Balancing Chemical Equations | Exercises Chemistry | Docsity - Free Printable

Let’s solve each chemical equation step by step. We need to balance them — that means making sure the number of atoms of each element is the same on both sides (left and right) of the arrow.

We’ll go one by one, carefully counting atoms and adjusting coefficients (the numbers in front of compounds) until everything balances.

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1. ___ K + ___ Br₂ → ___ KBr

Left: K = 1, Br = 2
Right: K = 1, Br = 1 → Not balanced!

Put a 2 in front of KBr:
→ K + Br₂ → 2KBr
Now Right: K=2, Br=2
Left: K=1 → put 2 in front of K
✔ 2K + Br₂ → 2KBr

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2. ___ Fe(OH)₃ → ___ Fe₂O₃ + ___ H₂O

Left: Fe=1, O=3, H=3
Right: Fe=2, O=3+1=4? Wait — Fe₂O₃ has 2Fe and 3O; H₂O has 2H and 1O.

Try putting 2 in front of Fe(OH)₃:
→ 2Fe(OH)₃ → Fe₂O₃ + H₂O
Left: Fe=2, O=6, H=6
Right: Fe=2, O=3 (from Fe₂O₃) + ? from H₂O → need 3 more O → so 3 H₂O gives 3O and 6H → perfect!
✔ 2Fe(OH)₃ → Fe₂O₃ + 3H₂O

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3. ___ Fe + ___ Cl₂ → ___ FeCl₃

Left: Fe=1, Cl=2
Right: Fe=1, Cl=3 → not balanced

Find LCM of 2 and 3 → 6
So make Cl = 6 on both sides:
→ Fe + 3Cl₂ → 2FeCl₃? Wait, now Fe: left=1, right=2 → fix Fe:
→ 2Fe + 3Cl₂ → 2FeCl₃
Check: Left: Fe=2, Cl=6; Right: Fe=2, Cl=6 ✔
✔ 2Fe + 3Cl₂ → 2FeCl₃

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4. ___ CaCO₃ → ___ CaO + ___ CO₂

Left: Ca=1, C=1, O=3
Right: Ca=1, C=1, O=1+2=3 → already balanced!
✔ CaCO₃ → CaO + CO₂

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5. ___ NaOH + ___ H₂SO₄ → ___ Na₂SO₄ + ___ H₂O

Left: Na=1, O=1+4=5? Wait — better count per compound.

NaOH: Na=1, O=1, H=1
H₂SO₄: H=2, S=1, O=4
Total left: Na=1, H=3, S=1, O=5

Right: Na₂SO₄ → Na=2, S=1, O=4
H₂O → H=2, O=1 → total right: Na=2, H=2, S=1, O=5

Need 2 Na on left → put 2 in front of NaOH:
→ 2NaOH + H₂SO₄ → Na₂SO₄ + H₂O
Left: Na=2, O=2+4=6, H=2+2=4, S=1
Right: Na=2, S=1, O=4+?= need 2 more O → so 2 H₂O → gives 4H and 2O → total O=6, H=4 ✔
✔ 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

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6. ___ Al + ___ CuSO₄ → ___ Al₂(SO₄)₃ + ___ Cu

Left: Al=1, Cu=1, S=1, O=4
Right: Al=2, S=3, O=12, Cu=1

Need 2 Al on left → 2Al
Need 3 SO₄ on left → 3CuSO₄ → then Cu=3 on left → so 3Cu on right
Check:
Left: Al=2, Cu=3, S=3, O=12
Right: Al=2, S=3, O=12, Cu=3 ✔
✔ 2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu

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7. ___ Zn + ___ HCl → ___ ZnCl₂ + ___ H₂

Left: Zn=1, H=1, Cl=1
Right: Zn=1, Cl=2, H=2 → need 2 HCl on left
→ Zn + 2HCl → ZnCl₂ + H₂
Check: Left: Zn=1, H=2, Cl=2; Right: Zn=1, Cl=2, H=2 ✔
✔ Zn + 2HCl → ZnCl₂ + H₂

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8. ___ C₃H₈ + ___ O₂ → ___ CO₂ + ___ H₂O

Left: C=3, H=8, O=?
Right: C=1, H=2, O=2+1=3? Per molecule.

Balance C first: 3 CO₂
Balance H: 8 H → need 4 H₂O (since each has 2H)
Now right: C=3, H=8, O= (3×2)+(4×1)=6+4=10
Left: O₂ → need 5 O₂ (since 5×2=10)
✔ C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

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9. ___ Fe + ___ O₂ → ___ Fe₂O₃

Left: Fe=1, O=2
Right: Fe=2, O=3

LCM for O: 6 → so 3O₂ and 2Fe₂O₃? Wait:

Try 4Fe + 3O₂ → 2Fe₂O₃
Left: Fe=4, O=6
Right: Fe=4, O=6 ✔
✔ 4Fe + 3O₂ → 2Fe₂O₃

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10. ___ Mg + ___ HCl → ___ MgCl₂ + ___ H₂

Left: Mg=1, H=1, Cl=1
Right: Mg=1, Cl=2, H=2 → need 2 HCl
→ Mg + 2HCl → MgCl₂ + H₂ ✔
✔ Mg + 2HCl → MgCl₂ + H₂

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11. ___ Na + ___ H₂O → ___ NaOH + ___ H₂

Left: Na=1, H=2, O=1
Right: Na=1, O=1, H=1+2=3? Wait — NaOH has H=1, H₂ has H=2 → total H=3 → too many.

