Balancing chemical equations worksheet with 15 problems for students to solve.
A worksheet titled "Balancing Equations" with 15 chemical equations to balance, including reactants and products, with blank lines for answers.
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Show Answer Key & Explanations
Step-by-step solution for: Daniel Davis: Physical Science: November 2008 Archives
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Show Answer Key & Explanations
Step-by-step solution for: Daniel Davis: Physical Science: November 2008 Archives
Let’s go through each equation one by one and balance them. Balancing means making sure the number of atoms of each element is the same on both sides of the arrow.
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1. CH₄ + O₂ → CO₂ + H₂O
Left: C=1, H=4, O=2
Right: C=1, H=2, O=3 (2 from CO₂ + 1 from H₂O)
→ Put a 2 in front of H₂O to get 4 H on right:
CH₄ + O₂ → CO₂ + 2H₂O
Now Right: H=4, O=4 (2 from CO₂ + 2 from 2H₂O)
Left: O=2 → need 4 O → put 2 in front of O₂
✔ Balanced: CH₄ + 2O₂ → CO₂ + 2H₂O
---
2. Na + I₂ → NaI
Left: Na=1, I=2
Right: Na=1, I=1
→ Put 2 in front of NaI:
Na + I₂ → 2NaI
Now Right: Na=2, I=2 → so put 2 in front of Na
✔ Balanced: 2Na + I₂ → 2NaI
---
3. N₂ + O₂ → N₂O
Left: N=2, O=2
Right: N=2, O=1
→ Put 2 in front of N₂O:
N₂ + O₂ → 2N₂O
Now Right: N=4, O=2 → Left has N=2 → put 2 in front of N₂
But now N=4 on left, O=2 on left → matches right? Wait:
Wait — if we do:
2N₂ + O₂ → 2N₂O
Left: N=4, O=2
Right: N=4, O=2 → YES!
✔ Balanced: 2N₂ + O₂ → 2N₂O
*(Note: Some might think it's N₂ + ½O₂ → N₂O, but we use whole numbers only.)*
---
4. N₂ + H₂ → NH₃
Left: N=2, H=2
Right: N=1, H=3
→ Put 2 in front of NH₃:
N₂ + H₂ → 2NH₃
Now Right: N=2, H=6 → so need 6 H on left → put 3 in front of H₂
✔ Balanced: N₂ + 3H₂ → 2NH₃
---
5. KI + Cl₂ → KCl + I₂
Left: K=1, I=1, Cl=2
Right: K=1, Cl=1, I=2
→ Put 2 in front of KI and 2 in front of KCl:
2KI + Cl₂ → 2KCl + I₂
Check:
Left: K=2, I=2, Cl=2
Right: K=2, Cl=2, I=2 → Good!
✔ Balanced: 2KI + Cl₂ → 2KCl + I₂
---
6. HCl + Ca(OH)₂ → CaCl₂ + H₂O
Left: H=1+2=3? Wait — Ca(OH)₂ has 2 OH → so 2 H and 2 O from that, plus HCl has 1 H and 1 Cl.
Better to count per atom:
Ca(OH)₂ = Ca, 2O, 2H
So total left: H from HCl + 2H from Ca(OH)₂ → depends on how many HCl.
Try:
Put 2 in front of HCl:
2HCl + Ca(OH)₂ → CaCl₂ + H₂O
Left: H=2+2=4, Cl=2, Ca=1, O=2
Right: Ca=1, Cl=2, H=2, O=1 → not enough H or O on right
→ Put 2 in front of H₂O:
2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O
Now Right: H=4, O=2 → matches left!
✔ Balanced: 2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O
---
7. KClO₃ → KCl + O₂
Left: K=1, Cl=1, O=3
Right: K=1, Cl=1, O=2
→ Need even oxygen on right. Try 2KClO₃:
2KClO₃ → ?
