This worksheet provides twenty distinct chemical reactions for students to practice balancing atoms on both sides of the equation.

Let's go through each of the 20 chemical equations and balance them step by step. The goal is to ensure that the number of atoms of each element is the same on both sides of the equation (conservation of mass).

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1) Na + O₂ → Na₂O

- Left: Na = 1, O = 2
- Right: Na = 2, O = 1
- Balance O: Use 1 O₂ → 2 Na₂O → need 4 Na on left
- So: 4Na + O₂ → 2Na₂O ✔

> Answer: 4Na + O₂ → 2Na₂O

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2) P + O₂ → P₂O₃

- Left: P = 1, O = 2
- Right: P = 2, O = 3
- Balance P: 2P
- Balance O: LCM of 2 and 3 is 6 → use 3O₂ (6 O), 2P₂O₃ (6 O)
- So: 4P + 3O₂ → 2P₂O₃ ✔

> Answer: 4P + 3O₂ → 2P₂O₃

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3) Sb₂S₃ + HCl → SbCl₃ + H₂S

- Left: Sb = 2, S = 3, H = 1, Cl = 1
- Right: Sb = 1, Cl = 3, H = 2, S = 1
- Balance Sb: 2SbCl₃ → need 2Sb on left → already have
- Balance S: 3H₂S → need 3S on right → match with Sb₂S₃
- So: Sb₂S₃ → 2SbCl₃ + 3H₂S
- Now balance Cl: 2SbCl₃ has 6 Cl → need 6 HCl
- H: 6 HCl → 6 H → gives 3H₂S (6 H) ✔

> Answer: Sb₂S₃ + 6HCl → 2SbCl₃ + 3H₂S

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4) NH₃ + H₂SO₄ → (NH₄)₂SO₄

- Left: N = 1, H = 3 + 2 = 5, S = 1, O = 4
- Right: N = 2, H = 8, S = 1, O = 4
- Need 2N on left → 2NH₃
- Then H: 2NH₃ → 6H, H₂SO₄ → 2H → total 8H → matches (NH₄)₂SO₄
- S and O balanced.

> Answer: 2NH₃ + H₂SO₄ → (NH₄)₂SO₄

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5) CuO + HCl → CuCl₂ + H₂O

- Left: Cu = 1, O = 1, H = 1, Cl = 1
- Right: Cu = 1, Cl = 2, H = 2, O = 1
- Cl and H unbalanced → need 2HCl
- So: CuO + 2HCl → CuCl₂ + H₂O ✔

> Answer: CuO + 2HCl → CuCl₂ + H₂O

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6) AgNO₃ + H₂S → Ag₂S + HNO₃

- Left: Ag = 1, N = 1, O = 3, H = 2, S = 1
- Right: Ag = 2, S = 1, H = 1, N = 1, O = 3
- Ag: need 2AgNO₃
- H: 2H₂S → 4H → but only 1 H in HNO₃ → need 2HNO₃
- So: 2AgNO₃ + H₂S → Ag₂S + 2HNO₃
- Check: Ag = 2, N = 2, O = 6, H = 2, S = 1 → right side: Ag₂S → Ag=2,S=1; 2HNO₃ → H=2,N=2,O=6 ✔

> Answer: 2AgNO₃ + H₂S → Ag₂S + 2HNO₃

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7) Cu + S → Cu₂S

- Left: Cu = 1, S = 1
- Right: Cu = 2, S = 1
- So: 2Cu + S → Cu₂S ✔

> Answer: 2Cu + S → Cu₂S

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8) Al + H₃PO₄ → H₂ + AlPO₄

- Left: Al = 1, H = 3, P = 1, O = 4
- Right: H = 2, Al = 1, P = 1, O = 4
- H mismatch: 3 vs 2 → need common multiple
- LCM of 3 and 2 is 6 → use 2H₃PO₄ → 6H, 2P, 8O
- Right: need 2AlPO₄ → 2Al, 2P, 8O
- H: 2H₃PO₄ → 6H → produces 3H₂
- Al: need 2Al on left
- So: 2Al + 2H₃PO₄ → 3H₂ + 2AlPO₄ ✔

> Answer: 2Al + 2H₃PO₄ → 3H₂ + 2AlPO₄

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9) NaNO₃ → NaNO₂ + O₂

- Left: Na = 1, N = 1, O = 3
- Right: Na = 1, N = 1, O = 2 + 2 = 4? Wait: NaNO₂ has 2 O, O₂ has 2 → total 4 O → too many
- But left has only 3 O → so not balanced
- Try: 2NaNO₃ → 2NaNO₂ + O₂
- Left: 2Na, 2N, 6O
- Right: 2Na, 2N, 4O (in 2NaNO₂) + 2O (in O₂) = 6O ✔

> Answer: 2NaNO₃ → 2NaNO₂ + O₂

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10) Mg(ClO₃)₂ → MgCl₂ + O₂

- Left: Mg = 1, Cl = 2, O = 6
- Right: Mg = 1, Cl = 2, O = 2 (only in O₂) → need 6 O → 3O₂
- So: Mg(ClO₃)₂ → MgCl₂ + 3O₂ ✔

> Answer: Mg(ClO₃)₂ → MgCl₂ + 3O₂

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11) H₂O₂ → H₂O + O₂

- Left: H = 2, O = 2
- Right: H = 2, O = 1 + 2 = 3 → unbalanced
- Try 2H₂O₂ → 2H₂O + O₂
- Left: H = 4, O = 4
- Right: 2H₂O → H=4, O=2; O₂ → O=2 → total O=4 ✔