Try 2Na + 2H₂O → 2NaOH + H₂
Left: Na=2, H=4, O=2
Right: Na=2, O=2, H=2 (from 2NaOH) + 2 (from H₂) = 4 ✔
✔ 2Na + 2H₂O → 2NaOH + H₂

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12. ___ CH₄ + ___ O₂ → ___ CO₂ + ___ H₂O

Left: C=1, H=4, O=?
Right: C=1, H=2, O=3?

Balance H: 4H → need 2 H₂O
Then right: O = 2 (from CO₂) + 2 (from 2H₂O) = 4 → so 2 O₂ on left
→ CH₄ + 2O₂ → CO₂ + 2H₂O ✔
✔ CH₄ + 2O₂ → CO₂ + 2H₂O

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13. ___ Al + ___ HCl → ___ AlCl₃ + ___ H₂

Left: Al=1, H=1, Cl=1
Right: Al=1, Cl=3, H=2

Need 3 Cl on left → 3 HCl → but then H=3, need even H for H₂ → try 6 HCl → gives 6H → 3 H₂
Then Al: need 2 Al to match 2 AlCl₃? Let's see:

2Al + 6HCl → 2AlCl₃ + 3H₂
Left: Al=2, H=6, Cl=6
Right: Al=2, Cl=6, H=6 ✔
✔ 2Al + 6HCl → 2AlCl₃ + 3H₂

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14. ___ Fe + ___ O₂ → ___ Fe₃O

Left: Fe=1, O=2
Right: Fe=3, O=4

Make Fe=3 on left → 3Fe
O: need 4 on left → 2 O₂ (since 2×2=4)
→ 3Fe + 2O₂ → Fe₃O₄ ✔
✔ 3Fe + 2O₂ → Fe₃O₄

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15. ___ NH₃ + ___ O₂ → ___ NO + ___ H₂O

Left: N=1, H=3, O=?
Right: N=1, O=1+1=2? Wait — NO has O=1, H₂O has O=1 → total O=2, H=2

Balance H: 3H on left → need multiple of 2 and 3 → LCM=6

Try 4NH₃ → H=12 → need 6 H₂O → H=12
Then N=4 → need 4 NO
O on right: 4 (from NO) + 6 (from H₂O) = 10 → so 5 O₂ on left (5×2=10)
→ 4NH₃ + 5O₂ → 4NO + 6H₂O ✔
✔ 4NH₃ + 5O₂ → 4NO + 6H₂O

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16. ___ C₂H₅OH + ___ O₂ → ___ CO₂ + ___ H₂O

Ethanol: C₂H₆O actually? Wait — C₂H₅OH is same as C₂H₆O.

Left: C=2, H=6, O=1 (in ethanol) + ? from O₂
Right: C=1, H=2, O=3?

Balance C: 2 CO₂
Balance H: 6H → 3 H₂O
Now right: O = 4 (from 2CO₂) + 3 (from 3H₂O) = 7
Left: ethanol has 1 O → so O₂ must provide 6 O → 3 O₂
→ C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O ✔
✔ C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O

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17. ___ Fe + ___ Cl₂ → ___ FeCl₂

Left: Fe=1, Cl=2
Right: Fe=1, Cl=2 → already balanced!
✔ Fe + Cl₂ → FeCl₂

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18. ___ Mg + ___ O₂ → ___ MgO

Left: Mg=1, O=2
Right: Mg=1, O=1 → need 2 MgO → then Mg=2 → so 2 Mg on left
→ 2Mg + O₂ → 2MgO ✔
✔ 2Mg + O₂ → 2MgO

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19. ___ Na + ___ Cl₂ → ___ NaCl

Left: Na=1, Cl=2
Right: Na=1, Cl=1 → need 2 NaCl → then Na=2 → so 2 Na on left
→ 2Na + Cl₂ → 2NaCl ✔
✔ 2Na + Cl₂ → 2NaCl

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20. ___ Al + ___ O₂ → ___ Al₂O₃

Left: Al=1, O=2
Right: Al=2, O=3

LCM for O: 6 → so 3 O₂ and 2 Al₂O₃? Then Al=4 on right → so 4 Al on left
→ 4Al + 3O₂ → 2Al₂O₃ ✔
✔ 4Al + 3O₂ → 2Al₂O₃

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Final Answer:
1. 2K + Br₂ → 2KBr
2. 2Fe(OH)₃ → Fe₂O₃ + 3H₂O
3. 2Fe + 3Cl₂ → 2FeCl₃
4. CaCO₃ → CaO + CO₂
5. 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
6. 2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu
7. Zn + 2HCl → ZnCl₂ + H₂
8. C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
9. 4Fe + 3O₂ → 2Fe₂O₃
10. Mg + 2HCl → MgCl₂ + H₂
11. 2Na + 2H₂O → 2NaOH + H₂
12. CH₄ + 2O₂ → CO₂ + 2H₂O
13. 2Al + 6HCl → 2AlCl₃ + 3H₂
14. 3Fe + 2O₂ → Fe₃O₄
15. 4NH₃ + 5O₂ → 4NO + 6H₂O
16. C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
17. Fe + Cl₂ → FeCl₂
18. 2Mg + O₂ → 2MgO
19. 2Na + Cl₂ → 2NaCl
20. 4Al + 3O₂ → 2Al₂O₃