Free Printable Law of Conservation of Mass Worksheets - Free Printable

To balance these chemical equations, we need to make sure that the number of atoms of each element is the same on both sides of the equation. This is based on the Law of Conservation of Mass, which says matter cannot be created or destroyed in a chemical reaction.

Let’s go through each equation one by one and balance them carefully.

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1. O₂ → O₃

We have 2 oxygen atoms on the left, 3 on the right.
Find least common multiple of 2 and 3 → 6.
So, 3 O₂ → 2 O₃
Check: Left = 6 O, Right = 6 O → Balanced.

✔ Answer: 3O₂ → 2O₃

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2. Zn + HCl → ZnCl₂ + H₂

Left: Zn=1, H=1, Cl=1
Right: Zn=1, Cl=2, H=2
Need 2 HCl to get 2 H and 2 Cl.

Zn + 2HCl → ZnCl₂ + H₂
Check: Zn=1, H=2, Cl=2 → Balanced.

✔ Answer: Zn + 2HCl → ZnCl₂ + H₂

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3. N₂ + H₂ → NH₃

Left: N=2, H=2
Right: N=1, H=3
Multiply NH₃ by 2 → 2NH₃ (N=2, H=6)
Now need 3 H₂ to get 6 H.

N₂ + 3H₂ → 2NH₃
Check: N=2, H=6 → Balanced.

✔ Answer: N₂ + 3H₂ → 2NH₃

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4. Al + Cr₂O₃ → Al₂O₃ + Cr

Left: Al=1, Cr=2, O=3
Right: Al=2, O=3, Cr=1
Need 2 Al on left → 2Al
Need 2 Cr on right → 2Cr

2Al + Cr₂O₃ → Al₂O₃ + 2Cr
Check: Al=2, Cr=2, O=3 → Balanced.

✔ Answer: 2Al + Cr₂O₃ → Al₂O₃ + 2Cr

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5. KClO₃ → KCl + O₂

Left: K=1, Cl=1, O=3
Right: K=1, Cl=1, O=2
Need even number of O on right → multiply O₂ by 3 → 3O₂ (O=6)
Then need 2 KClO₃ → 2K, 2Cl, 6O
Right: 2KCl + 3O₂ → 2K, 2Cl, 6O

2KClO₃ → 2KCl + 3O₂
Balanced.

✔ Answer: 2KClO₃ → 2KCl + 3O₂

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6. BF₃ + H₂O → B₂O₃ + HF

Left: B=1, F=3, H=2, O=1
Right: B=2, O=3, H=1, F=1
Need 2 BF₃ → B=2, F=6
Need 3 H₂O → H=6, O=3
Then need 6 HF → H=6, F=6

2BF₃ + 3H₂O → B₂O₃ + 6HF
Check: B=2, F=6, H=6, O=3 → Balanced.

✔ Answer: 2BF₃ + 3H₂O → B₂O₃ + 6HF

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7. PCl₅ + AsF₃ → PF₅ + AsCl₃

Left: P=1, Cl=5, As=1, F=3
Right: P=1, F=5, As=1, Cl=3
Need to balance Cl and F.

Try 3 PCl₅ → Cl=15 → need 5 AsCl₃ → As=5, Cl=15
Then 5 AsF₃ → F=15 → need 3 PF₅ → F=15, P=3

3PCl₅ + 5AsF₃ → 3PF₅ + 5AsCl₃
Check: P=3, Cl=15, As=5, F=15 → Balanced.

✔ Answer: 3PCl₅ + 5AsF₃ → 3PF₅ + 5AsCl₃

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8. K + O₂ → K₂O₂

Left: K=1, O=2
Right: K=2, O=2
Need 2 K on left.

2K + O₂ → K₂O₂
Balanced.

✔ Answer: 2K + O₂ → K₂O₂

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9. Fe + O₂ + H₂O → Fe(OH)₂

Left: Fe=1, O=2+1=3? Wait — O₂ has 2 O, H₂O has 1 O → total O=3? But product Fe(OH)₂ has 2 O and 2 H.

Actually, Fe(OH)₂ = Fe, 2O, 2H.

So we need 2 H from H₂O → so 1 H₂O gives 2H and 1O. But we also need another O for Fe(OH)₂? Let's think:

Fe(OH)₂ has: Fe, 2O, 2H.

From H₂O: provides 2H and 1O → still need 1 more O → from O₂? But O₂ gives 2O.

Better to write:

Fe + ½O₂ + H₂O → Fe(OH)₂ → but we want whole numbers.

Multiply by 2:

2Fe + O₂ + 2H₂O → 2Fe(OH)₂

Check: Left: Fe=2, O=2+2=4, H=4
Right: 2Fe, 4O, 4H → Balanced.

