Practice worksheet for balancing chemical equations with various reactions involving acids, bases, metals, and compounds.
Balancing Chemical Reactions worksheet with 25 equations to balance, featuring chemical formulas and blank spaces for coefficients.
PNG
850×1100
60.4 KB
Free · Personal Use
Quality Assured by Worksheets Library Team
Reviewed for educational accuracy and age-appropriateness
ID: #482308
⭐
Show Answer Key & Explanations
Step-by-step solution for: Balancing Chemical Equations - Worksheet 1 | Science 9th Grade
▼
Show Answer Key & Explanations
Step-by-step solution for: Balancing Chemical Equations - Worksheet 1 | Science 9th Grade
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.
We’ll use coefficients (numbers in front) to balance — never change the subscripts!
---
1. H₃PO₄ + KOH → K₃PO₄ + H₂O
Left: H=3+1=4, P=1, O=4+1=5, K=1
Right: K=3, P=1, O=4+1=5, H=2
Try putting 3 in front of KOH and 3 in front of H₂O:
→ H₃PO₄ + 3KOH → K₃PO₄ + 3H₂O
Check:
Left: H=3+3=6, K=3, O=4+3=7, P=1
Right: K=3, P=1, O=4+3=7, H=6 ✔
✔ Balanced: 1, 3, 1, 3
---
2. HCl + NaOH → NaCl + H₂O
Already balanced? Let’s check:
Left: H=1+1=2, Cl=1, Na=1, O=1
Right: Na=1, Cl=1, H=2, O=1 ✔
✔ Balanced: 1, 1, 1, 1
---
3. Na + NaNO₃ → Na₂O + N₂
Left: Na=1+1=2, N=1, O=3
Right: Na=2, O=1, N=2
Need more N on left → try 2 NaNO₃ → gives 2N, 6O
Now right needs 6O → so 6 Na₂O? That would be 12 Na — too many.
Try this:
Let’s set N₂ coefficient = 1 → need 2 N on left → so 2 NaNO₃
Then O from 2 NaNO₃ = 6 → so need 6 Na₂O? But that’s 12 Na on right.
Left: Na + 2 NaNO₃ → total Na = 1 + 2 = 3? Not enough.
Better approach:
Set NaNO₃ = 2 → N=2 → N₂=1
O=6 → so Na₂O must be 6 → Na=12 on right
So left: Na + 2 NaNO₃ → need 10 more Na → so Na = 10
Total left: Na=10 + 2 (from NaNO₃) = 12? Wait — NaNO₃ has 1 Na per molecule → 2 NaNO₃ = 2 Na
So total Na on left: x (from Na) + 2 (from NaNO₃) = x+2
On right: 6 Na₂O = 12 Na → so x+2=12 → x=10
Equation: 10Na + 2NaNO₃ → 6Na₂O + 1N₂
Check:
Left: Na=10+2=12, N=2, O=6
Right: Na=12, O=6, N=2 ✔
✔ Balanced: 10, 2, 6, 1
---
4. N₂ + O₂ → N₂O₅
Left: N=2, O=2
Right: N=2, O=5
Need even O on right → multiply N₂O₅ by 2 → N=4, O=10
Then left: N₂ → need 2 molecules → 2N₂ = 4N
O₂ → need 5 molecules → 5O₂ = 10O
Equation: 2N₂ + 5O₂ → 2N₂O₅
Check: N=4, O=10 on both sides ✔
✔ Balanced: 2, 5, 2
---
5. H₃PO₄ + Mg(OH)₂ → Mg₃(PO₄)₂ + H₂O
Right: Mg=3, P=2, O=8+?= wait — PO₄ is 4 O each → 2×4=8, plus OH in water? Better count all.
