Practice balancing chemical equations with this printable worksheet, ideal for chemistry students.
Balancing chemical equations worksheet with 14 incomplete reactions to solve, featuring reactants and products with blank coefficients.
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Show Answer Key & Explanations
Step-by-step solution for: Balancing equations worksheet | PDF
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Show Answer Key & Explanations
Step-by-step solution for: Balancing equations worksheet | PDF
Here's the complete solution to balancing all 14 chemical equations, with step-by-step explanations for each.
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1. _ H₂ + _ O₂ → _ H₂O
We need to balance hydrogen and oxygen.
- Left: H=2, O=2
- Right: H=2, O=1
To balance oxygen, put 2 in front of H₂O → now right has H=4, O=2
Now left needs 2 H₂ to get 4 H atoms.
✔ Balanced: 2 H₂ + 1 O₂ → 2 H₂O
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2. _ H₃PO₄ + _ KOH → _ K₃PO₄ + _ H₂O
This is an acid-base reaction.
- H₃PO₄ has 3 H⁺, so it needs 3 KOH to neutralize.
- Products: K₃PO₄ (needs 3 K) and H₂O (each KOH gives one H₂O).
Try: 1 H₃PO₄ + 3 KOH → 1 K₃PO₄ + 3 H₂O
Check atoms:
- H: 3+3=6 on left → 3×2=6 on right ✔
- P: 1=1 ✔
- O: 4+3=7 → 4+3=7 ✔
- K: 3=3 ✔
✔ Balanced: 1 H₃PO₄ + 3 KOH → 1 K₃PO₄ + 3 H₂O
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3. _ K + _ B₂O₃ → _ K₂O + _ B
Balance potassium and boron.
Left: K=1, B=2, O=3
Right: K=2, O=1, B=1
Start by balancing B: put 2 in front of B on right → now B=2
Now balance O: left has 3 O, right has 1 O per K₂O → need 3 K₂O → now K=6 on right
So need 6 K on left.
✔ Balanced: 6 K + 1 B₂O₃ → 3 K₂O + 2 B
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4. _ HCl + _ NaOH → _ NaCl + _ H₂O
Acid-base neutralization — already balanced as written.
Check:
- H: 1+1=2 → 2 in H₂O ✔
- Cl: 1=1 ✔
- Na: 1=1 ✔
- O: 1=1 ✔
✔ Balanced: 1 HCl + 1 NaOH → 1 NaCl + 1 H₂O
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5. _ Na + _ NaNO₃ → _ Na₂O + _ N₂
Sodium reduces nitrate to nitrogen gas.
Left: Na, N, O
Right: Na, O, N
Assume 2 NaNO₃ → gives 2 N, 6 O → need 3 Na₂O (to use 6 O and 6 Na)
But we also have Na metal on left.
Try: 10 Na + 2 NaNO₃ → 6 Na₂O + 1 N₂
Check:
- Na: 10 + 2 = 12 → 6×2=12 ✔
- N: 2 = 2 ✔
- O: 2×3=6 → 6×1=6 ✔
✔ Balanced: 10 Na + 2 NaNO₃ → 6 Na₂O + 1 N₂
*(Note: This is a redox reaction; sodium reduces nitrate to N₂ while being oxidized to Na₂O.)*
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6. _ C + _ S₈ → _ CS₂
Carbon disulfide formation.
S₈ has 8 sulfur atoms. Each CS₂ has 2 S atoms → need 4 CS₂ to use 8 S.
Then need 4 C atoms.
✔ Balanced: 4 C + 1 S₈ → 4 CS₂
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7. _ Na + _ O₂ → _ Na₂O₂
Sodium peroxide formation.
Na₂O₂ has 2 Na and 2 O.
Left: O₂ has 2 O → matches right.
Need 2 Na on left.
✔ Balanced: 2 Na + 1 O₂ → 1 Na₂O₂
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8. _ N₂ + _ O₂ → _ N₂O₄
Dinitrogen tetroxide.
N₂O₄ has 2 N and 4 O.
Left: N₂ gives 2 N → good. O₂ gives 2 O per molecule → need 2 O₂ to get 4 O.
✔ Balanced: 1 N₂ + 2 O₂ → 1 N₂O₄
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9. _ H₃PO₄ + _ Mg(OH)₂ → _ Mg₃(PO₄)₂ + _ H₂O
Acid-base reaction.