✔ Answer: 2Fe + O₂ + 2H₂O → 2Fe(OH)₂

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10. Al + HCl → AlCl₃ + H₂O

Wait — this looks wrong. Aluminum with HCl usually gives AlCl₃ and H₂, not H₂O. But the equation says H₂O? That would require oxygen, which isn’t on left.

Looking back at image: It says “Al + HCl → AlCl₃ + H₂O” — that can’t be balanced as written because no oxygen on left.

But wait — maybe it’s a typo? In many worksheets, it’s Al + HCl → AlCl₃ + H₂.

Check original image description: User said “Image contains...” and listed #10 as “Al + HCl → AlCl₃ + H₂O”

That must be a mistake. Because without oxygen source, you can’t make water.

Perhaps it’s supposed to be Al + HCl → AlCl₃ + H₂?

Let me assume that’s the case — otherwise it’s impossible.

If it’s Al + HCl → AlCl₃ + H₂:

Balance: Al=1, Cl=3 on right → need 3 HCl → H=3 → but H₂ needs even H → so 2Al + 6HCl → 2AlCl₃ + 3H₂

But if it’s really H₂O, then it’s invalid.

Given context, I think it’s a typo and should be H₂.

I’ll proceed with H₂.

So: 2Al + 6HCl → 2AlCl₃ + 3H₂

✔ Answer: 2Al + 6HCl → 2AlCl₃ + 3H₂
*(Note: Assuming H₂ instead of H₂O due to impossibility otherwise)*

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11. NH₃ + O₂ → N₂O₃ + O₂

Wait — O₂ appears on both sides? That’s odd. Probably meant to be NH₃ + O₂ → N₂O₃ + H₂O? Or something else?

Looking at image: “NH₃ + O₂ → N₂O₃ + O₂” — again, O₂ on both sides? Doesn’t make sense.

Perhaps it’s NH₃ + O₂ → N₂O₃ + H₂O? That would be standard.

Assume typo: NH₃ + O₂ → N₂O₃ + H₂O

Balance:

Left: N=1, H=3, O=2
Right: N=2, O=3+1=4, H=2

Multiply NH₃ by 2 → N=2, H=6
Then H₂O by 3 → H=6, O=3
N₂O₃ has O=3 → total O on right = 3+3=6
So O₂ needed: 3 molecules → O=6

2NH₃ + 3O₂ → N₂O₃ + 3H₂O
Check: N=2, H=6, O=6 → Balanced.

✔ Answer: 2NH₃ + 3O₂ → N₂O₃ + 3H₂O
*(Assuming H₂O was intended instead of O₂ on right)*

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12. K₂O₂ + H₂O → KOH + O₂

Left: K=2, O=2+1=3, H=2
Right: K=1, O=1+2=3, H=1

Need 2 KOH → K=2, O=2, H=2
Then O₂ must account for remaining O: left has 3 O, right has 2 O in KOH → need 1 O in O₂? But O₂ has 2 O.

Set: K₂O₂ + H₂O → 2KOH + ½O₂ → multiply by 2:

2K₂O₂ + 2H₂O → 4KOH + O₂
Check: Left: K=4, O=4+2=6, H=4
Right: K=4, O=4+2=6, H=4 → Balanced.

✔ Answer: 2K₂O₂ + 2H₂O → 4KOH + O₂

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13. Mn + CuCl → Cu + MnCl₂

Left: Mn=1, Cu=1, Cl=1
Right: Cu=1, Mn=1, Cl=2
Need 2 CuCl → Cl=2, Cu=2
Then need 2 Cu on right.

Mn + 2CuCl → 2Cu + MnCl₂
Check: Mn=1, Cu=2, Cl=2 → Balanced.

✔ Answer: Mn + 2CuCl → 2Cu + MnCl₂

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14. Mg(OH)₂ + H₃PO₄ → H₂O + Mg₃(PO₄)₂

Left: Mg=1, O=2+4=6? OH has O and H, PO₄ has P and O.

Better: Mg(OH)₂ = Mg, 2O, 2H
H₃PO₄ = 3H, P, 4O

Right: H₂O = 2H, 1O
Mg₃(PO₄)₂ = 3Mg, 2P, 8O

So need 3 Mg(OH)₂ → Mg=3, O=6, H=6
Need 2 H₃PO₄ → H=6, P=2, O=8
Total left: Mg=3, H=12, P=2, O=14

Right: Mg₃(PO₄)₂ = 3Mg, 2P, 8O
H₂O: need 6H₂O → H=12, O=6 → total O=8+6=14

So: 3Mg(OH)₂ + 2H₃PO₄ → 6H₂O + Mg₃(PO₄)₂
Check: Mg=3, H=12, P=2, O=6+8=14 → Balanced.

