Grade 8 Balancing Chemical Equations Worksheets 2024 - Free Printable

Let’s go step by step to balance each chemical equation. Balancing means making sure the number of atoms of each element is the same on both sides of the arrow.

We’ll use coefficients (numbers in front) to balance — never change subscripts!

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1. Zn + AgNO₃ → Zn(NO₃)₂ + Ag

- Left: Zn=1, Ag=1, N=1, O=3
- Right: Zn=1, Ag=1, N=2, O=6 → Not balanced!
- Put 2 in front of AgNO₃ → now left: Ag=2, N=2, O=6
- Then put 2 in front of Ag on right → balances Ag
✔ Balanced: 1 Zn + 2 AgNO₃ → 1 Zn(NO₃)₂ + 2 Ag

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2. N₂ + H₂ → NH₃

- Left: N=2, H=2
- Right: N=1, H=3 → Not balanced
- Try 2 NH₃ → right: N=2, H=6
- So need 3 H₂ on left → H=6
✔ Balanced: 1 N₂ + 3 H₂ → 2 NH₃

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3. NaCl + AgC₂H₃O₂ → NaC₂H₃O₂ + AgCl

Check atoms:
- Left: Na=1, Cl=1, Ag=1, C=2, H=3, O=2
- Right: same → already balanced!
✔ Balanced: 1 NaCl + 1 AgC₂H₃O₂ → 1 NaC₂H₃O₂ + 1 AgCl

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4. Mg(OH)₂ + H₃PO₄ → H₂O + Mg₃(PO₄)₂

Right has Mg=3, PO₄=2 → so need 3 Mg(OH)₂ and 2 H₃PO₄ on left
Left: Mg=3, OH=6, H from acid=6 → total H=12? Wait let’s count carefully.

Try:
3 Mg(OH)₂ + 2 H₃PO₄ → ? H₂O + 1 Mg₃(PO)₂

Mg: 3 = 3 ✔️
P: 2 = 2 ✔️
O from phosphate: 8 on right, plus water... better to count H and O together.

Left H: from 3 Mg(OH)₂ → 6H; from 2 H₃PO₄ → 6H → total H=12
Right: only in H₂O → so need 6 H₂O → gives 12H and 6O

Now check O:
Left: 3 Mg(OH)₂ → 6O; 2 H₃PO₄ → 8O → total 14O
Right: Mg₃(PO₄)₂ → 8O; 6 H₂O → 6O → total 14O ✔️

✔ Balanced: 3 Mg(OH)₂ + 2 H₃PO₄ → 6 H₂O + 1 Mg₃(PO₄)₂

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5. HNO₃ + Ni → Ni(NO₃)₂ + H₂

Ni goes from 0 to +2 → needs 2 NO₃⁻ → so 2 HNO₃? But then H=2 → makes H₂ ✔️

But wait: Ni(NO₃)₂ has 2 NO₃ → so need 2 HNO₃ for nitrate, but also need to provide H for H₂.

Actually: 2 HNO₃ → gives 2H and 2NO₃ → perfect for Ni(NO₃)₂ and H₂

But Ni is solid → becomes Ni²⁺ → so yes.

Wait — is this correct? Let’s write:

2 HNO₃ + Ni → Ni(NO₃)₂ + H₂

Atoms:
Left: H=2, N=2, O=6, Ni=1
Right: Ni=1, N=2, O=6, H=2 ✔️

✔ Balanced: 2 HNO₃ + 1 Ni → 1 Ni(NO₃)₂ + 1 H₂

*(Note: In reality, nickel doesn’t react with dilute nitric acid to give hydrogen gas — it gives nitrogen oxides — but for worksheet purposes, we’ll go with this.)*

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6. Ba(HCO₃)₂ → BaCO₃ + H₂O + CO₂

Decomposition. Count atoms:

Left: Ba=1, H=2, C=2, O=6
Right: BaCO₃ → Ba=1, C=1, O=3; H₂O → H=2, O=1; CO₂ → C=1, O=2 → total C=2, O=6, H=2 ✔️

✔ Balanced: 1 Ba(HCO₃)₂ → 1 BaCO₃ + 1 H₂O + 1 CO₂

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7. BaCl₂ + Na₂SO₄ → NaCl + BaSO₄

Left: Ba=1, Cl=2, Na=2, S=1, O=4
Right: NaCl → Na=1, Cl=1; BaSO₄ → Ba=1, S=1, O=4 → not balanced