Mg₃(PO₄)₂ → Mg=3, P=2, O=8
H₂O → H=2, O=1 per molecule
Left: H₃PO₄ → H=3, P=1, O=4
Mg(OH)₂ → Mg=1, O=2, H=2
To get 2 P on right → need 2 H₃PO₄ → H=6, P=2, O=8
To get 3 Mg → need 3 Mg(OH)₂ → Mg=3, O=6, H=6
Total left: H=6+6=12, O=8+6=14, P=2, Mg=3
Right: Mg₃(PO₄)₂ → Mg=3, P=2, O=8
Water: need to account for remaining H and O
H used: 12 → so 6 H₂O (since each has 2H) → H=12, O=6
Total O on right: 8 (from phosphate) + 6 (from water) = 14 ✔
Equation: 2H₃PO₄ + 3Mg(OH)₂ → 1Mg₃(PO₄)₂ + 6H₂O
✔ Balanced: 2, 3, 1, 6
---
6. NaOH + H₂CO₃ → Na₂CO₃ + H₂O
Left: Na=1, O=1+3=4, H=1+2=3, C=1
Right: Na=2, C=1, O=3+1=4, H=2
Need 2 Na on left → 2 NaOH → Na=2, O=2, H=2
Plus H₂CO₃ → H=2, C=1, O=3
Total left: Na=2, H=4, C=1, O=5
Right: Na₂CO₃ → Na=2, C=1, O=3
H₂O → if we put 2 → H=4, O=2 → total O=5 ✔
Equation: 2NaOH + 1H₂CO₃ → 1Na₂CO₃ + 2H₂O
✔ Balanced: 2, 1, 1, 2
---
7. H₂ + O₂ → H₂O₂
Left: H=2, O=2
Right: H=2, O=2 → already balanced!
✔ Balanced: 1, 1, 1
---
8. Na + O₂ → Na₂O
Left: Na=1, O=2
Right: Na=2, O=1
Need 2 Na on left → 2 Na
But O₂ has 2 O → right needs 2 O → so 2 Na₂O → Na=4, O=2
So left: 4 Na + O₂ → 2 Na₂O
Check: Na=4, O=2 on both sides ✔
✔ Balanced: 4, 1, 2
---
9. Al + S₈ → Al₂S₃
S₈ has 8 S atoms. Al₂S₃ has 3 S.
LCM of 8 and 3 is 24.
So S₈ × 3 = 24 S → Al₂S₃ × 8 = 24 S → Al=16
Left: Al=16, S=24
Right: Al=16, S=24
Equation: 16Al + 3S₈ → 8Al₂S₃
✔ Balanced: 16, 3, 8
---
10. Cs + N₂ → Cs₃N
Right: Cs=3, N=1
Left: N₂ → 2 N → so need 2 Cs₃N → Cs=6, N=2
Left: Cs=6, N₂=1 → 2 N
Equation: 6Cs + 1N₂ → 2Cs₃N
✔ Balanced: 6, 1, 2
---
11. Mg + Cl₂ → MgCl₂
Already balanced: Mg=1, Cl=2 on both sides.
✔ Balanced: 1, 1, 1
---
12. Rb + RbNO₃ → Rb₂O + N₂
Similar to #3.
Set N₂=1 → need 2 N → so 2 RbNO₃ → Rb=2, N=2, O=6
Right: Rb₂O → to get 6 O → 6 Rb₂O → Rb=12
Left: Rb + 2 RbNO₃ → total Rb = x + 2 = 12 → x=10
Equation: 10Rb + 2RbNO₃ → 6Rb₂O + 1N₂
Check: Rb=10+2=12, N=2, O=6 → right: Rb=12, O=6, N=2 ✔
✔ Balanced: 10, 2, 6, 1
---
13. C₆H₆ + O₂ → CO₂ + H₂O
Combustion reaction.
C₆H₆ → 6C, 6H
Right: CO₂ → 6C → 6 CO₂
H₂O → 6H → 3 H₂O
Oxygen: 6 CO₂ → 12 O, 3 H₂O → 3 O → total 15 O → so O₂ = 15/2 → not integer.
Multiply entire equation by 2:
2 C₆H₆ → 12C, 12H → 12 CO₂, 6 H₂O → O = 24 + 6 = 30 → O₂ = 15
Equation: 2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O
✔ Balanced: 2, 15, 12, 6
---
14. N₂ + H₂ → NH₃
Classic Haber process.
N₂ → 2N → need 2 NH₃ → H=6 → so 3 H₂
Equation: 1N₂ + 3H₂ → 2NH₃
✔ Balanced: 1, 3, 2
---
15. C + O₂ → CO₂
Already balanced.
✔ Balanced: 1, 1, 1
---
16. C₃H₈ + O₂ → CO₂ + H₂O
Propane combustion.
C₃H₈ → 3C, 8H
→ 3 CO₂, 4 H₂O (since 8H / 2 = 4)
Oxygen: 3×2=6 from CO₂, 4×1=4 from H₂O → total 10 O → O₂=5
Equation: 1C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
✔ Balanced: 1, 5, 3, 4
---
17. Li + AlCl₃ → LiCl + Al
Single displacement.
