This worksheet challenges students to balance 25 chemical equations categorized by reaction type, ranging from simple synthesis to complex combustion.

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

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 → 2 Ba
- Balanced: 2 Ba + O₂ → 2 BaO

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

3. Li + O₂ → Li₂O
- Left: 1 Li, 2 O
- Right: 2 Li, 1 O
- To balance O: put 2 Li₂O → 4 Li, 2 O
- Need 4 Li on left → 4 Li
- Balanced: 4 Li + O₂ → 2 Li₂O

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

5. FeCl₂ + Cl₂ → FeCl₃
- Left: 1 Fe, 2 Cl (from FeCl₂) + 2 Cl (from Cl₂) = 4 Cl
- Right: 1 Fe, 3 Cl
- To balance Cl: find LCM of 4 and 3 → 12
- Multiply FeCl₂ by 2 → 2 Fe, 4 Cl
- Multiply Cl₂ by 1 → 2 Cl → total Cl = 6? Wait — let’s rework.

Try:
- FeCl₂ → FeCl₃: Fe goes from +2 to +3 → oxidation
- Cl₂ provides Cl⁻

Let’s balance:
- Start with 2 FeCl₂ → 2 Fe, 4 Cl
- Need 2 FeCl₃ → 2 Fe, 6 Cl
- So need 2 more Cl → Cl₂ provides 2 Cl → 1 Cl₂
- But we already have 4 Cl from FeCl₂, plus 2 from Cl₂ → total 6 Cl → perfect
- So: 2 FeCl₂ + 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 not balanced → need even number of O on right
- 2 KClO₃ → 2 K, 2 Cl, 6 O
- Right: 2 KCl → 2 K, 2 Cl; O₂ → 3 O₂ gives 6 O
- Balanced: 2 KClO₃ → 2 KCl + 3 O₂

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

8. CuCO₃ → CuO + CO₂
- Left: 1 Cu, 1 C, 3 O
- Right: CuO → 1 Cu, 1 O; CO₂ → 1 C, 2 O → total: 1 Cu, 1 C, 3 O
- Already balanced!
- CuCO₃ → CuO + CO₂

9. AuBr₃ → Au + Br₂
- Left: 1 Au, 3 Br
- Right: 1 Au, 2 Br
- Need 3 Br → but Br₂ comes as pairs → use 3/2 Br₂
- But we want whole numbers → multiply all by 2
- 2 AuBr₃ → 2 Au + 3 Br₂
- Balanced: 2 AuBr₃ → 2 Au + 3 Br₂

10. UF₄ → U + F₂
- Left: 1 U, 4 F
- Right: 1 U, 2 F
- Need 4 F → 2 F₂
- Balanced: UF₄ → 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 has 2 H → good
- Balanced: Zn + 2 HCl → ZnCl₂ + H₂

12. Zn + CuSO₄ → ZnSO₄ + Cu
- All elements are 1:1 → already balanced
- Zn + CuSO₄ → ZnSO₄ + 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 Cu(NO₃)₂ → 1 Cu, 2 N, 6 O
- Ag: 2 Ag on right → 2 Ag
- So: Cu + 2 AgNO₃ → 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), 2 H (in H₂) → total 3 H
- Not balanced
- Try 2 K → 2 K
- 2 H₂O → 4 H, 2 O
- Products: 2 KOH → 2 K, 2 O, 2 H; H₂ → 2 H → total 4 H
- Left: 2 K, 4 H, 2 O
- Right: 2 K, 2 O, 4 H → balanced
- 2 K + 2 H₂O → 2 KOH + H₂

Wait: H₂ is diatomic → only one molecule needed for 2 H → yes
But 2 H₂O → 4 H → 2 H in KOH, 2 H in H₂ → correct
So: 2 K + 2 H₂O → 2 KOH + H₂

15. Al + CuCl₂ → AlCl₃ + Cu
- Left: 1 Al, 1 Cu, 2 Cl
- Right: 1 Al, 3 Cl, 1 Cu
- Cl mismatch
- LCM of 2 and 3 is 6
- So 3 CuCl₂ → 3 Cu, 6 Cl
- 2 AlCl₃ → 2 Al, 6 Cl
- So need 2 Al on left → 2 Al
- Cu: 3 Cu on right → 3 Cu on left
- Balanced: 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 → Na and Cl unbalanced
- Need 2 NaCl → 2 Na, 2 Cl
- So: BaCl₂ + Na₂SO₄ → 2 NaCl + 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 → not balanced
- Need 2 NH₄Cl → 2 N, 8 H, 2 Cl
- So: ZnCl₂ + (NH₄)₂S → 2 NH₄Cl + ZnS

18. NaOH + HCl → NaCl + H₂O
- All 1:1 → already balanced
- NaOH + HCl → NaCl + 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
- H₂S → 2 H, 1 S
- So: FeS + 2 HCl → FeCl₂ + 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
- Not balanced
- AlCl₃ → needs 3 NaCl → 3 Na, 3 Cl
- So need 3 NaOH → 3 Na, 3 O, 3 H
- Al(OH)₃ → 1 Al, 3 O, 3 H
- Balanced: AlCl₃ + 3 NaOH → Al(OH)₃ + 3 NaCl

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Combustion

General form: Hydrocarbon + O₂ → CO₂ + H₂O

21. CH₄ + O₂ → CO₂ + H₂O
- C: 1 = 1 → OK
- H: 4 → 2 H₂O → 4 H → OK
- O: Left: 2 O₂ → 4 O
Right: CO₂ → 2 O, 2 H₂O → 2 O → total 4 O
- So: CH₄ + 2 O₂ → CO₂ + 2 H₂O

22. C₂H₆ + O₂ → CO₂ + H₂O
- C: 2 → 2 CO₂
- H: 6 → 3 H₂O
- O: Right: 2 CO₂ → 4 O, 3 H₂O → 3 O → total 7 O → need 7/2 O₂
- Multiply by 2:
2 C₂H₆ → 4 C, 12 H
→ 4 CO₂, 6 H₂O
O: 4×2 + 6×1 = 8+6=14 O → need 7 O₂
- So: 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

23. C₃H₆ + O₂ → CO₂ + H₂O
- C: 3 → 3 CO₂
- H: 6 → 3 H₂O
- O: Right: 3×2 + 3×1 = 6+3=9 O → need 9/2 O₂
- Multiply by 2:
2 C₃H₆ → 6 C, 12 H
→ 6 CO₂, 6 H₂O
O: 6×2 + 6×1 = 12+6=18 O → 9 O₂
- So: 2 C₃H₆ + 9 O₂ → 6 CO₂ + 6 H₂O

24. C₄H₈ + O₂ → CO₂ + H₂O
- C: 4 → 4 CO₂
- H: 8 → 4 H₂O
- O: Right: 4×2 + 4×1 = 8+4=12 O → need 6 O₂
- So: C₄H₈ + 6 O₂ → 4 CO₂ + 4 H₂O

25. C₅H₁₀ + O₂ → CO₂ + H₂O
- C: 5 → 5 CO₂
- H: 10 → 5 H₂O
- O: Right: 5×2 + 5×1 = 10+5=15 O → need 15/2 O₂
- Multiply by 2:
2 C₅H₁₀ → 10 C, 20 H
→ 10 CO₂, 10 H₂O
O: 10×2 + 10×1 = 20+10=30 O → 15 O₂
- So: 2 C₅H₁₀ + 15 O₂ → 10 CO₂ + 10 H₂O

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✔ Final Balanced Equations:

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