Need 2 NaCl on right → then Na=2, Cl=2 ✔️

✔ Balanced: 1 BaCl₂ + 1 Na₂SO₄ → 2 NaCl + 1 BaSO₄

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8. Al₂(CO₃)₃ → Al₂O₃ + CO₂

Left: Al=2, C=3, O=9
Right: Al₂O₃ → Al=2, O=3; CO₂ → if 3 CO₂ → C=3, O=6 → total O=9 ✔️

✔ Balanced: 1 Al₂(CO₃)₃ → 1 Al₂O₃ + 3 CO₂

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9. Ca + H₂O → Ca(OH)₂ + H₂

Left: Ca=1, H=2, O=1
Right: Ca=1, O=2, H=4 → not balanced

Need 2 H₂O on left → H=4, O=2
Then right: Ca(OH)₂ has O=2, H=2; H₂ has H=2 → total H=4 ✔️

✔ Balanced: 1 Ca + 2 H₂O → 1 Ca(OH)₂ + 1 H₂

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10. LiHCO₃ → Li₂CO₃ + H₂O + CO₂

Left: Li=1, H=1, C=1, O=3
Right: Li₂CO₃ → Li=2 → so need 2 LiHCO₃ on left

2 LiHCO₃ → Li=2, H=2, C=2, O=6
Right: Li₂CO₃ → Li=2, C=1, O=3; H₂O → H=2, O=1; CO₂ → C=1, O=2 → total C=2, O=6, H=2 ✔️

✔ Balanced: 2 LiHCO₃ → 1 Li₂CO₃ + 1 H₂O + 1 CO₂

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11. N₂ + O₂ → N₂O₅

Left: N=2, O=2
Right: N=2, O=5 → not balanced

Find LCM of O: 2 and 5 → 10

So 5 O₂ → 10 O atoms
Then 2 N₂O₅ → 4N and 10O → so need 2 N₂ on left

✔ Balanced: 2 N₂ + 5 O₂ → 2 N₂O₅

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12. MgBr₂ + KOH → KBr + Mg(OH)₂

Left: Mg=1, Br=2, K=1, O=1, H=1
Right: KBr → K=1, Br=1; Mg(OH)₂ → Mg=1, O=2, H=2 → not balanced

Need 2 KBr → so 2 K on right → need 2 KOH on left
Then left: K=2, O=2, H=2 → matches Mg(OH)₂

✔ Balanced: 1 MgBr₂ + 2 KOH → 2 KBr + 1 Mg(OH)₂

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13. Mn + CuCl → Cu + MnCl₂

Left: Mn=1, Cu=1, Cl=1
Right: Cu=1, Mn=1, Cl=2 → not balanced

Need 2 CuCl on left → Cl=2, Cu=2
Then right: need 2 Cu

✔ Balanced: 1 Mn + 2 CuCl → 2 Cu + 1 MnCl₂

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14. Zn + S₈ → ZnS

S₈ has 8 S atoms → so need 8 ZnS on right → then 8 Zn on left

✔ Balanced: 8 Zn + 1 S₈ → 8 ZnS

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15. NaOH + H₂SO₄ → H₂O + Na₂SO₄

Right: Na=2 → so need 2 NaOH on left
Left: Na=2, O=2+4=6? Wait:

2 NaOH → Na=2, O=2, H=2
H₂SO₄ → H=2, S=1, O=4 → total H=4, O=6, S=1, Na=2

Right: Na₂SO₄ → Na=2, S=1, O=4; H₂O → if 2 H₂O → H=4, O=2 → total O=6 ✔️

✔ Balanced: 2 NaOH + 1 H₂SO₄ → 2 H₂O + 1 Na₂SO₄

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16. K + H₂O → KOH + H₂

Left: K=1, H=2, O=1
Right: K=1, O=1, H=1+2=3? KOH has H=1, H₂ has H=2 → total H=3 → too many

Try 2 K + 2 H₂O → 2 KOH + H₂

Left: K=2, H=4, O=2
Right: 2 KOH → K=2, O=2, H=2; H₂ → H=2 → total H=4 ✔️

✔ Balanced: 2 K + 2 H₂O → 2 KOH + 1 H₂

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17. C₅H₁₂ + O₂ → H₂O + CO₂

Combustion. Balance C first: 5 CO₂
Balance H: 12 H → 6 H₂O
Now O: right = 5×2 + 6×1 = 10+6=16 O → so 8 O₂ on left

✔ Balanced: 1 C₅H₁₂ + 8 O₂ → 6 H₂O + 5 CO₂

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18. KOH + H₂CO₃ → H₂O + K₂CO₃

Right: K=2 → so 2 KOH on left
Left: K=2, O=2+3=5? Wait:

