Free Printable Balancing Equations Worksheets - Free Printable
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Step-by-step solution for: Free Printable Balancing Equations Worksheets
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Show Answer Key & Explanations
Step-by-step solution for: Free Printable Balancing Equations Worksheets
Let's solve each of these chemical equations by balancing them. Balancing a chemical equation means ensuring the number of atoms of each element is the same on both sides (reactants and products), following the Law of Conservation of Mass.
We'll go through each one step by step, showing how to balance them.
---
- Left: H=2, O=1+3=4, S=1
- Right: H=2, S=1, O=4
✔ Already balanced!
Answer:
1 H₂O + 1 SO₃ → 1 H₂SO₄
---
Unbalanced:
- C: 2 on left, 1 on right → need 2 CO₂
- H: 4 on left, 2 on right → need 2 H₂O
Now:
C₂H₄ + O₂ → 2CO₂ + 2H₂O
Now count O:
- Right: 2×2 (from CO₂) + 2×1 (from H₂O) = 4 + 2 = 6 O
- Left: O₂ has 2 O → need 3 O₂
Answer:
1 C₂H₄ + 3 O₂ → 2 CO₂ + 2 H₂O
---
This is a decomposition reaction.
Left: Pb=1, S=1, O=4
Right: PbSO₃ has Pb=1, S=1, O=3; O₂ has 2 O → total O = 5 ✘
Wait — that doesn’t balance.
But PbSO₄ → PbSO₃ + ½O₂ would work:
PbSO₄ → PbSO₃ + ½O₂
Multiply entire equation by 2:
2 PbSO₄ → 2 PbSO₃ + O₂
Check:
- Pb: 2 = 2 ✔
- S: 2 = 2 ✔
- O: 8 → 6 (in 2 PbSO₃) + 2 (in O₂) = 8 ✔
Answer:
2 PbSO₄ → 2 PbSO₃ + 1 O₂
---
Wait — this is wrong! Pb is lead, but we have P (phosphorus). The product should be PBr₃, not PbBr₃.
So correct reaction:
P₄ + Br₂ → PBr₃
Now balance:
- P₄ has 4 P → need 4 PBr₃
- Each PBr₃ has 3 Br → 4 × 3 = 12 Br needed
- Br₂ provides 2 Br per molecule → need 6 Br₂
Answer:
1 P₄ + 6 Br₂ → 4 PBr₃
---
Mg: 1 on left, 1 on right
O: 2 on left, 1 on right → need 2 MgO
So:
Mg + O₂ → 2MgO → now Mg: 1 vs 2 → need 2 Mg
Answer:
2 Mg + 1 O₂ → 2 MgO
---
Propane combustion.
C₃H₈ → 3 CO₂ (for carbon)
C₃H₈ → 4 H₂O (since 8 H → 4 H₂O)
Now:
C₃H₈ + O₂ → 3 CO₂ + 4 H₂O
Count O:
- Right: 3×2 + 4×1 = 6 + 4 = 10 O
- So need 5 O₂ (since each O₂ has 2 O)
Answer:
1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
---
Nitrogen in ammonia forms nitrogen gas.
N: 1 on left, 2 on right → need 2 NH₃
H: 2×3 = 6 H → need 3 H₂O
Now:
2 NH₃ + O₂ → N₂ + 3 H₂O
O: 3 O on right → need 3/2 O₂ → multiply whole equation by 2:
4 NH₃ + 3 O₂ → 2 N₂ + 6 H₂O
Check:
- N: 4 = 4 ✔
- H: 12 = 12 ✔
- O: 6 = 6 ✔
Answer:
4 NH₃ + 3 O₂ → 2 N₂ + 6 H₂O
---
Hydrogen peroxide formation.
H₂O + O₂ → H₂O₂
Left: H=2, O=1+2=3
Right: H=2, O=2 → not balanced
But H₂O₂ has 2 H and 2 O. We need more oxygen.
Actually, this reaction isn't straightforward. But let's try:
H₂O + ½O₂ → H₂O₂ → multiply by 2:
2 H₂O + O₂ → 2 H₂O₂
Check:
- H: 4 = 4 ✔
- O: 2 + 2 = 4 → right: 4 ✔
Answer:
2 H₂O + 1 O₂ → 2 H₂O₂
---
Double displacement.
