 O
- Then O₂ → 1 O₂
- Ag: 4 Ag on left → 4 Ag on right
- Answer:
2 Ag₂O → 4 Ag + 1 O₂

8. CuCO₃ → CuO + CO₂
- Already balanced:
Left: 1 Cu, 1 C, 3 O
Right: 1 Cu, 1 O (from CuO) + 1 C, 2 O (from CO₂) = 1 Cu, 1 C, 3 O
- Answer:
1 CuCO₃ → 1 CuO + 1 CO₂

9. AuBr₃ → Au + Br₂
- Left: 1 Au, 3 Br
- Right: 1 Au, 2 Br
- Need even number of Br → use 2 AuBr₃ → 2 Au, 6 Br
- Then 3 Br₂ → 6 Br
- Answer:
2 AuBr₃ → 2 Au + 3 Br₂

10. UF₄ → U + F₂
- Left: 1 U, 4 F
- Right: 1 U, 2 F
- Need 2 F₂ → 4 F
- Answer:
1 UF₄ → 1 U + 2 F₂

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Single Replacement

11. Zn + HCl → ZnCl₂ + H₂
- Left: 1 Zn, 1 H, 1 Cl
- Right: 1 Zn, 2 Cl, 2 H
- Need 2 HCl → 2 H, 2 Cl
- Then H₂ → 2 H
- Answer:
1 Zn + 2 HCl → 1 ZnCl₂ + 1 H₂

12. Zn + CuSO₄ → ZnSO₄ + Cu
- Already balanced: 1 Zn, 1 Cu, 1 S, 4 O on both sides
- Answer:
1 Zn + 1 CuSO₄ → 1 ZnSO₄ + 1 Cu

13. Cu + AgNO₃ → Cu(NO₃)₂ + Ag
- Left: 1 Cu, 1 Ag, 1 N, 3 O
- Right: 1 Cu, 2 N, 6 O, 1 Ag
- Need 2 AgNO₃ → 2 Ag, 2 N, 6 O
- Then 2 Ag on right
- Answer:
1 Cu + 2 AgNO₃ → 1 Cu(NO₃)₂ + 2 Ag

14. K + H₂O → KOH + H₂
- Left: 1 K, 2 H, 1 O
- Right: 1 K, 1 O, 1 H (in KOH), plus 2 H in H₂ → total 3 H
- Not balanced
- Try 2 K → 2 K
- 2 H₂O → 4 H, 2 O
- 2 KOH → 2 K, 2 O, 2 H
- H₂ → 2 H
- Total H: 2 (from KOH) + 2 (from H₂) = 4 H → matches
- Answer:
2 K + 2 H₂O → 2 KOH + 1 H₂

15. Al + CuCl₂ → AlCl₃ + Cu
- Left: 1 Al, 1 Cu, 2 Cl
- Right: 1 Al, 3 Cl, 1 Cu
- Balance Cl: LCM of 2 and 3 is 6
- Use 3 CuCl₂ → 3 Cu, 6 Cl
- 2 AlCl₃ → 2 Al, 6 Cl
- Then 2 Al on left
- 3 Cu on right
- Answer:
2 Al + 3 CuCl₂ → 2 AlCl₃ + 3 Cu

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Double Replacement

16. BaCl₂ + Na₂SO₄ → NaCl + BaSO₄
- Left: 1 Ba, 2 Cl, 2 Na, 1 S, 4 O
- Right: 1 Na, 1 Cl, 1 Ba, 1 S, 4 O
- Need 2 NaCl → 2 Na, 2 Cl
- Answer:
1 BaCl₂ + 1 Na₂SO₄ → 2 NaCl + 1 BaSO₄

17. ZnCl₂ + (NH₄)₂S → NH₄Cl + ZnS
- Left: 1 Zn, 2 Cl, 2 N, 8 H, 1 S
- Right: 1 N, 4 H, 1 Cl, 1 Zn, 1 S
- Need 2 NH₄Cl → 2 N, 8 H, 2 Cl
- Answer:
1 ZnCl₂ + 1 (NH₄)₂S → 2 NH₄Cl + 1 ZnS

18. NaOH + HCl → NaCl + H₂O
- Already balanced:
1 Na, 1 O, 1 H (from NaOH), 1 H, 1 Cl → all match
- Answer:
1 NaOH + 1 HCl → 1 NaCl + 1 H₂O

19. FeS + HCl → FeCl₂ + H₂S
- Left: 1 Fe, 1 S, 1 H, 1 Cl
- Right: 1 Fe, 2 Cl, 2 H, 1 S
- Need 2 HCl → 2 H, 2 Cl
- Then H₂S → 2 H, 1 S
- Answer:
1 FeS + 2 HCl → 1 FeCl₂ + 1 H₂S

