Balancing Chemical Equations Easy Introduction Problems 8th 10th ... - Free Printable
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Step-by-step solution for: Balancing Chemical Equations Easy Introduction Problems 8th 10th ...
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Show Answer Key & Explanations
Step-by-step solution for: Balancing Chemical Equations Easy Introduction Problems 8th 10th ...
Let’s go through each equation one by one and balance them step by step. 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. N₂ + H₂ → NH₃
Left: 2 N, 2 H
Right: 1 N, 3 H
We need to make N and H equal on both sides.
Try putting a 2 in front of NH₃ → now right has 2 N and 6 H.
So left needs 6 H → put 3 in front of H₂ (3×2=6 H)
N₂ already gives 2 N → matches.
✔ Balanced: N₂ + 3H₂ → 2NH₃
---
2. S₈ + O₂ → SO₃
Left: 8 S, 2 O
Right: 1 S, 3 O
We have 8 S on left → need 8 SO₃ on right → that’s 8 S and 24 O.
O₂ on left → need 12 O₂ to get 24 O (12×2=24)
✔ Balanced: S₈ + 12O₂ → 8SO₃
---
3. HgO → Hg + O₂
Left: 1 Hg, 1 O
Right: 1 Hg, 2 O
Need even number of O on left → try 2 HgO → 2 Hg, 2 O
Then right: 2 Hg and 1 O₂ → perfect.
✔ 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 → gives 2 H and 2 Cl → matches right.
✔ Balanced: Zn + 2HCl → ZnCl₂ + H₂
---
5. SiCl₄ + H₂O → H₄SiO + 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
Total right: 5 H, 1 Si, 4 O, 1 Cl
Left: SiCl₄ = 1 Si, 4 Cl; H₂O = 2 H, 1 O → so we need more water and more HCl.
Try 4 H₂O → gives 8 H, 4 O
Now left: 1 Si, 4 Cl, 8 H, 4 O
Right: H₄SiO₄ uses 4 H, 4 O, 1 Si → leaves 4 H and 4 Cl → which makes 4 HCl
✔ 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 → total 3 H? 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 + 2 H₂O → left: 2 Na, 4 H, 2 O
Right: 2 NaOH = 2 Na, 2 O, 2 H → plus H₂ = 2 H → total 4 H → good!
✔ Balanced: 2Na + 2H₂O → 2NaOH + H₂
---
7. H₃PO₄ → H₄P₂O₇ + 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
So we need 2 H₃PO₄ on left → 6 H, 2 P, 8 O → matches right.
✔ Balanced: 2H₃PO₄ → H₄P₂O₇ + H₂O
---
8. Si₂H₃ + O₂ → SiO₂ + H₂O
Left: 2 Si, 3 H, 2 O
Right: 1 Si, 2 O from SiO₂; 2 H, 1 O from H₂O → total per set: 1 Si, 2 H, 3 O
We have 2 Si on left → need 2 SiO₂ on right → 2 Si, 4 O
H: 3 on left → need multiple of 2 on right? Try 3 H₂O → 6 H, 3 O → too many.
Wait — let’s find LCM for H: left has 3 H, right has 2 H per H₂O → LCM is 6.
So multiply Si₂H₃ by 2 → 4 Si, 6 H
Then right: 4 SiO₂ (for 4 Si) → 4 Si, 8 O
And 3 H₂O → 6 H, 3 O → total O on right: 8+3=11 → odd? Problem.
Wait — better way:
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 → so a must be even → try a=2
Then c=4, d=3
O: 2b = 2*4 + 3 = 8+3=11 → b=5.5 → not integer.
Try a=4 → c=8, d=6
O: 2b = 2*8 + 6 = 16+6=22 → b=11
Perfect!
✔ Balanced: 4Si₂H₃ + 11O₂ → 8SiO₂ + 6H₂O
---
9. Al(OH)₃ + H₂SO₄ → Al₂(SO₄)₃ + H₂O
Left: Al, 3O, 3H from hydroxide + 2H, 1S, 4O from acid → messy.
Better: look at products.
Al₂(SO₄)₃ has 2 Al, 3 SO₄ groups.