Na: 1 → 1 → OK
Br: 1 → 2 → need 2 NaBr
Ca: 1 → 1 → OK
F: 2 → 1 → need 2 NaF
So:
2 NaBr + CaF₂ → 2 NaF + CaBr₂
Check:
- Na: 2 = 2 ✔
- Br: 2 = 2 ✔
- Ca: 1 = 1 ✔
- F: 2 = 2 ✔
Answer:
2 NaBr + 1 CaF₂ → 2 NaF + 1 CaBr₂
---
Synthesis of ammonia.
N₂ → 2 NH₃ → so need 2 NH₃
Each NH₃ has 3 H → 2×3 = 6 H → need 3 H₂
Answer:
1 N₂ + 3 H₂ → 2 NH₃
---
Decomposition of potassium chlorate.
K: 1 = 1 ✔
Cl: 1 = 1 ✔
O: 3 on left → need 3/2 O₂ → so write:
2 KClO₃ → 2 KCl + 3 O₂
Check:
- K: 2 = 2 ✔
- Cl: 2 = 2 ✔
- O: 6 = 6 ✔
Answer:
2 KClO₃ → 2 KCl + 3 O₂
---
Methane combustion.
CH₄ → CO₂ (C balanced)
CH₄ → 2 H₂O (4 H → 2 H₂O)
Now:
CH₄ + O₂ → CO₂ + 2 H₂O
O: right = 2 (CO₂) + 2 (H₂O) = 4 → need 2 O₂
Answer:
1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O
---
Single replacement.
Al: 1 = 1 ✔
Br: 3 → need 3 KBr
K: 3 → need 3 K
Answer:
1 AlBr₃ + 3 K → 3 KBr + 1 Al
---
Same as #4.
P₄ has 4 P → need 4 PBr₃
Each PBr₃ has 3 Br → 4×3 = 12 Br → need 6 Br₂
Answer:
1 P₄ + 6 Br₂ → 4 PBr₃
---
Single replacement.
Li: 1 → 1 ✔
Cl: 1 → 2 (in Cl₂) → need 2 LiCl
Br: 2 → 2 → need 2 LiBr
So:
2 LiCl + Br₂ → 2 LiBr + Cl₂
Check:
- Li: 2 = 2 ✔
- Cl: 2 = 2 ✔
- Br: 2 = 2 ✔
Answer:
2 LiCl + 1 Br₂ → 2 LiBr + 1 Cl₂
---
## ✔ Final Balanced Equations:
1. 1 H₂O + 1 SO₃ → 1 H₂SO₄
2. 1 C₂H₄ + 3 O₂ → 2 CO₂ + 2 H₂O
3. 2 PbSO₄ → 2 PbSO₃ + 1 O₂
4. 1 P₄ + 6 Br₂ → 4 PBr₃
5. 2 Mg + 1 O₂ → 2 MgO
6. 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
7. 4 NH₃ + 3 O₂ → 2 N₂ + 6 H₂O
8. 2 H₂O + 1 O₂ → 2 H₂O₂
9. 2 NaBr + 1 CaF₂ → 2 NaF + 1 CaBr₂
10. 1 N₂ + 3 H₂ → 2 NH₃
11. 2 KClO₃ → 2 KCl + 3 O₂
12. 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O
13. 1 AlBr₃ + 3 K → 3 KBr + 1 Al
14. 1 P₄ + 6 Br₂ → 4 PBr₃
15. 2 LiCl + 1 Br₂ → 2 LiBr + 1 Cl₂
---
- Start with elements that appear in only one compound on each side.
- Save elements like O and H for last (they often appear in multiple compounds).
- Use coefficients (whole numbers) to balance atoms.
- Check all atoms after balancing.
Let me know if you'd like this formatted as a printable worksheet!
We'll go through each one step by step, showing how to balance them.
---
1. H₂O + SO₃ → H₂SO₄
- Left: H=2, O=1+3=4, S=1
- Right: H=2, S=1, O=4
✔ Already balanced!