Left: K=2, Cl=2, O=6
→ So right: 2KCl + 3O₂ (since 3×2=6 O)
✔ Balanced: 2KClO₃ → 2KCl + 3O₂
---
8. K₃PO₄ + HCl → KCl + H₃PO₄
Left: K=3, P=1, O=4, H=1, Cl=1
Right: K=1, Cl=1, H=3, P=1, O=4
→ Need 3 K on right → put 3 in front of KCl
→ Then Cl=3 on right → so put 3 in front of HCl
→ Now H=3 on left → matches H₃PO₄ on right
✔ Balanced: K₃PO₄ + 3HCl → 3KCl + H₃PO₄
---
9. S + O₂ → SO₃
Left: S=1, O=2
Right: S=1, O=3
→ Need common multiple for O: LCM of 2 and 3 is 6
→ So 3O₂ (gives 6 O) and 2SO₃ (needs 6 O)
Then S: 2 on right → so 2S on left
✔ Balanced: 2S + 3O₂ → 2SO₃
---
10. KI + Pb(NO₃)₂ → KNO₃ + PbI₂
Left: K=1, I=1, Pb=1, N=2, O=6
Right: K=1, N=1, O=3, Pb=1, I=2
→ Iodine: 2 on right → need 2 KI on left
→ Then K=2 on left → need 2 KNO₃ on right
→ Then N=2 on right → matches Pb(NO₃)₂ which has 2 N
✔ Balanced: 2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂
---
11. CaSO₄ + AlBr₃ → CaBr₂ + Al₂(SO₄)₃
Left: Ca=1, S=1, O=4, Al=1, Br=3
Right: Ca=1, Br=2, Al=2, S=3, O=12
→ Aluminum: 2 on right → need 2 AlBr₃ on left → then Br=6
→ Calcium bromide: need 3 CaBr₂ to get 6 Br → so 3 Ca on left → need 3 CaSO₄
→ Then S=3, O=12 on left → matches Al₂(SO₄)₃ on right
✔ Balanced: 3CaSO₄ + 2AlBr₃ → 3CaBr₂ + Al₂(SO₄)₃
---
12. H₂O₂ → H₂O + O₂
Left: H=2, O=2
Right: H=2, O=1+2=3 → too many O on right
→ Try 2H₂O₂:
2H₂O₂ → ?
Left: H=4, O=4
→ Right: need 4 H → so 2H₂O → gives H=4, O=2
→ Remaining O: 4-2=2 → so 1 O₂
✔ Balanced: 2H₂O₂ → 2H₂O + O₂
---
13. Na + H₂O → NaOH + H₂
Left: Na=1, H=2, O=1
Right: Na=1, O=1, H=1+2=3 → too many H on right
→ Try 2Na:
2Na + H₂O → ?
Still H mismatch.
Try 2Na + 2H₂O → 2NaOH + H₂
Left: Na=2, H=4, O=2
Right: Na=2, O=2, H=2+2=4 → yes!
✔ Balanced: 2Na + 2H₂O → 2NaOH + H₂
---
14. C₂H₆ + O₂ → CO₂ + H₂O
Left: C=2, H=6, O=2
Right: C=1, H=2, O=3
→ Put 2 in front of CO₂ → C=2
→ Put 3 in front of H₂O → H=6
→ Now O on right: 2×2 + 3×1 = 4+3=7 → odd number → hard with O₂
Multiply all by 2 to make O even:
Start over:
C₂H₆ + O₂ → 2CO₂ + 3H₂O
O right: 4 + 3 = 7 → so need 7/2 O₂ → multiply entire equation by 2:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Check:
Left: C=4, H=12, O=14
Right: C=4, H=12, O=8+6=14 → perfect!
✔ Balanced: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
---
15. Mg(NO₃)₂ + K₃PO₄ → Mg₃(PO₄)₂ + KNO₃
Left: Mg=1, N=2, O=6+4=10? Wait — better break down:
Mg(NO₃)₂ = Mg, 2N, 6O
K₃PO₄ = 3K, P, 4O
Total left per formula unit: varies.