20. AlCl₃ + NaOH → Al(OH)₃ + NaCl
- Left: 1 Al, 3 Cl, 1 Na, 1 O, 1 H
- Right: 1 Al, 3 O, 3 H, 1 Na, 1 Cl
- Balance Cl: 3 NaCl → 3 Cl
- So need 3 NaOH → 3 Na, 3 O, 3 H
- Al(OH)₃ → 3 OH groups → 3 O, 3 H
- All good
- Answer:
1 AlCl₃ + 3 NaOH → 1 Al(OH)₃ + 3 NaCl

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Combustion Reactions

General form: Hydrocarbon + O₂ → CO₂ + H₂O

21. CH₄ + O₂ → CO₂ + H₂O
- Left: 1 C, 4 H, 2 O
- Right: 1 C, 2 O, 2 H, 1 O → total 3 O, 2 H
- Balance H: 2 H₂O → 4 H, 2 O
- Now right: 1 C, 2 O (from CO₂) + 2 O (from H₂O) = 4 O
- So need 2 O₂ → 4 O
- Answer:
1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O

22. C₂H₆ + O₂ → CO₂ + H₂O
- Left: 2 C, 6 H
- Right: 1 C, 2 H
- Need 2 CO₂ → 2 C, 4 O
- 3 H₂O → 6 H, 3 O
- Total O on right: 4 + 3 = 7 O → so need 7/2 O₂ → not integer
- Multiply entire equation by 2:
2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O
- Answer:
2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

23. C₃H₆ + O₂ → CO₂ + H₂O
- Left: 3 C, 6 H
- Right: 1 C, 2 H
- Need 3 CO₂ → 3 C, 6 O
- 3 H₂O → 6 H, 3 O
- Total O: 6 + 3 = 9 O → need 9/2 O₂ → multiply by 2
- 2 C₃H₆ + 9 O₂ → 6 CO₂ + 6 H₂O
- Answer:
2 C₃H₆ + 9 O₂ → 6 CO₂ + 6 H₂O

24. C₄H₈ + O₂ → CO₂ + H₂O
- Left: 4 C, 8 H
- Need 4 CO₂ → 4 C, 8 O
- 4 H₂O → 8 H, 4 O
- Total O: 8 + 4 = 12 → need 6 O₂
- Answer:
1 C₄H₈ + 6 O₂ → 4 CO₂ + 4 H₂O

25. C₅H₁₀ + O₂ → CO₂ + H₂O
- Left: 5 C, 10 H
- Need 5 CO₂ → 5 C, 10 O
- 5 H₂O → 10 H, 5 O
- Total O: 10 + 5 = 15 → need 15/2 O₂ → multiply by 2
- 2 C₅H₁₀ + 15 O₂ → 10 CO₂ + 10 H₂O
- Answer:
2 C₅H₁₀ + 15 O₂ → 10 CO₂ + 10 H₂O

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✔ Final Balanced Equations:

| # | Equation |
|---|--------|
| 1 | 2 Ba + 1 O₂ → 2 BaO |
| 2 | 1 C + 2 S → 1 CS₂ |
| 3 | 4 Li + 1 O₂ → 2 Li₂O |
| 4 | 3 Mg + 1 N₂ → 1 Mg₃N₂ |
| 5 | 2 FeCl₂ + 1 Cl₂ → 2 FeCl₃ |
| 6 | 2 KClO₃ → 2 KCl + 3 O₂ |
| 7 | 2 Ag₂O → 4 Ag + 1 O₂ |
| 8 | 1 CuCO₃ → 1 CuO + 1 CO₂ |
| 9 | 2 AuBr₃ → 2 Au + 3 Br₂ |
|10 | 1 UF₄ → 1 U + 2 F₂ |
|11 | 1 Zn + 2 HCl → 1 ZnCl₂ + 1 H₂ |
|12 | 1 Zn + 1 CuSO₄ → 1 ZnSO₄ + 1 Cu |
|13 | 1 Cu + 2 AgNO₃ → 1 Cu(NO₃)₂ + 2 Ag |
|14 | 2 K + 2 H₂O → 2 KOH + 1 H₂ |
|15 | 2 Al + 3 CuCl₂ → 2 AlCl₃ + 3 Cu |
|16 | 1 BaCl₂ + 1 Na₂SO₄ → 2 NaCl + 1 BaSO₄ |
|17 | 1 ZnCl₂ + 1 (NH₄)₂S → 2 NH₄Cl + 1 ZnS |
|18 | 1 NaOH + 1 HCl → 1 NaCl + 1 H₂O |
|19 | 1 FeS + 2 HCl → 1 FeCl₂ + 1 H₂S |
|20 | 1 AlCl₃ + 3 NaOH → 1 Al(OH)₃ + 3 NaCl |
|21 | 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O |
|22 | 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O |
|23 | 2 C₃H₆ + 9 O₂ → 6 CO₂ + 6 H₂O |
|24 | 1 C₄H₈ + 6 O₂ → 4 CO₂ + 4 H₂O |
|25 | 2 C₅H₁₀ + 15 O₂ → 10 CO₂ + 10 H₂O |

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✔ All equations are now balanced.
Each one satisfies the Law of Conservation of Mass — atoms of each element are equal on both sides.

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