49 Balancing Chemical Equations Worksheets [with Answers] - Free Printable
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Step-by-step solution for: 49 Balancing Chemical Equations Worksheets [with Answers]
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Show Answer Key & Explanations
Step-by-step solution for: 49 Balancing Chemical Equations Worksheets [with Answers]
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.
We’ll write coefficients (numbers in front) to make atoms match — never change subscripts!
---
1. N₂ + H₂ → NH₃
Left: 2 N, 2 H
Right: 1 N, 3 H
→ Put 2 in front of NH₃: now right has 2 N, 6 H
→ So left needs 3 H₂ (since 3×2=6 H)
→ N₂ already gives 2 N
✔ Balanced: N₂ + 3H₂ → 2NH₃
---
2. S₈ + O₂ → SO₃
Left: 8 S, 2 O
Right: 1 S, 3 O
→ Need 8 SO₃ on right → 8 S, 24 O
→ So need 12 O₂ on left (12×2=24 O)
✔ Balanced: S₈ + 12O₂ → 8SO₃
---
3. HgO → Hg + O₂
Left: 1 Hg, 1 O
Right: 1 Hg, 2 O
→ Need 2 HgO on left → 2 Hg, 2 O
→ Then right needs 2 Hg and 1 O₂
✔ Balanced: 2HgO → 2Hg + O₂
---
4. Zn + HCl → ZnCl₂ + H₂
Left: 1 Zn, 1 H, 1 Cl
Right: 1 Zn, 2 Cl, 2 H
→ Need 2 HCl on left → 2 H, 2 Cl
→ Right already has ZnCl₂ and H₂
✔ Balanced: Zn + 2HCl → ZnCl₂ + H₂
---
5. SiCl₄ + H₂O → H₄SiO₄ + HCl
Left: 1 Si, 4 Cl, 2 H, 1 O
Right: 1 Si, 4 H, 4 O, 1 H, 1 Cl → total 5 H, 4 O, 1 Cl? Wait — let's count properly:
H₄SiO₄ = 4 H, 1 Si, 4 O
HCl = 1 H, 1 Cl
So right: 5 H, 1 Si, 4 O, 1 Cl
Left: SiCl₄ = 1 Si, 4 Cl; H₂O = 2 H, 1 O
→ Need 4 HCl on right → so 4 Cl from left → good
→ But then right has 4 H from HCl + 4 H from H₄SiO₄ = 8 H
→ Left: need 4 H₂O → 8 H, 4 O
→ Right: H₄SiO₄ has 4 O → perfect
✔ Balanced: SiCl₄ + 4H₂O → H₄SiO₄ + 4HCl
---
6. Na + H₂O → NaOH + H₂
Left: 1 Na, 2 H, 1 O
Right: 1 Na, 1 O, 1 H (from NaOH) + 2 H (from H₂) = 3 H total? Wait:
NaOH = 1 Na, 1 O, 1 H
H₂ = 2 H
Total right: 1 Na, 1 O, 3 H
Left: H₂O has 2 H → not enough
→ Try 2 Na on left → 2 Na
→ Then 2 NaOH on right → 2 Na, 2 O, 2 H
→ Plus H₂ → 2 more H → total 4 H on right
→ Left: need 2 H₂O → 4 H, 2 O → matches
✔ Balanced: 2Na + 2H₂O → 2NaOH + H₂
Wait — check H: left 4 H, right: 2 from NaOH + 2 from H₂ = 4 → yes
O: 2 on each side
Na: 2 on each side
Actually, standard balanced form is:
2Na + 2H₂O → 2NaOH + H₂
Yes.
---
7. H₃PO₄ → HP₂O₇ + H₂O
Left: 3 H, 1 P, 4 O
Right: H₄P₂O₇ = 4 H, 2 P, 7 O; H₂O = 2 H, 1 O → total 6 H, 2 P, 8 O
→ Need 2 H₃PO₄ on left → 6 H, 2 P, 8 O → matches right
✔ Balanced: 2H₃PO₄ → H₄P₂O₇ + H₂O
Check: left 6H, 2P, 8O; right: H₄P₂O₇ (4H,2P,7O) + H₂O (2H,1O) = 6H,2P,8O → yes
---
8. Si₂H₃ + O₂ → SiO₂ + H₂O
Left: 2 Si, 3 H, 2 O
Right: SiO₂ = 1 Si, 2 O; H₂O = 2 H, 1 O
→ Need 2 SiO₂ on right → 2 Si, 4 O
→ Need 3/2 H₂O for 3 H? Better use whole numbers.