Right: Mg₃(PO₄)₂ = 3Mg, 2P, 8O
KNO₃ = K, N, 3O
→ Need 3 Mg on right → so 3 Mg(NO₃)₂ on left → then N=6, O from nitrate=18
→ Need 2 PO₄ on right → so 2 K₃PO₄ on left → then K=6, P=2
→ Right: Mg₃(PO₄)₂ + ? KNO₃ → need 6 K → so 6 KNO₃ → then N=6, O=18 from KNO₃ + 8 from phosphate = 26? Let’s calculate properly.
Left:
3 Mg(NO₃)₂ → 3Mg, 6N, 18O
2 K₃PO₄ → 6K, 2P, 8O
Total left: Mg=3, N=6, O=26, K=6, P=2
Right:
Mg₃(PO)₂ → 3Mg, 2P, 8O
6 KNO₃ → 6K, 6N, 18O
Total right: Mg=3, P=2, O=8+18=26, K=6, N=6 → matches!
✔ Balanced: 3Mg(NO₃)₂ + 2K₃PO₄ → Mg₃(PO₄)₂ + 6KNO₃
---
Final Answer:
1. CH₄ + 2O₂ → CO₂ + 2H₂O
2. 2Na + I₂ → 2NaI
3. 2N₂ + O₂ → 2N₂O
4. N₂ + 3H₂ → 2NH₃
5. 2KI + Cl₂ → 2KCl + I₂
6. 2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O
7. 2KClO₃ → 2KCl + 3O₂
8. K₃PO₄ + 3HCl → 3KCl + H₃PO₄
9. 2S + 3O₂ → 2SO₃
10. 2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂
11. 3CaSO₄ + 2AlBr₃ → 3CaBr₂ + Al₂(SO₄)₃
12. 2H₂O₂ → 2H₂O + O₂
13. 2Na + 2H₂O → 2NaOH + H₂
14. 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
15. 3Mg(NO₃)₂ + 2K₃PO₄ → Mg(PO₄)₂ + 6KNO₃
---
1. CH₄ + O₂ → CO₂ + H₂O
Left: C=1, H=4, O=2
Right: C=1, H=2, O=3 (2 from CO₂ + 1 from H₂O)
→ Put a 2 in front of H₂O to get 4 H on right:
CH₄ + O₂ → CO₂ + 2H₂O
Now Right: H=4, O=4 (2 from CO₂ + 2 from 2H₂O)
Left: O=2 → need 4 O → put 2 in front of O₂
✔ Balanced: CH₄ + 2O₂ → CO₂ + 2H₂O
---
2. Na + I₂ → NaI
Left: Na=1, I=2
Right: Na=1, I=1
→ Put 2 in front of NaI:
Na + I₂ → 2NaI
Now Right: Na=2, I=2 → so put 2 in front of Na
✔ Balanced: 2Na + I₂ → 2NaI
---
3. N₂ + O₂ → N₂O
Left: N=2, O=2
Right: N=2, O=1
→ Put 2 in front of N₂O:
N₂ + O₂ → 2N₂O
Now Right: N=4, O=2 → Left has N=2 → put 2 in front of N₂
But now N=4 on left, O=2 on left → matches right? Wait:
Wait — if we do:
2N₂ + O₂ → 2N₂O
Left: N=4, O=2
Right: N=4, O=2 → YES!