So need 2 Al(OH)₃ and 3 H₂SO₄ on left.
Left: 2 Al, 6 OH, 6 H, 3 SO₄
Right: Al₂(SO₄)₃ + ? H₂O
The H and OH will form water: 6 H + 6 OH → 6 H₂O
Check:
Left: 2 Al, 3 S, 6 O from sulfate + 6 O from hydroxide? Wait — better count atoms.
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
Plus 6 H₂O = 12 H, 6 O → total 2 Al, 3 S, 18 O, 12 H → matches.
✔ Balanced: 2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O
---
10. Fe + O₂ → Fe₂O₃
Left: 1 Fe, 2 O
Right: 2 Fe, 3 O
LCM for Fe: 2 → so 2 Fe on left? But then O: 2 vs 3 → LCM 6.
Try 4 Fe + 3 O₂ → 4 Fe, 6 O
Right: 2 Fe₂O₃ = 4 Fe, 6 O → perfect.
✔ Balanced: 4Fe + 3O₂ → 2Fe₂O₃
---
11. Fe₂(SO₄)₃ + KOH → K₂SO₄ + Fe(OH)₃
Left: 2 Fe, 3 SO₄, 1 K, 1 O, 1 H
Right: 2 K, 1 SO₄, 1 Fe, 3 O, 3 H
Need 3 K₂SO₄ on right → 6 K, 3 SO₄
So left: need 6 KOH → 6 K, 6 O, 6 H
Right: Fe(OH)₃ → need 2 Fe(OH)₃ → 2 Fe, 6 O, 6 H
Check:
Left: Fe₂(SO₄)₃ = 2 Fe, 3 S, 12 O
6 KOH = 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: 1 Fe, 2 S, 2 O
Right: 2 Fe, 3 O, 1 S, 2 O → total 2 Fe, 1 S, 5 O? Not matching.
Need 2 Fe on right → so 2 FeS₂ on left → 2 Fe, 4 S
Then right: need 4 SO₂ → 4 S, 8 O
Plus Fe₂O₃ → 2 Fe, 3 O → total O on right: 8+3=11
Left: O₂ → need 11/2 → so multiply all by 2.
Try 4 FeS₂ + ? O₂ → 2 Fe₂O₃ + 8 SO₂
Left: 4 Fe, 8 S
Right: 4 Fe, 8 S, O: 2*3 + 8*2 = 6+16=22 O → so 11 O₂
✔ Balanced: 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
---
13. Al + FeO → Al₂O₃ + Fe
Left: 1 Al, 1 Fe, 1 O
Right: 2 Al, 3 O, 1 Fe
Need 2 Al on left → 2 Al
Need 3 O on left → 3 FeO → 3 Fe, 3 O
Right: Al₂O₃ + 3 Fe
Check: left: 2 Al, 3 Fe, 3 O → right: 2 Al, 3 O, 3 Fe → good.
✔ Balanced: 2Al + 3FeO → Al₂O₃ + 3Fe
---
14. Na₂CO₃ + HCl → NaCl + H₂O + CO₂
Left: 2 Na, 1 C, 3 O, 1 H, 1 Cl
Right: 1 Na, 1 Cl, 2 H, 1 O, 1 C, 2 O → total 1 Na, 1 Cl, 2 H, 3 O, 1 C
Need 2 NaCl on right → 2 Na, 2 Cl
So left: need 2 HCl → 2 H, 2 Cl
Now left: 2 Na, 1 C, 3 O, 2 H, 2 Cl
Right: 2 NaCl + 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 left: 2 K
✔ 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
Left: O₂ → need 5 O₂ → 10 O
✔ Balanced: P₄ + 5O₂ → 2P₂O₅
---
17. C₂H₂ + O₂ → CO₂ + H₂O
Left: 2 C, 2 H, 2 O
Right: 1 C, 2 O, 2 H, 1 O → total 1 C, 2 H, 3 O
Need 2 CO₂ on right → 2 C, 4 O
Plus H₂O → 2 H, 1 O → total 2 C, 2 H, 5 O
Left: O₂ → need 5/2 → so multiply all by 2.