Multiply everything by 2 later.
Try:
Si₂H₃ + ? O₂ → 2SiO₂ + ? H₂O
H: 3 on left → need 3/2 H₂O → so multiply all by 2:
2Si₂H₃ + ? O₂ → 4SiO₂ + 3H₂O
Now H: 6 on left → 6 on right (3×2)
Si: 4 on each
O: right = 4×2 + 3×1 = 8+3=11 → so need 11/2 O₂ → multiply all by 2 again?
Better approach:
Set coefficients:
a Si₂H₃ + b O₂ → c SiO₂ + d H₂O
Si: 2a = c
H: 3a = 2d
O: 2b = 2c + d
From Si: c = 2a
From H: d = (3a)/2
Plug into O: 2b = 2*(2a) + (3a)/2 = 4a + 1.5a = 5.5a → b = 2.75a
Multiply all by 4 to eliminate decimals:
a=4 → c=8, d=6, b=11
✔ Balanced: 4Si₂H₃ + 11O₂ → 8SiO₂ + 6H₂O
Check:
Left: Si=8, H=12, O=22
Right: Si=8, O=16+6=22, H=12 → yes
---
9. Al(OH)₃ + H₂SO₄ → Al₂(SO₄)₃ + H₂O
Left: Al, 3O, 3H from hydroxide + 2H, S, 4O from acid → messy
Better: treat OH and SO₄ as groups if possible, but let’s count atoms.
Al(OH)₃ = 1 Al, 3 O, 3 H
H₂SO₄ = 2 H, 1 S, 4 O
Right: Al₂(SO₄)₃ = 2 Al, 3 S, 12 O
H₂O = 2 H, 1 O
Need 2 Al on left → 2 Al(OH)₃
Need 3 SO₄ on right → 3 H₂SO₄ on left
Left: 2 Al(OH)₃ → 2 Al, 6 O, 6 H
3 H₂SO₄ → 6 H, 3 S, 12 O
Total left: 2 Al, 3 S, 18 O, 12 H
Right: Al₂(SO₄)₃ → 2 Al, 3 S, 12 O
Need water: remaining H and O → 12 H and 6 O left → that’s 6 H₂O
Check: 6 H₂O = 12 H, 6 O → total right O: 12 + 6 = 18 → matches
✔ Balanced: 2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O
---
10. Fe + O₂ → Fe₂O₃
Left: 1 Fe, 2 O
Right: 2 Fe, 3 O
→ Need 2 Fe on left → 2 Fe
→ Need 3/2 O₂ → so multiply by 2:
4 Fe + 3 O₂ → 2 Fe₂O₃
Check: left 4 Fe, 6 O; right 4 Fe, 6 O → yes
✔ Balanced: 4Fe + 3O₂ → 2Fe₂O₃
---
11. Fe₂(SO₄)₃ + KOH → K₂SO₄ + Fe(OH)₃
Left: Fe₂(SO₄)₃ = 2 Fe, 3 S, 12 O
KOH = 1 K, 1 O, 1 H
Right: K₂SO₄ = 2 K, 1 S, 4 O
Fe(OH)₃ = 1 Fe, 3 O, 3 H
Need 2 Fe on right → 2 Fe(OH)₃
Need 3 SO₄ on right → 3 K₂SO₄ → so 6 K on right → need 6 KOH on left
Left: Fe₂(SO₄)₃ + 6 KOH → 2 Fe, 3 S, 12 O + 6 K, 6 O, 6 H → total 2 Fe, 3 S, 18 O, 6 K, 6 H
Right: 3 K₂SO₄ = 6 K, 3 S, 12 O
2 Fe(OH)₃ = 2 Fe, 6 O, 6 H → total 2 Fe, 3 S, 18 O, 6 K, 6 H → matches
✔ Balanced: Fe₂(SO₄)₃ + 6KOH → 3K₂SO₄ + 2Fe(OH)₃
---
12. FeS₂ + O₂ → Fe₂O₃ + SO₂
Left: Fe, 2 S, 2 O
Right: Fe₂O₃ = 2 Fe, 3 O; SO₂ = 1 S, 2 O
Need 2 Fe on left → 2 FeS₂ → 2 Fe, 4 S
Then right: need 2 Fe₂O₃? No — only 2 Fe total → so 1 Fe₂O₃? But that requires 2 Fe → ok