2 KOH → K=2, O=2, H=2
H₂CO₃ → H=2, C=1, O=3 → total H=4, O=5, C=1, K=2

Right: K₂CO₃ → K=2, C=1, O=3; H₂O → if 2 H₂O → H=4, O=2 → total O=5 ✔️

✔ Balanced: 2 KOH + 1 H₂CO₃ → 2 H₂O + 1 K₂CO₃

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19. C₄H₈O₂ + O₂ → H₂O + CO₂

Combustion. Balance C: 4 CO₂
Balance H: 8 H → 4 H₂O
Now O: left: from C₄H₈O₂ → 2 O; from O₂ → ?
Right: 4 CO₂ → 8 O; 4 H₂O → 4 O → total 12 O
So need 10 O from O₂ → 5 O₂

Check: left O: 2 + 10 = 12 ✔️

✔ Balanced: 1 C₄H₈O₂ + 5 O₂ → 4 H₂O + 4 CO₂

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20. Al + S₈ → Al₂S₃

S₈ has 8 S → Al₂S₃ has 3 S per formula → LCM of 8 and 3 is 24

So 3 S₈ → 24 S → need 8 Al₂S₃ → which requires 16 Al

✔ Balanced: 16 Al + 3 S₈ → 8 Al₂S₃

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Now, for the “Types of Reaction” part — since you didn’t ask to explain or teach, I’ll just list the types as commonly taught:

The 5 main types are:
1. Synthesis (combination)
2. Decomposition
3. Single Replacement (displacement)
4. Double Replacement (metathesis)
5. Combustion

And for each reaction above, here’s the type:

1. Single replacement (Zn replaces Ag)
2. Synthesis
3. Double replacement
4. Double replacement (acid-base neutralization)
5. Single replacement (metal-acid)
6. Decomposition
7. Double replacement
8. Decomposition
9. Single replacement
10. Decomposition
11. Synthesis
12. Double replacement
13. Single replacement
14. Synthesis
15. Double replacement (neutralization)
16. Single replacement
17. Combustion
18. Double replacement (neutralization)
19. Combustion
20. Synthesis

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Final Answer:

Balanced Equations:

1. 1 Zn + 2 AgNO₃ → 1 Zn(NO₃)₂ + 2 Ag
2. 1 N₂ + 3 H₂ → 2 NH₃
3. 1 NaCl + 1 AgC₂H₃O₂ → 1 NaC₂H₃O₂ + 1 AgCl
4. 3 Mg(OH)₂ + 2 H₃PO₄ → 6 H₂O + 1 Mg₃(PO₄)₂
5. 2 HNO₃ + 1 Ni → 1 Ni(NO₃)₂ + 1 H₂
6. 1 Ba(HCO₃)₂ → 1 BaCO₃ + 1 H₂O + 1 CO₂
7. 1 BaCl₂ + 1 Na₂SO₄ → 2 NaCl + 1 BaSO₄
8. 1 Al₂(CO₃)₃ → 1 Al₂O₃ + 3 CO₂
9. 1 Ca + 2 H₂O → 1 Ca(OH)₂ + 1 H₂
10. 2 LiHCO₃ → 1 Li₂CO₃ + 1 H₂O + 1 CO₂
11. 2 N₂ + 5 O₂ → 2 N₂O₅
12. 1 MgBr₂ + 2 KOH → 2 KBr + 1 Mg(OH)₂
13. 1 Mn + 2 CuCl → 2 Cu + 1 MnCl₂
14. 8 Zn + 1 S₈ → 8 ZnS
15. 2 NaOH + 1 H₂SO₄ → 2 H₂O + 1 Na₂SO₄
16. 2 K + 2 H₂O → 2 KOH + 1 H₂
17. 1 C₅H₁₂ + 8 O₂ → 6 H₂O + 5 CO₂
18. 2 KOH + 1 H₂CO₃ → 2 H₂O + 1 K₂CO₃
19. 1 C₄H₈O₂ + 5 O₂ → 4 H₂O + 4 CO₂
20. 16 Al + 3 S₈ → 8 Al₂S₃

Reaction Types:

1. Single Replacement
2. Synthesis
3. Double Replacement
4. Double Replacement
5. Single Replacement
6. Decomposition
7. Double Replacement
8. Decomposition
9. Single Replacement
10. Decomposition
11. Synthesis
12. Double Replacement
13. Single Replacement
14. Synthesis
15. Double Replacement
16. Single Replacement
17. Combustion
18. Double Replacement
19. Combustion
20. Synthesis