2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O
Check:
Left: 4 C, 4 H, 10 O
Right: 4 C, 8 O from CO₂ + 2 H, 2 O from H₂O → total 4 C, 4 H, 10 O → good.
✔ 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
---
1. N₂ + H₂ → NH₃
Left: 2 N, 2 H
Right: 1 N, 3 H
We need to make N and H equal on both sides.
Try putting a 2 in front of NH₃ → now right has 2 N and 6 H.
So left needs 6 H → put 3 in front of H₂ (3×2=6 H)
N₂ already gives 2 N → matches.
✔ Balanced: N₂ + 3H₂ → 2NH₃
---
2. S₈ + O₂ → SO₃
Left: 8 S, 2 O
Right: 1 S, 3 O
We have 8 S on left → need 8 SO₃ on right → that’s 8 S and 24 O.
O₂ on left → need 12 O₂ to get 24 O (12×2=24)
✔ Balanced: S₈ + 12O₂ → 8SO₃
---
3. HgO → Hg + O₂
Left: 1 Hg, 1 O
Right: 1 Hg, 2 O
Need even number of O on left → try 2 HgO → 2 Hg, 2 O
Then right: 2 Hg and 1 O₂ → perfect.
✔ 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 → gives 2 H and 2 Cl → matches right.
✔ Balanced: Zn + 2HCl → ZnCl₂ + H₂
---
5. SiCl₄ + H₂O → H₄SiO + 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
Total right: 5 H, 1 Si, 4 O, 1 Cl
Left: SiCl₄ = 1 Si, 4 Cl; H₂O = 2 H, 1 O → so we need more water and more HCl.
Try 4 H₂O → gives 8 H, 4 O
Now left: 1 Si, 4 Cl, 8 H, 4 O
Right: H₄SiO₄ uses 4 H, 4 O, 1 Si → leaves 4 H and 4 Cl → which makes 4 HCl
✔ 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 → total 3 H? 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 + 2 H₂O → left: 2 Na, 4 H, 2 O
Right: 2 NaOH = 2 Na, 2 O, 2 H → plus H₂ = 2 H → total 4 H → good!
✔ Balanced: 2Na + 2H₂O → 2NaOH + H₂
---
7. H₃PO₄ → H₄P₂O₇ + 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
So we need 2 H₃PO₄ on left → 6 H, 2 P, 8 O → matches right.
✔ Balanced: 2H₃PO₄ → H₄P₂O₇ + H₂O
---
8. Si₂H₃ + O₂ → SiO₂ + H₂O
Left: 2 Si, 3 H, 2 O
Right: 1 Si, 2 O from SiO₂; 2 H, 1 O from H₂O → total per set: 1 Si, 2 H, 3 O
We have 2 Si on left → need 2 SiO₂ on right → 2 Si, 4 O
H: 3 on left → need multiple of 2 on right? Try 3 H₂O → 6 H, 3 O → too many.
Wait — let’s find LCM for H: left has 3 H, right has 2 H per H₂O → LCM is 6.
So multiply Si₂H₃ by 2 → 4 Si, 6 H
Then right: 4 SiO₂ (for 4 Si) → 4 Si, 8 O
And 3 H₂O → 6 H, 3 O → total O on right: 8+3=11 → odd? Problem.
Wait — better way:
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 → so a must be even → try a=2
Then c=4, d=3
O: 2b = 2*4 + 3 = 8+3=11 → b=5.5 → not integer.
Try a=4 → c=8, d=6
O: 2b = 2*8 + 6 = 16+6=22 → b=11
Perfect!
✔ Balanced: 4Si₂H₃ + 11O₂ → 8SiO₂ + 6H₂O
---
9. Al(OH)₃ + H₂SO₄ → Al₂(SO₄)₃ + H₂O
Left: Al, 3O, 3H from hydroxide + 2H, 1S, 4O from acid → messy.
Better: look at products.
Al₂(SO₄)₃ has 2 Al, 3 SO₄ groups.
So need 2 Al(OH)₃ and 3 H₂SO₄ on left.