Mg₃(PO₄)₂ requires 3 Mg and 2 PO₄.
Each H₃PO₄ gives 1 PO₄ → need 2 H₃PO₄.
Each Mg(OH)₂ gives 1 Mg → need 3 Mg(OH)₂.
Now check H and O:
Left H: 2×3 + 3×2 = 6 + 6 = 12
Right: H₂O → need 6 H₂O to get 12 H.
O: left: 2×4 + 3×2 = 8 + 6 = 14
Right: Mg₃(PO₄)₂ has 8 O, 6 H₂O has 6 O → total 14 ✔
✔ Balanced: 2 H₃PO₄ + 3 Mg(OH)₂ → 1 Mg₃(PO₄)₂ + 6 H₂O
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10. _ NaOH + _ H₂CO₃ → _ Na₂CO₃ + _ H₂O
Acid-base reaction.
H₂CO₃ has 2 H → needs 2 NaOH.
Products: Na₂CO₃ (has 2 Na) and H₂O.
Left: 2 NaOH → 2 Na, 2 O, 2 H
H₂CO₃ → 2 H, 1 C, 3 O
Total left: Na=2, H=4, C=1, O=5
Right: Na₂CO₃ → 2 Na, 1 C, 3 O
H₂O → need 2 H₂O → 4 H, 2 O → total O=5 ✔
✔ Balanced: 2 NaOH + 1 H₂CO₃ → 1 Na₂CO₃ + 2 H₂O
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11. _ KOH + _ HBr → _ KBr + _ H₂O
Simple acid-base.
Already balanced as written.
✔ Balanced: 1 KOH + 1 HBr → 1 KBr + 1 H₂O
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12. _ H₂ + _ O₂ → _ H₂O₂
Hydrogen peroxide formation.
H₂O₂ has 2 H and 2 O.
Left: H₂ gives 2 H, O₂ gives 2 O → perfect match.
✔ Balanced: 1 H₂ + 1 O₂ → 1 H₂O₂
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13. _ Na + _ O₂ → _ Na₂O
Sodium oxide.
Na₂O has 2 Na and 1 O.
O₂ has 2 O → need 2 Na₂O to use 2 O → then need 4 Na on left.
✔ Balanced: 4 Na + 1 O₂ → 2 Na₂O
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14. _ Al(OH)₃ + _ H₂CO₃ → _ Al₂(CO₃)₃ + _ H₂O
Acid-base reaction.
Al₂(CO₃)₃ requires 2 Al and 3 CO₃.
Each Al(OH)₃ gives 1 Al → need 2 Al(OH)₃.
Each H₂CO₃ gives 1 CO₃ → need 3 H₂CO₃.
Now check H and O:
Left H: 2×3 + 3×2 = 6 + 6 = 12
→ need 6 H₂O on right → 12 H ✔
O: left: 2×3 + 3×3 = 6 + 9 = 15
Right: Al₂(CO₃)₃ has 9 O, 6 H₂O has 6 O → 15 ✔
✔ Balanced: 2 Al(OH)₃ + 3 H₂CO₃ → 1 Al₂(CO₃)₃ + 6 H₂O
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## ✔ Final Answer — All Balanced Equations:
1. 2 H₂ + 1 O₂ → 2 H₂O
2. 1 H₃PO₄ + 3 KOH → 1 K₃PO₄ + 3 H₂O
3. 6 K + 1 B₂O₃ → 3 K₂O + 2 B
4. 1 HCl + 1 NaOH → 1 NaCl + 1 H₂O
5. 10 Na + 2 NaNO₃ → 6 Na₂O + 1 N₂
6. 4 C + 1 S₈ → 4 CS₂
7. 2 Na + 1 O₂ → 1 Na₂O₂
8. 1 N₂ + 2 O₂ → 1 N₂O₄
9. 2 H₃PO₄ + 3 Mg(OH)₂ → 1 Mg₃(PO₄)₂ + 6 H₂O
10. 2 NaOH + 1 H₂CO₃ → 1 Na₂CO₃ + 2 H₂O
11. 1 KOH + 1 HBr → 1 KBr + 1 H₂O
12. 1 H₂ + 1 O₂ → 1 H₂O₂
13. 4 Na + 1 O₂ → 2 Na₂O
14. 2 Al(OH)₃ + 3 H₂CO₃ → 1 Al₂(CO₃)₃ + 6 H₂O
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Let me know if you’d like to see the oxidation states or reaction types for any of these!