AlCl₃ → 3 Cl → so 3 LiCl → need 3 Li
Equation: 3Li + 1AlCl₃ → 3LiCl + 1Al
✔ Balanced: 3, 1, 3, 1
---
18. Rb + P → Rb₃P
Phosphorus is P₄ usually, but here it's written as P → assume atomic? But product is Rb₃P → so P=1
Left: Rb=?, P=1
Right: Rb=3, P=1
So Rb=3
Equation: 3Rb + 1P → 1Rb₃P
But phosphorus is typically P₄ — however, since problem writes "P", we'll go with it.
If it were P₄, we’d adjust — but as written, it’s fine.
✔ Balanced: 3, 1, 1
Wait — actually, in reality, phosphorus is P₄, but the problem says “P”, so we treat it as monatomic for balancing purposes.
Alternatively, maybe typo? But let’s follow what’s written.
Actually, looking at #20: Rb + S₈ → Rb₂S — so they specify S₈, meaning elemental form.
Here it’s just “P” — probably should be P₄? But problem says “P”.
To be safe, let’s assume it’s P (monatomic) as written.
So: 3Rb + 1P → 1Rb₃P
✔ Balanced: 3, 1, 1
But let me double-check common practice — usually phosphorus is P₄. However, since the problem didn’t write P₄, we’ll proceed as is.
Actually, looking ahead to #20: S₈ is specified, so for consistency, perhaps P should be P₄? But it’s not written.
I think we have to go with what’s given: “P”
So answer is 3,1,1
But let’s see if it makes sense — Rb₃P implies P has -3 charge, so one P atom.
Okay.
✔ Balanced: 3, 1, 1
---
19. Al(OH)₃ + H₂SO₄ → Al₂(SO₄)₃ + H₂O
Left: Al=1, O=3+4=7? Wait — Al(OH)₃: Al, 3O, 3H
H₂SO₄: 2H, S, 4O
Right: Al₂(SO₄)₃ → Al=2, S=3, O=12
H₂O: H=2, O=1
Need 2 Al → so 2 Al(OH)₃ → Al=2, O=6, H=6
Need 3 SO₄ → so 3 H₂SO₄ → H=6, S=3, O=12
Total left: Al=2, H=6+6=12, S=3, O=6+12=18
Right: Al₂(SO₄)₃ → Al=2, S=3, O=12
H₂O: need 6 H₂O → H=12, O=6 → total O=18 ✔
Equation: 2Al(OH)₃ + 3H₂SO₄ → 1Al₂(SO₄)₃ + 6H₂O
✔ Balanced: 2, 3, 1, 6
---
20. Rb + S₈ → Rb₂S
S₈ has 8 S atoms. Rb₂S has 1 S.
So need 8 Rb₂S → Rb=16, S=8
Left: S₈=1 → S=8
Rb=16
Equation: 16Rb + 1S₈ → 8Rb₂S
✔ Balanced: 16, 1, 8
---
21. CaSO₄ + KOH → Ca(OH)₂ + K₂SO₄
Double displacement.
Left: Ca=1, S=1, O=4+1=5? CaSO₄: Ca,S,4O; KOH: K,O,H
Right: Ca(OH)₂: Ca,2O,2H; K₂SO₄: 2K,S,4O
Need 2 K on right → so 2 KOH → K=2, O=2, H=2
Left: CaSO₄ + 2 KOH → Ca=1, S=1, O=4+2=6, K=2, H=2
Right: Ca(OH)₂ → Ca=1, O=2, H=2; K₂SO₄ → K=2, S=1, O=4 → total O=6 ✔
Equation: 1CaSO₄ + 2KOH → 1Ca(OH)₂ + 1K₂SO₄
✔ Balanced: 1, 2, 1, 1
---
22. FeBr₃ + Na → Fe + NaBr
Single displacement.