Left: 2 Al, 6 OH, 6 H, 3 SO₄
Right: Al₂(SO₄)₃ + ? H₂O
The H and OH will form water: 6 H + 6 OH → 6 H₂O
Check:
Left: 2 Al, 3 S, 6 O from sulfate + 6 O from hydroxide? Wait — better count atoms.
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
Plus 6 H₂O = 12 H, 6 O → total 2 Al, 3 S, 18 O, 12 H → matches.
✔ Balanced: 2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O
---
10. Fe + O₂ → Fe₂O₃
Left: 1 Fe, 2 O
Right: 2 Fe, 3 O
LCM for Fe: 2 → so 2 Fe on left? But then O: 2 vs 3 → LCM 6.
Try 4 Fe + 3 O₂ → 4 Fe, 6 O
Right: 2 Fe₂O₃ = 4 Fe, 6 O → perfect.
✔ Balanced: 4Fe + 3O₂ → 2Fe₂O₃
---
11. Fe₂(SO₄)₃ + KOH → K₂SO₄ + Fe(OH)₃
Left: 2 Fe, 3 SO₄, 1 K, 1 O, 1 H
Right: 2 K, 1 SO₄, 1 Fe, 3 O, 3 H
Need 3 K₂SO₄ on right → 6 K, 3 SO₄
So left: need 6 KOH → 6 K, 6 O, 6 H
Right: Fe(OH)₃ → need 2 Fe(OH)₃ → 2 Fe, 6 O, 6 H
Check:
Left: Fe₂(SO₄)₃ = 2 Fe, 3 S, 12 O
6 KOH = 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: 1 Fe, 2 S, 2 O
Right: 2 Fe, 3 O, 1 S, 2 O → total 2 Fe, 1 S, 5 O? Not matching.
Need 2 Fe on right → so 2 FeS₂ on left → 2 Fe, 4 S
Then right: need 4 SO₂ → 4 S, 8 O
Plus Fe₂O₃ → 2 Fe, 3 O → total O on right: 8+3=11
Left: O₂ → need 11/2 → so multiply all by 2.
Try 4 FeS₂ + ? O₂ → 2 Fe₂O₃ + 8 SO₂
Left: 4 Fe, 8 S
Right: 4 Fe, 8 S, O: 2*3 + 8*2 = 6+16=22 O → so 11 O₂
✔ Balanced: 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
---
13. Al + FeO → Al₂O₃ + Fe
Left: 1 Al, 1 Fe, 1 O
Right: 2 Al, 3 O, 1 Fe
Need 2 Al on left → 2 Al
Need 3 O on left → 3 FeO → 3 Fe, 3 O
Right: Al₂O₃ + 3 Fe
Check: left: 2 Al, 3 Fe, 3 O → right: 2 Al, 3 O, 3 Fe → good.
✔ Balanced: 2Al + 3FeO → Al₂O₃ + 3Fe
---
14. Na₂CO₃ + HCl → NaCl + H₂O + CO₂
Left: 2 Na, 1 C, 3 O, 1 H, 1 Cl
Right: 1 Na, 1 Cl, 2 H, 1 O, 1 C, 2 O → total 1 Na, 1 Cl, 2 H, 3 O, 1 C
Need 2 NaCl on right → 2 Na, 2 Cl
So left: need 2 HCl → 2 H, 2 Cl
Now left: 2 Na, 1 C, 3 O, 2 H, 2 Cl
Right: 2 NaCl + 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 left: 2 K
✔ 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
Left: O₂ → need 5 O₂ → 10 O
✔ Balanced: P₄ + 5O₂ → 2P₂O₅
---
17. C₂H₂ + O₂ → CO₂ + H₂O
Left: 2 C, 2 H, 2 O
Right: 1 C, 2 O, 2 H, 1 O → total 1 C, 2 H, 3 O
Need 2 CO₂ on right → 2 C, 4 O
Plus H₂O → 2 H, 1 O → total 2 C, 2 H, 5 O
Left: O₂ → need 5/2 → so multiply all by 2.
2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O
Check:
Left: 4 C, 4 H, 10 O
Right: 4 C, 8 O from CO₂ + 2 H, 2 O from H₂O → total 4 C, 4 H, 10 O → good.
✔ 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 chemistry balancing chemical equations worksheet answers.