Answer:
1 H₂O + 1 SO₃ → 1 H₂SO₄
---
2. C₂H₄ + O₂ → CO₂ + H₂O
Unbalanced:
- C: 2 on left, 1 on right → need 2 CO₂
- H: 4 on left, 2 on right → need 2 H₂O
Now:
C₂H₄ + O₂ → 2CO₂ + 2H₂O
Now count O:
- Right: 2×2 (from CO₂) + 2×1 (from H₂O) = 4 + 2 = 6 O
- Left: O₂ has 2 O → need 3 O₂
Answer:
1 C₂H₄ + 3 O₂ → 2 CO₂ + 2 H₂O
---
3. PbSO₄ → PbSO₃ + O₂
This is a decomposition reaction.
Left: Pb=1, S=1, O=4
Right: PbSO₃ has Pb=1, S=1, O=3; O₂ has 2 O → total O = 5 ✘
Wait — that doesn’t balance.
But PbSO₄ → PbSO₃ + ½O₂ would work:
PbSO₄ → PbSO₃ + ½O₂
Multiply entire equation by 2:
2 PbSO₄ → 2 PbSO₃ + O₂
Check:
- Pb: 2 = 2 ✔
- S: 2 = 2 ✔
- O: 8 → 6 (in 2 PbSO₃) + 2 (in O₂) = 8 ✔
Answer:
2 PbSO₄ → 2 PbSO₃ + 1 O₂
---
4. P₄ + Br₂ → PbBr₃
Wait — this is wrong! Pb is lead, but we have P (phosphorus). The product should be PBr₃, not PbBr₃.
So correct reaction:
P₄ + Br₂ → PBr₃
Now balance:
- P₄ has 4 P → need 4 PBr₃
- Each PBr₃ has 3 Br → 4 × 3 = 12 Br needed
- Br₂ provides 2 Br per molecule → need 6 Br₂
Answer:
1 P₄ + 6 Br₂ → 4 PBr₃
---
5. Mg + O₂ → MgO
Mg: 1 on left, 1 on right
O: 2 on left, 1 on right → need 2 MgO
So:
Mg + O₂ → 2MgO → now Mg: 1 vs 2 → need 2 Mg
Answer:
2 Mg + 1 O₂ → 2 MgO
---
6. C₃H₈ + O₂ → CO₂ + H₂O
Propane combustion.
C₃H₈ → 3 CO₂ (for carbon)
C₃H₈ → 4 H₂O (since 8 H → 4 H₂O)
Now:
C₃H₈ + O₂ → 3 CO₂ + 4 H₂O
Count O:
- Right: 3×2 + 4×1 = 6 + 4 = 10 O
- So need 5 O₂ (since each O₂ has 2 O)
Answer:
1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
---
7. NH₃ + O₂ → N₂ + H₂O
Nitrogen in ammonia forms nitrogen gas.
N: 1 on left, 2 on right → need 2 NH₃
H: 2×3 = 6 H → need 3 H₂O
Now:
2 NH₃ + O₂ → N₂ + 3 H₂O
O: 3 O on right → need 3/2 O₂ → multiply whole equation by 2:
4 NH₃ + 3 O₂ → 2 N₂ + 6 H₂O
Check:
- N: 4 = 4 ✔
- H: 12 = 12 ✔
- O: 6 = 6 ✔
Answer:
4 NH₃ + 3 O₂ → 2 N₂ + 6 H₂O
---
8. H₂O + O₂ → H₂O₂
Hydrogen peroxide formation.
H₂O + O₂ → H₂O₂
Left: H=2, O=1+2=3
Right: H=2, O=2 → not balanced
But H₂O₂ has 2 H and 2 O. We need more oxygen.
Actually, this reaction isn't straightforward. But let's try:
H₂O + ½O₂ → H₂O₂ → multiply by 2:
2 H₂O + O₂ → 2 H₂O₂
Check:
- H: 4 = 4 ✔
- O: 2 + 2 = 4 → right: 4 ✔
Answer:
2 H₂O + 1 O₂ → 2 H₂O₂
---
9. NaBr + CaF₂ → NaF + CaBr₂
Double displacement.