FeBr₃ → 3 Br → so 3 NaBr → need 3 Na
Equation: 1FeBr₃ + 3Na → 1Fe + 3NaBr
✔ Balanced: 1, 3, 1, 3
---
23. Al₄C₃ + H₂O → CH₄ + Al(OH)₃
Left: Al=4, C=3, H=2, O=1
Right: CH₄: C=1, H=4; Al(OH)₃: Al=1, O=3, H=3
Need 3 CH₄ → C=3, H=12
Need 4 Al(OH)₃ → Al=4, O=12, H=12
Total right: H=12+12=24, O=12
Left: Al₄C₃ + x H₂O → H=2x, O=x
Set 2x=24 → x=12 → O=12 ✔
Equation: 1Al₄C₃ + 12H₂O → 3CH₄ + 4Al(OH)₃
Check: Al=4, C=3, H=24, O=12 on both sides ✔
✔ Balanced: 1, 12, 3, 4
---
24. Ba + HNO₃ → H₂ + Ba(NO₃)₂
Left: Ba=1, H=1, N=1, O=3
Right: H₂=2H, Ba(NO₃)₂: Ba=1, N=2, O=6
Need 2 H on left → 2 HNO₃ → H=2, N=2, O=6
Ba=1 on both sides
Equation: 1Ba + 2HNO₃ → 1H₂ + 1Ba(NO₃)₂
Check: Ba=1, H=2, N=2, O=6 on both sides ✔
✔ Balanced: 1, 2, 1, 1
---
25. NH₄OH + HBr → H₂O + NH₄Br
Acid-base reaction.
Left: N=1, H=4+1+1=6? NH₄OH: N,5H,O; HBr: H,Br
Better: NH₄OH → N, 5H, O
HBr → H, Br
Total left: N=1, H=6, O=1, Br=1
Right: H₂O → H=2, O=1; NH₄Br → N=1, H=4, Br=1 → total H=6, O=1, N=1, Br=1 ✔
Already balanced.
✔ Balanced: 1, 1, 1, 1
---
Now compiling all answers:
1. 1, 3, 1, 3
2. 1, 1, 1, 1
3. 10, 2, 6, 1
4. 2, 5, 2
5. 2, 3, 1, 6
6. 2, 1, 1, 2
7. 1, 1, 1
8. 4, 1, 2
9. 16, 3, 8
10. 6, 1, 2
11. 1, 1, 1
12. 10, 2, 6, 1
13. 2, 15, 12, 6
14. 1, 3, 2
15. 1, 1, 1
16. 1, 5, 3, 4
17. 3, 1, 3, 1
18. 3, 1, 1
19. 2, 3, 1, 6
20. 16, 1, 8
21. 1, 2, 1, 1
22. 1, 3, 1, 3
23. 1, 12, 3, 4
24. 1, 2, 1, 1
25. 1, 1, 1, 1
Final Answer:
1. 1, 3, 1, 3
2. 1, 1, 1, 1
3. 10, 2, 6, 1
4. 2, 5, 2
5. 2, 3, 1, 6
6. 2, 1, 1, 2
7. 1, 1, 1
8. 4, 1, 2
9. 16, 3, 8
10. 6, 1, 2
11. 1, 1, 1
12. 10, 2, 6, 1
13. 2, 15, 12, 6
14. 1, 3, 2
15. 1, 1, 1
16. 1, 5, 3, 4
17. 3, 1, 3, 1
18. 3, 1, 1
19. 2, 3, 1, 6
20. 16, 1, 8
21. 1, 2, 1, 1
22. 1, 3, 1, 3
23. 1, 12, 3, 4
24. 1, 2, 1, 1
25. 1, 1, 1, 1
We’ll use coefficients (numbers in front) to balance — never change the subscripts!
---
1. H₃PO₄ + KOH → K₃PO₄ + H₂O
Left: H=3+1=4, P=1, O=4+1=5, K=1
Right: K=3, P=1, O=4+1=5, H=2
Try putting 3 in front of KOH and 3 in front of H₂O:
→ H₃PO₄ + 3KOH → K₃PO₄ + 3H₂O
Check:
Left: H=3+3=6, K=3, O=4+3=7, P=1
Right: K=3, P=1, O=4+3=7, H=6 ✔
✔ Balanced: 1, 3, 1, 3
---
2. HCl + NaOH → NaCl + H₂O
Already balanced? Let’s check:
Left: H=1+1=2, Cl=1, Na=1, O=1
Right: Na=1, Cl=1, H=2, O=1 ✔
✔ Balanced: 1, 1, 1, 1
---
3. Na + NaNO₃ → Na₂O + N₂
Left: Na=1+1=2, N=1, O=3
Right: Na=2, O=1, N=2
Need more N on left → try 2 NaNO₃ → gives 2N, 6O
Now right needs 6O → so 6 Na₂O? That would be 12 Na — too many.