> Answer: 2H₂O₂ → 2H₂O + O₂

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12) BaO₂ → BaO + O₂

- Left: Ba = 1, O = 2
- Right: Ba = 1, O = 1 + 2 = 3 → too many
- Try: 2BaO₂ → 2BaO + O₂
- Left: Ba = 2, O = 4
- Right: 2BaO → O=2, O₂ → O=2 → total O=4 ✔

> Answer: 2BaO₂ → 2BaO + O₂

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13) Pb(NO₃)₂ + KCl → PbCl₂ + KNO₃

- Left: Pb = 1, N = 2, O = 6, K = 1, Cl = 1
- Right: Pb = 1, Cl = 2, K = 1, N = 1, O = 3
- Need 2KNO₃ → so 2K, 2N, 6O
- So need 2KCl → 2K, 2Cl
- So: Pb(NO₃)₂ + 2KCl → PbCl₂ + 2KNO₃ ✔

> Answer: Pb(NO₃)₂ + 2KCl → PbCl₂ + 2KNO₃

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14) P + O₂ → P₂O₅

- Left: P = 1, O = 2
- Right: P = 2, O = 5
- Need 4P → 2P₂O₅ → 4P, 10O
- O₂ needed: 5O₂ → 10O
- So: 4P + 5O₂ → 2P₂O₅ ✔

> Answer: 4P + 5O₂ → 2P₂O₅

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15) NH₄NO₂ → N₂ + H₂O

- Left: N = 2, H = 4, O = 2
- Right: N = 2, H = 2, O = 1 → unbalanced
- H and O: need 2H₂O → H = 4, O = 2
- So: NH₄NO₂ → N₂ + 2H₂O ✔

> Answer: NH₄NO₂ → N₂ + 2H₂O

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16) H₂ + N₂ → NH₃

- Left: H = 2, N = 2
- Right: N = 1, H = 3
- LCM: 6H → use 3H₂ (6H), 2N₂ (4N)? No — need N₂ → 1N₂ → 2N
- Want 2N on right → 2NH₃ → N = 2, H = 6
- So need 3H₂ → 6H
- So: 3H₂ + N₂ → 2NH₃ ✔

> Answer: 3H₂ + N₂ → 2NH₃

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17) Cl₂ + KBr → KCl + Br₂

- Left: Cl = 2, K = 1, Br = 1
- Right: K = 1, Cl = 1, Br = 2
- Cl: need 2KCl → 2Cl
- Br: need 2KBr → 2Br → Br₂
- K: 2KBr → 2K → 2KCl
- So: Cl₂ + 2KBr → 2KCl + Br₂ ✔

> Answer: Cl₂ + 2KBr → 2KCl + Br₂

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18) BaCl₂ + (NH₄)₂CO₃ → BaCO₃ + NH₄Cl

- Left: Ba = 1, Cl = 2, N = 2, H = 8, C = 1, O = 3
- Right: Ba = 1, C = 1, O = 3, N = 1, H = 4, Cl = 1
- Need 2NH₄Cl → 2N, 8H, 2Cl
- So: BaCl₂ + (NH₄)₂CO₃ → BaCO₃ + 2NH₄Cl ✔

> Answer: BaCl₂ + (NH₄)₂CO₃ → BaCO₃ + 2NH₄Cl

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19) MgCO₃ + HCl → MgCl₂ + CO₂ + H₂O

- Left: Mg = 1, C = 1, O = 3, H = 1, Cl = 1
- Right: Mg = 1, Cl = 2, C = 1, O = 2 + 1 = 3, H = 2
- H and Cl: need 2HCl → 2H, 2Cl
- H₂O needs 2H → OK
- So: MgCO₃ + 2HCl → MgCl₂ + CO₂ + H₂O ✔

> Answer: MgCO₃ + 2HCl → MgCl₂ + CO₂ + H₂O

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20) H₂CO₃ → H₂O + CO₂

- Left: H = 2, C = 1, O = 3
- Right: H = 2, O = 1 + 2 = 3, C = 1
- Already balanced ✔

> Answer: H₂CO₃ → H₂O + CO₂

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## ✔ Final Balanced Equations:

1. 4Na + O₂ → 2Na₂O
2. 4P + 3O₂ → 2P₂O₃
3. Sb₂S₃ + 6HCl → 2SbCl₃ + 3H₂S
4. 2NH₃ + H₂SO₄ → (NH₄)₂SO₄
5. CuO + 2HCl → CuCl₂ + H₂O
6. 2AgNO₃ + H₂S → Ag₂S + 2HNO₃
7. 2Cu + S → Cu₂S
8. 2Al + 2H₃PO₄ → 3H₂ + 2AlPO₄
9. 2NaNO₃ → 2NaNO₂ + O₂
10. Mg(ClO₃)₂ → MgCl₂ + 3O₂
11. 2H₂O₂ → 2H₂O + O₂
12. 2BaO₂ → 2BaO + O₂
13. Pb(NO₃)₂ + 2KCl → PbCl₂ + 2KNO₃
14. 4P + 5O₂ → 2P₂O₅
15. NH₄NO₂ → N₂ + 2H₂O
16. 3H₂ + N₂ → 2NH₃
17. Cl₂ + 2KBr → 2KCl + Br₂
18. BaCl₂ + (NH₄)₂CO₃ → BaCO₃ + 2NH₄Cl
19. MgCO₃ + 2HCl → MgCl₂ + CO₂ + H₂O
20. H₂CO₃ → H₂O + CO₂

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📌 Summary:

Balancing chemical equations involves adjusting coefficients so that the number of atoms of each element is equal on both sides. Common strategies include:
- Start with elements that appear in only one compound on each side.
- Balance polyatomic ions as a group if they remain intact.
- Save hydrogen and oxygen for last (especially in combustion or acid-base reactions).
- Use the least common multiple when needed.

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