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1. _ H₂ + _ O₂ → _ H₂O
We need to balance hydrogen and oxygen.
- Left: H=2, O=2
- Right: H=2, O=1
To balance oxygen, put 2 in front of H₂O → now right has H=4, O=2
Now left needs 2 H₂ to get 4 H atoms.
✔ Balanced: 2 H₂ + 1 O₂ → 2 H₂O
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2. _ H₃PO₄ + _ KOH → _ K₃PO₄ + _ H₂O
This is an acid-base reaction.
- H₃PO₄ has 3 H⁺, so it needs 3 KOH to neutralize.
- Products: K₃PO₄ (needs 3 K) and H₂O (each KOH gives one H₂O).
Try: 1 H₃PO₄ + 3 KOH → 1 K₃PO₄ + 3 H₂O
Check atoms:
- H: 3+3=6 on left → 3×2=6 on right ✔
- P: 1=1 ✔
- O: 4+3=7 → 4+3=7 ✔
- K: 3=3 ✔
✔ Balanced: 1 H₃PO₄ + 3 KOH → 1 K₃PO₄ + 3 H₂O
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3. _ K + _ B₂O₃ → _ K₂O + _ B
Balance potassium and boron.
Left: K=1, B=2, O=3
Right: K=2, O=1, B=1
Start by balancing B: put 2 in front of B on right → now B=2
Now balance O: left has 3 O, right has 1 O per K₂O → need 3 K₂O → now K=6 on right
So need 6 K on left.
✔ Balanced: 6 K + 1 B₂O₃ → 3 K₂O + 2 B
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4. _ HCl + _ NaOH → _ NaCl + _ H₂O
Acid-base neutralization — already balanced as written.
Check:
- H: 1+1=2 → 2 in H₂O ✔
- Cl: 1=1 ✔
- Na: 1=1 ✔
- O: 1=1 ✔
✔ Balanced: 1 HCl + 1 NaOH → 1 NaCl + 1 H₂O
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5. _ Na + _ NaNO₃ → _ Na₂O + _ N₂
Sodium reduces nitrate to nitrogen gas.
Left: Na, N, O
Right: Na, O, N
Assume 2 NaNO₃ → gives 2 N, 6 O → need 3 Na₂O (to use 6 O and 6 Na)
But we also have Na metal on left.
Try: 10 Na + 2 NaNO₃ → 6 Na₂O + 1 N₂
Check:
- Na: 10 + 2 = 12 → 6×2=12 ✔
- N: 2 = 2 ✔
- O: 2×3=6 → 6×1=6 ✔
✔ Balanced: 10 Na + 2 NaNO₃ → 6 Na₂O + 1 N₂
*(Note: This is a redox reaction; sodium reduces nitrate to N₂ while being oxidized to Na₂O.)*
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6. _ C + _ S₈ → _ CS₂
Carbon disulfide formation.
S₈ has 8 sulfur atoms. Each CS₂ has 2 S atoms → need 4 CS₂ to use 8 S.
Then need 4 C atoms.
✔ Balanced: 4 C + 1 S₈ → 4 CS₂
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7. _ Na + _ O₂ → _ Na₂O₂
Sodium peroxide formation.
Na₂O₂ has 2 Na and 2 O.
Left: O₂ has 2 O → matches right.
Need 2 Na on left.
✔ Balanced: 2 Na + 1 O₂ → 1 Na₂O₂
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8. _ N₂ + _ O₂ → _ N₂O₄
Dinitrogen tetroxide.
N₂O₄ has 2 N and 4 O.
Left: N₂ gives 2 N → good. O₂ gives 2 O per molecule → need 2 O₂ to get 4 O.
✔ Balanced: 1 N₂ + 2 O₂ → 1 N₂O₄
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9. _ H₃PO₄ + _ Mg(OH)₂ → _ Mg₃(PO₄)₂ + _ H₂O
Acid-base reaction.
Mg₃(PO₄)₂ requires 3 Mg and 2 PO₄.
Each H₃PO₄ gives 1 PO₄ → need 2 H₃PO₄.
Each Mg(OH)₂ gives 1 Mg → need 3 Mg(OH)₂.