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

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15. Ba(HCO₃)₂ → BaCO₃ + H₂O + CO₂

Left: Ba=1, H=2, C=2, O=6
Right: Ba=1, C=1+1=2, H=2, O=3+1+2=6

Already balanced!

Ba(HCO₃)₂ → BaCO₃ + H₂O + CO₂
Check: Ba=1, H=2, C=2, O=6 → Balanced.

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

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16. Zn + S₈ → ZnS

S₈ has 8 S atoms. Each ZnS has 1 S. So need 8 ZnS → 8 Zn and 8 S.

8Zn + S₈ → 8ZnS
Check: Zn=8, S=8 → Balanced.

✔ Answer: 8Zn + S₈ → 8ZnS

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17. LiHCO₃ → Li₂CO₃ + H₂O + CO₂

Left: Li=1, H=1, C=1, O=3
Right: Li=2, C=1+1=2, H=2, O=3+1+2=6

Need 2 LiHCO₃ → Li=2, H=2, C=2, O=6
Right: Li₂CO₃ + H₂O + CO₂ → Li=2, C=2, H=2, O=3+1+2=6

Balanced.

✔ Answer: 2LiHCO₃ → Li₂CO₃ + H₂O + CO₂

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18. N₂ + O₂ → N₂O₅

Left: N=2, O=2
Right: N=2, O=5
Need 5/2 O₂ → multiply by 2:

2N₂ + 5O₂ → 2N₂O₅
Check: N=4, O=10 → Right: N=4, O=10 → Balanced.

✔ Answer: 2N₂ + 5O₂ → 2N₂O₅

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19. C₅H₁₂ + O₂ → H₂O + CO₂

Combustion: C₅H₁₂ + O₂ → CO₂ + H₂O

Left: C=5, H=12
Right: C=1, H=2, O=3? Need to balance.

C₅H₁₂ → 5CO₂ (C=5)
C₅H₁₂ → 6H₂O (H=12)
O in products: 5*2 + 6*1 = 10 + 6 = 16 O → so 8 O₂

C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O
Check: C=5, H=12, O=16 → Balanced.

✔ Answer: C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O

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20. C₄H₈O₂ + O₂ → H₂O + CO₂

Combustion: C₄H₈O₂ + O₂ → CO₂ + H₂O

Left: C=4, H=8, O=2
Right: C=1, H=2, O=3

Need 4 CO₂ → C=4
Need 4 H₂O → H=8
O in products: 4*2 + 4*1 = 8 + 4 = 12 O
Left has 2 O from C₄H₈O₂, so need 10 more O → 5 O₂

C₄H₈O₂ + 5O₂ → 4CO₂ + 4H₂O
Check: C=4, H=8, O=2+10=12 → Right: 8+4=12 → Balanced.

✔ Answer: C₄H₈O₂ + 5O₂ → 4CO₂ + 4H₂O

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Final Answers:

1. 3O₂ → 2O₃
2. Zn + 2HCl → ZnCl₂ + H₂
3. N₂ + 3H₂ → 2NH₃
4. 2Al + Cr₂O₃ → Al₂O₃ + 2Cr
5. 2KClO₃ → 2KCl + 3O₂
6. 2BF₃ + 3H₂O → B₂O₃ + 6HF
7. 3PCl₅ + 5AsF₃ → 3PF₅ + 5AsCl₃
8. 2K + O₂ → K₂O₂
9. 2Fe + O₂ + 2H₂O → 2Fe(OH)₂
10. 2Al + 6HCl → 2AlCl₃ + 3H₂ *(assuming H₂ instead of H₂O)*
11. 2NH₃ + 3O₂ → N₂O₃ + 3H₂O *(assuming H₂O instead of O₂ on right)*
12. 2K₂O₂ + 2H₂O → 4KOH + O₂
13. Mn + 2CuCl → 2Cu + MnCl₂
14. 3Mg(OH)₂ + 2H₃PO₄ → 6H₂O + Mg₃(PO₄)₂
15. Ba(HCO₃)₂ → BaCO₃ + H₂O + CO₂
16. 8Zn + S₈ → 8ZnS
17. 2LiHCO₃ → Li₂CO₃ + H₂O + CO₂
18. 2N₂ + 5O₂ → 2N₂O₅
19. C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O
20. C₄H₈O₂ + 5O₂ → 4CO₂ + 4H₂O

Final Answer:
[All balanced equations as above]