But S: 4 S on left → need 4 SO₂ on right
Now O: right: Fe₂O₃ has 3 O, 4 SO₂ has 8 O → total 11 O → so need 11/2 O₂ on left
Multiply all by 2:
4 FeS₂ + 11 O₂ → 2 Fe₂O₃ + 8 SO₂
Check:
Left: Fe=4, S=8, O=22
Right: Fe=4, S=8, O=6+16=22 → yes
✔ Balanced: 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
---
13. Al + FeO → Al₂O₃ + Fe
Left: Al, Fe, O
Right: 2 Al, 3 O, Fe
Need 2 Al on left → 2 Al
Need 3 O on left → 3 FeO → 3 Fe, 3 O
Then right: Al₂O₃ + 3 Fe
✔ Balanced: 2Al + 3FeO → Al₂O₃ + 3Fe
---
14. Na₂CO₃ + HCl → NaCl + H₂O + CO₂
Left: 2 Na, 1 C, 3 O, 1 H, 1 Cl
Right: NaCl = 1 Na, 1 Cl; H₂O = 2 H, 1 O; CO₂ = 1 C, 2 O
Need 2 NaCl on right → so 2 HCl on left → 2 H, 2 Cl
Left: Na₂CO₃ + 2HCl → 2 Na, 1 C, 3 O, 2 H, 2 Cl
Right: 2NaCl + H₂O + CO₂ → 2 Na, 2 Cl, 2 H, 1 O + 1 C, 2 O → total 2 Na, 2 Cl, 2 H, 3 O, 1 C → matches
✔ Balanced: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂
---
15. K + Br₂ → KBr
Left: 1 K, 2 Br
Right: 1 K, 1 Br
→ Need 2 KBr on right → 2 K, 2 Br
→ So 2 K on left
✔ Balanced: 2K + Br₂ → 2KBr
---
16. P₄ + O₂ → P₂O₅
Left: 4 P, 2 O
Right: 2 P, 5 O
→ Need 2 P₂O₅ on right → 4 P, 10 O
→ So need 5 O₂ on left
✔ Balanced: P₄ + 5O₂ → 2P₂O₅
---
17. C₂H₂ + O₂ → CO₂ + H₂O
Left: 2 C, 2 H, 2 O
Right: CO₂ = 1 C, 2 O; H₂O = 2 H, 1 O
Need 2 CO₂ on right → 2 C, 4 O
Need 1 H₂O → 2 H, 1 O → total O: 5 → so need 5/2 O₂ on left
Multiply by 2:
2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
Check:
Left: C=4, H=4, O=10
Right: C=4, O=8+2=10, H=4 → yes
✔ Balanced: 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
---
Final Answer:
1. N₂ + 3H₂ → 2NH₃
2. S₈ + 12O₂ → 8SO₃
3. 2HgO → 2Hg + O₂
4. Zn + 2HCl → ZnCl₂ + H₂
5. SiCl₄ + 4H₂O → H₄SiO₄ + 4HCl
6. 2Na + 2H₂O → 2NaOH + H₂
7. 2H₃PO₄ → H₄P₂O₇ + H₂O
8. 4Si₂H₃ + 11O₂ → 8SiO₂ + 6H₂O
9. 2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O
10. 4Fe + 3O₂ → 2Fe₂O₃
11. Fe₂(SO₄)₃ + 6KOH → 3K₂SO₄ + 2Fe(OH)₃
12. 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
13. 2Al + 3FeO → Al₂O₃ + 3Fe
14. Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂
15. 2K + Br₂ → 2KBr
16. P₄ + 5O₂ → 2P₂O₅
17. 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
We’ll write coefficients (numbers in front) to make atoms match — never change subscripts!