Now check H and O:
Left H: 2×3 + 3×2 = 6 + 6 = 12
Right: H₂O → need 6 H₂O to get 12 H.
O: left: 2×4 + 3×2 = 8 + 6 = 14
Right: Mg₃(PO₄)₂ has 8 O, 6 H₂O has 6 O → total 14 ✔
✔ Balanced: 2 H₃PO₄ + 3 Mg(OH)₂ → 1 Mg₃(PO₄)₂ + 6 H₂O
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10. _ NaOH + _ H₂CO₃ → _ Na₂CO₃ + _ H₂O
Acid-base reaction.
H₂CO₃ has 2 H → needs 2 NaOH.
Products: Na₂CO₃ (has 2 Na) and H₂O.
Left: 2 NaOH → 2 Na, 2 O, 2 H
H₂CO₃ → 2 H, 1 C, 3 O
Total left: Na=2, H=4, C=1, O=5
Right: Na₂CO₃ → 2 Na, 1 C, 3 O
H₂O → need 2 H₂O → 4 H, 2 O → total O=5 ✔
✔ Balanced: 2 NaOH + 1 H₂CO₃ → 1 Na₂CO₃ + 2 H₂O
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11. _ KOH + _ HBr → _ KBr + _ H₂O
Simple acid-base.
Already balanced as written.
✔ Balanced: 1 KOH + 1 HBr → 1 KBr + 1 H₂O
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12. _ H₂ + _ O₂ → _ H₂O₂
Hydrogen peroxide formation.
H₂O₂ has 2 H and 2 O.
Left: H₂ gives 2 H, O₂ gives 2 O → perfect match.
✔ Balanced: 1 H₂ + 1 O₂ → 1 H₂O₂
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13. _ Na + _ O₂ → _ Na₂O
Sodium oxide.
Na₂O has 2 Na and 1 O.
O₂ has 2 O → need 2 Na₂O to use 2 O → then need 4 Na on left.
✔ Balanced: 4 Na + 1 O₂ → 2 Na₂O
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14. _ Al(OH)₃ + _ H₂CO₃ → _ Al₂(CO₃)₃ + _ H₂O
Acid-base reaction.
Al₂(CO₃)₃ requires 2 Al and 3 CO₃.
Each Al(OH)₃ gives 1 Al → need 2 Al(OH)₃.
Each H₂CO₃ gives 1 CO₃ → need 3 H₂CO₃.
Now check H and O:
Left H: 2×3 + 3×2 = 6 + 6 = 12
→ need 6 H₂O on right → 12 H ✔
O: left: 2×3 + 3×3 = 6 + 9 = 15
Right: Al₂(CO₃)₃ has 9 O, 6 H₂O has 6 O → 15 ✔
✔ Balanced: 2 Al(OH)₃ + 3 H₂CO₃ → 1 Al₂(CO₃)₃ + 6 H₂O
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## ✔ Final Answer — All Balanced Equations:
1. 2 H₂ + 1 O₂ → 2 H₂O
2. 1 H₃PO₄ + 3 KOH → 1 K₃PO₄ + 3 H₂O
3. 6 K + 1 B₂O₃ → 3 K₂O + 2 B
4. 1 HCl + 1 NaOH → 1 NaCl + 1 H₂O
5. 10 Na + 2 NaNO₃ → 6 Na₂O + 1 N₂
6. 4 C + 1 S₈ → 4 CS₂
7. 2 Na + 1 O₂ → 1 Na₂O₂
8. 1 N₂ + 2 O₂ → 1 N₂O₄
9. 2 H₃PO₄ + 3 Mg(OH)₂ → 1 Mg₃(PO₄)₂ + 6 H₂O
10. 2 NaOH + 1 H₂CO₃ → 1 Na₂CO₃ + 2 H₂O
11. 1 KOH + 1 HBr → 1 KBr + 1 H₂O
12. 1 H₂ + 1 O₂ → 1 H₂O₂
13. 4 Na + 1 O₂ → 2 Na₂O
14. 2 Al(OH)₃ + 3 H₂CO₃ → 1 Al₂(CO₃)₃ + 6 H₂O
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Let me know if you’d like to see the oxidation states or reaction types for any of these!
Parent Tip: Review the logic above to help your child master the concept of balancing equations worksheet chemistry.