Try this:
Let’s set N₂ coefficient = 1 → need 2 N on left → so 2 NaNO₃
Then O from 2 NaNO₃ = 6 → so need 6 Na₂O? But that’s 12 Na on right.
Left: Na + 2 NaNO₃ → total Na = 1 + 2 = 3? Not enough.
Better approach:
Set NaNO₃ = 2 → N=2 → N₂=1
O=6 → so Na₂O must be 6 → Na=12 on right
So left: Na + 2 NaNO₃ → need 10 more Na → so Na = 10
Total left: Na=10 + 2 (from NaNO₃) = 12? Wait — NaNO₃ has 1 Na per molecule → 2 NaNO₃ = 2 Na
So total Na on left: x (from Na) + 2 (from NaNO₃) = x+2
On right: 6 Na₂O = 12 Na → so x+2=12 → x=10
Equation: 10Na + 2NaNO₃ → 6Na₂O + 1N₂
Check:
Left: Na=10+2=12, N=2, O=6
Right: Na=12, O=6, N=2 ✔
✔ Balanced: 10, 2, 6, 1
---
4. N₂ + O₂ → N₂O₅
Left: N=2, O=2
Right: N=2, O=5
Need even O on right → multiply N₂O₅ by 2 → N=4, O=10
Then left: N₂ → need 2 molecules → 2N₂ = 4N
O₂ → need 5 molecules → 5O₂ = 10O
Equation: 2N₂ + 5O₂ → 2N₂O₅
Check: N=4, O=10 on both sides ✔
✔ Balanced: 2, 5, 2
---
5. H₃PO₄ + Mg(OH)₂ → Mg₃(PO₄)₂ + H₂O
Right: Mg=3, P=2, O=8+?= wait — PO₄ is 4 O each → 2×4=8, plus OH in water? Better count all.
Mg₃(PO₄)₂ → Mg=3, P=2, O=8
H₂O → H=2, O=1 per molecule
Left: H₃PO₄ → H=3, P=1, O=4
Mg(OH)₂ → Mg=1, O=2, H=2
To get 2 P on right → need 2 H₃PO₄ → H=6, P=2, O=8
To get 3 Mg → need 3 Mg(OH)₂ → Mg=3, O=6, H=6
Total left: H=6+6=12, O=8+6=14, P=2, Mg=3
Right: Mg₃(PO₄)₂ → Mg=3, P=2, O=8
Water: need to account for remaining H and O
H used: 12 → so 6 H₂O (since each has 2H) → H=12, O=6
Total O on right: 8 (from phosphate) + 6 (from water) = 14 ✔
Equation: 2H₃PO₄ + 3Mg(OH)₂ → 1Mg₃(PO₄)₂ + 6H₂O
✔ Balanced: 2, 3, 1, 6
---
6. NaOH + H₂CO₃ → Na₂CO₃ + H₂O
Left: Na=1, O=1+3=4, H=1+2=3, C=1
Right: Na=2, C=1, O=3+1=4, H=2
Need 2 Na on left → 2 NaOH → Na=2, O=2, H=2
Plus H₂CO₃ → H=2, C=1, O=3
Total left: Na=2, H=4, C=1, O=5
Right: Na₂CO₃ → Na=2, C=1, O=3
H₂O → if we put 2 → H=4, O=2 → total O=5 ✔
Equation: 2NaOH + 1H₂CO₃ → 1Na₂CO₃ + 2H₂O
✔ Balanced: 2, 1, 1, 2
---
7. H₂ + O₂ → H₂O₂
Left: H=2, O=2
Right: H=2, O=2 → already balanced!
✔ Balanced: 1, 1, 1
---
8. Na + O₂ → Na₂O
Left: Na=1, O=2
Right: Na=2, O=1
Need 2 Na on left → 2 Na
But O₂ has 2 O → right needs 2 O → so 2 Na₂O → Na=4, O=2
So left: 4 Na + O₂ → 2 Na₂O
Check: Na=4, O=2 on both sides ✔
✔ Balanced: 4, 1, 2
---
9. Al + S₈ → Al₂S₃
S₈ has 8 S atoms. Al₂S₃ has 3 S.
LCM of 8 and 3 is 24.