Na: 1 → 1 → OK
Br: 1 → 2 → need 2 NaBr
Ca: 1 → 1 → OK
F: 2 → 1 → need 2 NaF
So:
2 NaBr + CaF₂ → 2 NaF + CaBr₂
Check:
- Na: 2 = 2 ✔
- Br: 2 = 2 ✔
- Ca: 1 = 1 ✔
- F: 2 = 2 ✔
Answer:
2 NaBr + 1 CaF₂ → 2 NaF + 1 CaBr₂
---
10. N₂ + H₂ → NH₃
Synthesis of ammonia.
N₂ → 2 NH₃ → so need 2 NH₃
Each NH₃ has 3 H → 2×3 = 6 H → need 3 H₂
Answer:
1 N₂ + 3 H₂ → 2 NH₃
---
11. KClO₃ → KCl + O₂
Decomposition of potassium chlorate.
K: 1 = 1 ✔
Cl: 1 = 1 ✔
O: 3 on left → need 3/2 O₂ → so write:
2 KClO₃ → 2 KCl + 3 O₂
Check:
- K: 2 = 2 ✔
- Cl: 2 = 2 ✔
- O: 6 = 6 ✔
Answer:
2 KClO₃ → 2 KCl + 3 O₂
---
12. CH₄ + O₂ → CO₂ + H₂O
Methane combustion.
CH₄ → CO₂ (C balanced)
CH₄ → 2 H₂O (4 H → 2 H₂O)
Now:
CH₄ + O₂ → CO₂ + 2 H₂O
O: right = 2 (CO₂) + 2 (H₂O) = 4 → need 2 O₂
Answer:
1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O
---
13. AlBr₃ + K → KBr + Al
Single replacement.
Al: 1 = 1 ✔
Br: 3 → need 3 KBr
K: 3 → need 3 K
Answer:
1 AlBr₃ + 3 K → 3 KBr + 1 Al
---
14. P₄ + Br₂ → PBr₃
Same as #4.
P₄ has 4 P → need 4 PBr₃
Each PBr₃ has 3 Br → 4×3 = 12 Br → need 6 Br₂
Answer:
1 P₄ + 6 Br₂ → 4 PBr₃
---
15. LiCl + Br₂ → LiBr + Cl₂
Single replacement.
Li: 1 → 1 ✔
Cl: 1 → 2 (in Cl₂) → need 2 LiCl
Br: 2 → 2 → need 2 LiBr
So:
2 LiCl + Br₂ → 2 LiBr + Cl₂
Check:
- Li: 2 = 2 ✔
- Cl: 2 = 2 ✔
- Br: 2 = 2 ✔
Answer:
2 LiCl + 1 Br₂ → 2 LiBr + 1 Cl₂
---
## ✔ Final Balanced Equations:
1. 1 H₂O + 1 SO₃ → 1 H₂SO₄
2. 1 C₂H₄ + 3 O₂ → 2 CO₂ + 2 H₂O
3. 2 PbSO₄ → 2 PbSO₃ + 1 O₂
4. 1 P₄ + 6 Br₂ → 4 PBr₃
5. 2 Mg + 1 O₂ → 2 MgO
6. 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
7. 4 NH₃ + 3 O₂ → 2 N₂ + 6 H₂O
8. 2 H₂O + 1 O₂ → 2 H₂O₂
9. 2 NaBr + 1 CaF₂ → 2 NaF + 1 CaBr₂
10. 1 N₂ + 3 H₂ → 2 NH₃
11. 2 KClO₃ → 2 KCl + 3 O₂
12. 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O
13. 1 AlBr₃ + 3 K → 3 KBr + 1 Al
14. 1 P₄ + 6 Br₂ → 4 PBr₃
15. 2 LiCl + 1 Br₂ → 2 LiBr + 1 Cl₂
---
🔍 Summary Tips for Balancing:
- Start with elements that appear in only one compound on each side.
- Save elements like O and H for last (they often appear in multiple compounds).
- Use coefficients (whole numbers) to balance atoms.
- Check all atoms after balancing.
Let me know if you'd like this formatted as a printable worksheet!
Parent Tip: Review the logic above to help your child master the concept of act practice worksheet.