---
1. N₂ + H₂ → NH₃
Left: 2 N, 2 H
Right: 1 N, 3 H
→ Put 2 in front of NH₃: now right has 2 N, 6 H
→ So left needs 3 H₂ (since 3×2=6 H)
→ N₂ already gives 2 N
✔ Balanced: N₂ + 3H₂ → 2NH₃
---
2. S₈ + O₂ → SO₃
Left: 8 S, 2 O
Right: 1 S, 3 O
→ Need 8 SO₃ on right → 8 S, 24 O
→ So need 12 O₂ on left (12×2=24 O)
✔ Balanced: S₈ + 12O₂ → 8SO₃
---
3. HgO → Hg + O₂
Left: 1 Hg, 1 O
Right: 1 Hg, 2 O
→ Need 2 HgO on left → 2 Hg, 2 O
→ Then right needs 2 Hg and 1 O₂
✔ Balanced: 2HgO → 2Hg + O₂
---
4. Zn + HCl → ZnCl₂ + H₂
Left: 1 Zn, 1 H, 1 Cl
Right: 1 Zn, 2 Cl, 2 H
→ Need 2 HCl on left → 2 H, 2 Cl
→ Right already has ZnCl₂ and H₂
✔ Balanced: Zn + 2HCl → ZnCl₂ + H₂
---
5. SiCl₄ + H₂O → H₄SiO₄ + HCl
Left: 1 Si, 4 Cl, 2 H, 1 O
Right: 1 Si, 4 H, 4 O, 1 H, 1 Cl → total 5 H, 4 O, 1 Cl? Wait — let's count properly:
H₄SiO₄ = 4 H, 1 Si, 4 O
HCl = 1 H, 1 Cl
So right: 5 H, 1 Si, 4 O, 1 Cl
Left: SiCl₄ = 1 Si, 4 Cl; H₂O = 2 H, 1 O
→ Need 4 HCl on right → so 4 Cl from left → good
→ But then right has 4 H from HCl + 4 H from H₄SiO₄ = 8 H
→ Left: need 4 H₂O → 8 H, 4 O
→ Right: H₄SiO₄ has 4 O → perfect
✔ Balanced: SiCl₄ + 4H₂O → H₄SiO₄ + 4HCl
---
6. Na + H₂O → NaOH + H₂
Left: 1 Na, 2 H, 1 O
Right: 1 Na, 1 O, 1 H (from NaOH) + 2 H (from H₂) = 3 H total? Wait:
NaOH = 1 Na, 1 O, 1 H
H₂ = 2 H
Total right: 1 Na, 1 O, 3 H
Left: H₂O has 2 H → not enough
→ Try 2 Na on left → 2 Na
→ Then 2 NaOH on right → 2 Na, 2 O, 2 H
→ Plus H₂ → 2 more H → total 4 H on right
→ Left: need 2 H₂O → 4 H, 2 O → matches
✔ Balanced: 2Na + 2H₂O → 2NaOH + H₂
Wait — check H: left 4 H, right: 2 from NaOH + 2 from H₂ = 4 → yes
O: 2 on each side
Na: 2 on each side
Actually, standard balanced form is:
2Na + 2H₂O → 2NaOH + H₂
Yes.
---
7. H₃PO₄ → HP₂O₇ + H₂O
Left: 3 H, 1 P, 4 O
Right: H₄P₂O₇ = 4 H, 2 P, 7 O; H₂O = 2 H, 1 O → total 6 H, 2 P, 8 O
→ Need 2 H₃PO₄ on left → 6 H, 2 P, 8 O → matches right
✔ Balanced: 2H₃PO₄ → H₄P₂O₇ + H₂O
Check: left 6H, 2P, 8O; right: H₄P₂O₇ (4H,2P,7O) + H₂O (2H,1O) = 6H,2P,8O → yes
---
8. Si₂H₃ + O₂ → SiO₂ + H₂O
Left: 2 Si, 3 H, 2 O
Right: SiO₂ = 1 Si, 2 O; H₂O = 2 H, 1 O
→ Need 2 SiO₂ on right → 2 Si, 4 O
→ Need 3/2 H₂O for 3 H? Better use whole numbers.