So S₈ × 3 = 24 S → Al₂S₃ × 8 = 24 S → Al=16
Left: Al=16, S=24
Right: Al=16, S=24
Equation: 16Al + 3S₈ → 8Al₂S₃
✔ Balanced: 16, 3, 8
---
10. Cs + N₂ → Cs₃N
Right: Cs=3, N=1
Left: N₂ → 2 N → so need 2 Cs₃N → Cs=6, N=2
Left: Cs=6, N₂=1 → 2 N
Equation: 6Cs + 1N₂ → 2Cs₃N
✔ Balanced: 6, 1, 2
---
11. Mg + Cl₂ → MgCl₂
Already balanced: Mg=1, Cl=2 on both sides.
✔ Balanced: 1, 1, 1
---
12. Rb + RbNO₃ → Rb₂O + N₂
Similar to #3.
Set N₂=1 → need 2 N → so 2 RbNO₃ → Rb=2, N=2, O=6
Right: Rb₂O → to get 6 O → 6 Rb₂O → Rb=12
Left: Rb + 2 RbNO₃ → total Rb = x + 2 = 12 → x=10
Equation: 10Rb + 2RbNO₃ → 6Rb₂O + 1N₂
Check: Rb=10+2=12, N=2, O=6 → right: Rb=12, O=6, N=2 ✔
✔ Balanced: 10, 2, 6, 1
---
13. C₆H₆ + O₂ → CO₂ + H₂O
Combustion reaction.
C₆H₆ → 6C, 6H
Right: CO₂ → 6C → 6 CO₂
H₂O → 6H → 3 H₂O
Oxygen: 6 CO₂ → 12 O, 3 H₂O → 3 O → total 15 O → so O₂ = 15/2 → not integer.
Multiply entire equation by 2:
2 C₆H₆ → 12C, 12H → 12 CO₂, 6 H₂O → O = 24 + 6 = 30 → O₂ = 15
Equation: 2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O
✔ Balanced: 2, 15, 12, 6
---
14. N₂ + H₂ → NH₃
Classic Haber process.
N₂ → 2N → need 2 NH₃ → H=6 → so 3 H₂
Equation: 1N₂ + 3H₂ → 2NH₃
✔ Balanced: 1, 3, 2
---
15. C + O₂ → CO₂
Already balanced.
✔ Balanced: 1, 1, 1
---
16. C₃H₈ + O₂ → CO₂ + H₂O
Propane combustion.
C₃H₈ → 3C, 8H
→ 3 CO₂, 4 H₂O (since 8H / 2 = 4)
Oxygen: 3×2=6 from CO₂, 4×1=4 from H₂O → total 10 O → O₂=5
Equation: 1C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
✔ Balanced: 1, 5, 3, 4
---
17. Li + AlCl₃ → LiCl + Al
Single displacement.
AlCl₃ → 3 Cl → so 3 LiCl → need 3 Li
Equation: 3Li + 1AlCl₃ → 3LiCl + 1Al
✔ Balanced: 3, 1, 3, 1
---
18. Rb + P → Rb₃P
Phosphorus is P₄ usually, but here it's written as P → assume atomic? But product is Rb₃P → so P=1
Left: Rb=?, P=1
Right: Rb=3, P=1
So Rb=3
Equation: 3Rb + 1P → 1Rb₃P
But phosphorus is typically P₄ — however, since problem writes "P", we'll go with it.
If it were P₄, we’d adjust — but as written, it’s fine.
✔ Balanced: 3, 1, 1
Wait — actually, in reality, phosphorus is P₄, but the problem says “P”, so we treat it as monatomic for balancing purposes.
Alternatively, maybe typo? But let’s follow what’s written.
Actually, looking at #20: Rb + S₈ → Rb₂S — so they specify S₈, meaning elemental form.
Here it’s just “P” — probably should be P₄? But problem says “P”.
To be safe, let’s assume it’s P (monatomic) as written.
So: 3Rb + 1P → 1Rb₃P
✔ Balanced: 3, 1, 1
But let me double-check common practice — usually phosphorus is P₄. However, since the problem didn’t write P₄, we’ll proceed as is.
Actually, looking ahead to #20: S₈ is specified, so for consistency, perhaps P should be P₄? But it’s not written.
I think we have to go with what’s given: “P”
So answer is 3,1,1
But let’s see if it makes sense — Rb₃P implies P has -3 charge, so one P atom.
Okay.