✔ Balanced: 2N₂ + O₂ → 2N₂O
*(Note: Some might think it's N₂ + ½O₂ → N₂O, but we use whole numbers only.)*
---
4. N₂ + H₂ → NH₃
Left: N=2, H=2
Right: N=1, H=3
→ Put 2 in front of NH₃:
N₂ + H₂ → 2NH₃
Now Right: N=2, H=6 → so need 6 H on left → put 3 in front of H₂
✔ Balanced: N₂ + 3H₂ → 2NH₃
---
5. KI + Cl₂ → KCl + I₂
Left: K=1, I=1, Cl=2
Right: K=1, Cl=1, I=2
→ Put 2 in front of KI and 2 in front of KCl:
2KI + Cl₂ → 2KCl + I₂
Check:
Left: K=2, I=2, Cl=2
Right: K=2, Cl=2, I=2 → Good!
✔ Balanced: 2KI + Cl₂ → 2KCl + I₂
---
6. HCl + Ca(OH)₂ → CaCl₂ + H₂O
Left: H=1+2=3? Wait — Ca(OH)₂ has 2 OH → so 2 H and 2 O from that, plus HCl has 1 H and 1 Cl.
Better to count per atom:
Ca(OH)₂ = Ca, 2O, 2H
So total left: H from HCl + 2H from Ca(OH)₂ → depends on how many HCl.
Try:
Put 2 in front of HCl:
2HCl + Ca(OH)₂ → CaCl₂ + H₂O
Left: H=2+2=4, Cl=2, Ca=1, O=2
Right: Ca=1, Cl=2, H=2, O=1 → not enough H or O on right
→ Put 2 in front of H₂O:
2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O
Now Right: H=4, O=2 → matches left!
✔ Balanced: 2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O
---
7. KClO₃ → KCl + O₂
Left: K=1, Cl=1, O=3
Right: K=1, Cl=1, O=2
→ Need even oxygen on right. Try 2KClO₃:
2KClO₃ → ?
Left: K=2, Cl=2, O=6
→ So right: 2KCl + 3O₂ (since 3×2=6 O)
✔ Balanced: 2KClO₃ → 2KCl + 3O₂
---
8. K₃PO₄ + HCl → KCl + H₃PO₄
Left: K=3, P=1, O=4, H=1, Cl=1
Right: K=1, Cl=1, H=3, P=1, O=4
→ Need 3 K on right → put 3 in front of KCl
→ Then Cl=3 on right → so put 3 in front of HCl
→ Now H=3 on left → matches H₃PO₄ on right
✔ Balanced: K₃PO₄ + 3HCl → 3KCl + H₃PO₄
---
9. S + O₂ → SO₃
Left: S=1, O=2
Right: S=1, O=3
→ Need common multiple for O: LCM of 2 and 3 is 6
→ So 3O₂ (gives 6 O) and 2SO₃ (needs 6 O)
Then S: 2 on right → so 2S on left
✔ Balanced: 2S + 3O₂ → 2SO₃
---
10. KI + Pb(NO₃)₂ → KNO₃ + PbI₂
Left: K=1, I=1, Pb=1, N=2, O=6
Right: K=1, N=1, O=3, Pb=1, I=2
→ Iodine: 2 on right → need 2 KI on left
→ Then K=2 on left → need 2 KNO₃ on right
→ Then N=2 on right → matches Pb(NO₃)₂ which has 2 N
✔ Balanced: 2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂
---
11. CaSO₄ + AlBr₃ → CaBr₂ + Al₂(SO₄)₃
Left: Ca=1, S=1, O=4, Al=1, Br=3
Right: Ca=1, Br=2, Al=2, S=3, O=12
→ Aluminum: 2 on right → need 2 AlBr₃ on left → then Br=6
→ Calcium bromide: need 3 CaBr₂ to get 6 Br → so 3 Ca on left → need 3 CaSO₄
→ Then S=3, O=12 on left → matches Al₂(SO₄)₃ on right
✔ Balanced: 3CaSO₄ + 2AlBr₃ → 3CaBr₂ + Al₂(SO₄)₃
---
12. H₂O₂ → H₂O + O₂
Left: H=2, O=2
Right: H=2, O=1+2=3 → too many O on right
→ Try 2H₂O₂:
2H₂O₂ → ?