Multiply everything by 2 later.
Try:
Si₂H₃ + ? O₂ → 2SiO₂ + ? H₂O
H: 3 on left → need 3/2 H₂O → so multiply all by 2:
2Si₂H₃ + ? O₂ → 4SiO₂ + 3H₂O
Now H: 6 on left → 6 on right (3×2)
Si: 4 on each
O: right = 4×2 + 3×1 = 8+3=11 → so need 11/2 O₂ → multiply all by 2 again?
Better approach:
Set coefficients:
a Si₂H₃ + b O₂ → c SiO₂ + d H₂O
Si: 2a = c
H: 3a = 2d
O: 2b = 2c + d
From Si: c = 2a
From H: d = (3a)/2
Plug into O: 2b = 2*(2a) + (3a)/2 = 4a + 1.5a = 5.5a → b = 2.75a
Multiply all by 4 to eliminate decimals:
a=4 → c=8, d=6, b=11
✔ Balanced: 4Si₂H₃ + 11O₂ → 8SiO₂ + 6H₂O
Check:
Left: Si=8, H=12, O=22
Right: Si=8, O=16+6=22, H=12 → yes
---
9. Al(OH)₃ + H₂SO₄ → Al₂(SO₄)₃ + H₂O
Left: Al, 3O, 3H from hydroxide + 2H, S, 4O from acid → messy
Better: treat OH and SO₄ as groups if possible, but let’s count atoms.
Al(OH)₃ = 1 Al, 3 O, 3 H
H₂SO₄ = 2 H, 1 S, 4 O
Right: Al₂(SO₄)₃ = 2 Al, 3 S, 12 O
H₂O = 2 H, 1 O
Need 2 Al on left → 2 Al(OH)₃
Need 3 SO₄ on right → 3 H₂SO₄ on left
Left: 2 Al(OH)₃ → 2 Al, 6 O, 6 H
3 H₂SO₄ → 6 H, 3 S, 12 O
Total left: 2 Al, 3 S, 18 O, 12 H
Right: Al₂(SO₄)₃ → 2 Al, 3 S, 12 O
Need water: remaining H and O → 12 H and 6 O left → that’s 6 H₂O
Check: 6 H₂O = 12 H, 6 O → total right O: 12 + 6 = 18 → matches
✔ Balanced: 2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O
---
10. Fe + O₂ → Fe₂O₃
Left: 1 Fe, 2 O
Right: 2 Fe, 3 O
→ Need 2 Fe on left → 2 Fe
→ Need 3/2 O₂ → so multiply by 2:
4 Fe + 3 O₂ → 2 Fe₂O₃
Check: left 4 Fe, 6 O; right 4 Fe, 6 O → yes
✔ Balanced: 4Fe + 3O₂ → 2Fe₂O₃
---
11. Fe₂(SO₄)₃ + KOH → K₂SO₄ + Fe(OH)₃
Left: Fe₂(SO₄)₃ = 2 Fe, 3 S, 12 O
KOH = 1 K, 1 O, 1 H
Right: K₂SO₄ = 2 K, 1 S, 4 O
Fe(OH)₃ = 1 Fe, 3 O, 3 H
Need 2 Fe on right → 2 Fe(OH)₃
Need 3 SO₄ on right → 3 K₂SO₄ → so 6 K on right → need 6 KOH on left
Left: Fe₂(SO₄)₃ + 6 KOH → 2 Fe, 3 S, 12 O + 6 K, 6 O, 6 H → total 2 Fe, 3 S, 18 O, 6 K, 6 H
Right: 3 K₂SO₄ = 6 K, 3 S, 12 O
2 Fe(OH)₃ = 2 Fe, 6 O, 6 H → total 2 Fe, 3 S, 18 O, 6 K, 6 H → matches
✔ Balanced: Fe₂(SO₄)₃ + 6KOH → 3K₂SO₄ + 2Fe(OH)₃
---
12. FeS₂ + O₂ → Fe₂O₃ + SO₂
Left: Fe, 2 S, 2 O
Right: Fe₂O₃ = 2 Fe, 3 O; SO₂ = 1 S, 2 O
Need 2 Fe on left → 2 FeS₂ → 2 Fe, 4 S
Then right: need 2 Fe₂O₃? No — only 2 Fe total → so 1 Fe₂O₃? But that requires 2 Fe → ok
But S: 4 S on left → need 4 SO₂ on right
Now O: right: Fe₂O₃ has 3 O, 4 SO₂ has 8 O → total 11 O → so need 11/2 O₂ on left
Multiply all by 2:
4 FeS₂ + 11 O₂ → 2 Fe₂O₃ + 8 SO₂
Check:
Left: Fe=4, S=8, O=22
Right: Fe=4, S=8, O=6+16=22 → yes
✔ Balanced: 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
---
13. Al + FeO → Al₂O₃ + Fe
Left: Al, Fe, O
Right: 2 Al, 3 O, Fe
Need 2 Al on left → 2 Al
Need 3 O on left → 3 FeO → 3 Fe, 3 O
Then right: Al₂O₃ + 3 Fe
✔ Balanced: 2Al + 3FeO → Al₂O₃ + 3Fe
---
14. Na₂CO₃ + HCl → NaCl + H₂O + CO₂
Left: 2 Na, 1 C, 3 O, 1 H, 1 Cl
Right: NaCl = 1 Na, 1 Cl; H₂O = 2 H, 1 O; CO₂ = 1 C, 2 O
Need 2 NaCl on right → so 2 HCl on left → 2 H, 2 Cl
Left: Na₂CO₃ + 2HCl → 2 Na, 1 C, 3 O, 2 H, 2 Cl
Right: 2NaCl + H₂O + CO₂ → 2 Na, 2 Cl, 2 H, 1 O + 1 C, 2 O → total 2 Na, 2 Cl, 2 H, 3 O, 1 C → matches
✔ Balanced: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂
---
15. K + Br₂ → KBr
Left: 1 K, 2 Br
Right: 1 K, 1 Br
→ Need 2 KBr on right → 2 K, 2 Br
→ So 2 K on left
✔ Balanced: 2K + Br₂ → 2KBr
---
16. P₄ + O₂ → P₂O₅
Left: 4 P, 2 O
Right: 2 P, 5 O
→ Need 2 P₂O₅ on right → 4 P, 10 O
→ So need 5 O₂ on left
✔ Balanced: P₄ + 5O₂ → 2P₂O₅
---
17. C₂H₂ + O₂ → CO₂ + H₂O
Left: 2 C, 2 H, 2 O
Right: CO₂ = 1 C, 2 O; H₂O = 2 H, 1 O
Need 2 CO₂ on right → 2 C, 4 O
Need 1 H₂O → 2 H, 1 O → total O: 5 → so need 5/2 O₂ on left
Multiply by 2:
2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
Check:
Left: C=4, H=4, O=10
Right: C=4, O=8+2=10, H=4 → yes
✔ Balanced: 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
---
Final Answer:
1. N₂ + 3H₂ → 2NH₃
2. S₈ + 12O₂ → 8SO₃
3. 2HgO → 2Hg + O₂
4. Zn + 2HCl → ZnCl₂ + H₂
5. SiCl₄ + 4H₂O → H₄SiO₄ + 4HCl
6. 2Na + 2H₂O → 2NaOH + H₂
7. 2H₃PO₄ → H₄P₂O₇ + H₂O
8. 4Si₂H₃ + 11O₂ → 8SiO₂ + 6H₂O
9. 2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O
10. 4Fe + 3O₂ → 2Fe₂O₃
11. Fe₂(SO₄)₃ + 6KOH → 3K₂SO₄ + 2Fe(OH)₃
12. 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
13. 2Al + 3FeO → Al₂O₃ + 3Fe
14. Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂
15. 2K + Br₂ → 2KBr
16. P₄ + 5O₂ → 2P₂O₅
17. 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
Parent Tip: Review the logic above to help your child master the concept of worksheet on balancing equations.