Left: H=4, O=4
→ Right: need 4 H → so 2H₂O → gives H=4, O=2
→ Remaining O: 4-2=2 → so 1 O₂
✔ Balanced: 2H₂O₂ → 2H₂O + O₂
---
13. Na + H₂O → NaOH + H₂
Left: Na=1, H=2, O=1
Right: Na=1, O=1, H=1+2=3 → too many H on right
→ Try 2Na:
2Na + H₂O → ?
Still H mismatch.
Try 2Na + 2H₂O → 2NaOH + H₂
Left: Na=2, H=4, O=2
Right: Na=2, O=2, H=2+2=4 → yes!
✔ Balanced: 2Na + 2H₂O → 2NaOH + H₂
---
14. C₂H₆ + O₂ → CO₂ + H₂O
Left: C=2, H=6, O=2
Right: C=1, H=2, O=3
→ Put 2 in front of CO₂ → C=2
→ Put 3 in front of H₂O → H=6
→ Now O on right: 2×2 + 3×1 = 4+3=7 → odd number → hard with O₂
Multiply all by 2 to make O even:
Start over:
C₂H₆ + O₂ → 2CO₂ + 3H₂O
O right: 4 + 3 = 7 → so need 7/2 O₂ → multiply entire equation by 2:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Check:
Left: C=4, H=12, O=14
Right: C=4, H=12, O=8+6=14 → perfect!
✔ Balanced: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
---
15. Mg(NO₃)₂ + K₃PO₄ → Mg₃(PO₄)₂ + KNO₃
Left: Mg=1, N=2, O=6+4=10? Wait — better break down:
Mg(NO₃)₂ = Mg, 2N, 6O
K₃PO₄ = 3K, P, 4O
Total left per formula unit: varies.
Right: Mg₃(PO₄)₂ = 3Mg, 2P, 8O
KNO₃ = K, N, 3O
→ Need 3 Mg on right → so 3 Mg(NO₃)₂ on left → then N=6, O from nitrate=18
→ Need 2 PO₄ on right → so 2 K₃PO₄ on left → then K=6, P=2
→ Right: Mg₃(PO₄)₂ + ? KNO₃ → need 6 K → so 6 KNO₃ → then N=6, O=18 from KNO₃ + 8 from phosphate = 26? Let’s calculate properly.
Left:
3 Mg(NO₃)₂ → 3Mg, 6N, 18O
2 K₃PO₄ → 6K, 2P, 8O
Total left: Mg=3, N=6, O=26, K=6, P=2
Right:
Mg₃(PO)₂ → 3Mg, 2P, 8O
6 KNO₃ → 6K, 6N, 18O
Total right: Mg=3, P=2, O=8+18=26, K=6, N=6 → matches!
✔ Balanced: 3Mg(NO₃)₂ + 2K₃PO₄ → Mg₃(PO₄)₂ + 6KNO₃
---
Final Answer:
1. CH₄ + 2O₂ → CO₂ + 2H₂O
2. 2Na + I₂ → 2NaI
3. 2N₂ + O₂ → 2N₂O
4. N₂ + 3H₂ → 2NH₃
5. 2KI + Cl₂ → 2KCl + I₂
6. 2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O
7. 2KClO₃ → 2KCl + 3O₂
8. K₃PO₄ + 3HCl → 3KCl + H₃PO₄
9. 2S + 3O₂ → 2SO₃
10. 2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂
11. 3CaSO₄ + 2AlBr₃ → 3CaBr₂ + Al₂(SO₄)₃
12. 2H₂O₂ → 2H₂O + O₂
13. 2Na + 2H₂O → 2NaOH + H₂
14. 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
15. 3Mg(NO₃)₂ + 2K₃PO₄ → Mg(PO₄)₂ + 6KNO₃
Parent Tip: Review the logic above to help your child master the concept of physical science if8767 worksheet.