✔ Balanced: 3, 1, 1
---
19. Al(OH)₃ + H₂SO₄ → Al₂(SO₄)₃ + H₂O
Left: Al=1, O=3+4=7? Wait — Al(OH)₃: Al, 3O, 3H
H₂SO₄: 2H, S, 4O
Right: Al₂(SO₄)₃ → Al=2, S=3, O=12
H₂O: H=2, O=1
Need 2 Al → so 2 Al(OH)₃ → Al=2, O=6, H=6
Need 3 SO₄ → so 3 H₂SO₄ → H=6, S=3, O=12
Total left: Al=2, H=6+6=12, S=3, O=6+12=18
Right: Al₂(SO₄)₃ → Al=2, S=3, O=12
H₂O: need 6 H₂O → H=12, O=6 → total O=18 ✔
Equation: 2Al(OH)₃ + 3H₂SO₄ → 1Al₂(SO₄)₃ + 6H₂O
✔ Balanced: 2, 3, 1, 6
---
20. Rb + S₈ → Rb₂S
S₈ has 8 S atoms. Rb₂S has 1 S.
So need 8 Rb₂S → Rb=16, S=8
Left: S₈=1 → S=8
Rb=16
Equation: 16Rb + 1S₈ → 8Rb₂S
✔ Balanced: 16, 1, 8
---
21. CaSO₄ + KOH → Ca(OH)₂ + K₂SO₄
Double displacement.
Left: Ca=1, S=1, O=4+1=5? CaSO₄: Ca,S,4O; KOH: K,O,H
Right: Ca(OH)₂: Ca,2O,2H; K₂SO₄: 2K,S,4O
Need 2 K on right → so 2 KOH → K=2, O=2, H=2
Left: CaSO₄ + 2 KOH → Ca=1, S=1, O=4+2=6, K=2, H=2
Right: Ca(OH)₂ → Ca=1, O=2, H=2; K₂SO₄ → K=2, S=1, O=4 → total O=6 ✔
Equation: 1CaSO₄ + 2KOH → 1Ca(OH)₂ + 1K₂SO₄
✔ Balanced: 1, 2, 1, 1
---
22. FeBr₃ + Na → Fe + NaBr
Single displacement.
FeBr₃ → 3 Br → so 3 NaBr → need 3 Na
Equation: 1FeBr₃ + 3Na → 1Fe + 3NaBr
✔ Balanced: 1, 3, 1, 3
---
23. Al₄C₃ + H₂O → CH₄ + Al(OH)₃
Left: Al=4, C=3, H=2, O=1
Right: CH₄: C=1, H=4; Al(OH)₃: Al=1, O=3, H=3
Need 3 CH₄ → C=3, H=12
Need 4 Al(OH)₃ → Al=4, O=12, H=12
Total right: H=12+12=24, O=12
Left: Al₄C₃ + x H₂O → H=2x, O=x
Set 2x=24 → x=12 → O=12 ✔
Equation: 1Al₄C₃ + 12H₂O → 3CH₄ + 4Al(OH)₃
Check: Al=4, C=3, H=24, O=12 on both sides ✔
✔ Balanced: 1, 12, 3, 4
---
24. Ba + HNO₃ → H₂ + Ba(NO₃)₂
Left: Ba=1, H=1, N=1, O=3
Right: H₂=2H, Ba(NO₃)₂: Ba=1, N=2, O=6
Need 2 H on left → 2 HNO₃ → H=2, N=2, O=6
Ba=1 on both sides
Equation: 1Ba + 2HNO₃ → 1H₂ + 1Ba(NO₃)₂
Check: Ba=1, H=2, N=2, O=6 on both sides ✔
✔ Balanced: 1, 2, 1, 1
---
25. NH₄OH + HBr → H₂O + NH₄Br
Acid-base reaction.
Left: N=1, H=4+1+1=6? NH₄OH: N,5H,O; HBr: H,Br
Better: NH₄OH → N, 5H, O
HBr → H, Br
Total left: N=1, H=6, O=1, Br=1
Right: H₂O → H=2, O=1; NH₄Br → N=1, H=4, Br=1 → total H=6, O=1, N=1, Br=1 ✔
Already balanced.
✔ Balanced: 1, 1, 1, 1
---
Now compiling all answers:
1. 1, 3, 1, 3
2. 1, 1, 1, 1
3. 10, 2, 6, 1
4. 2, 5, 2
5. 2, 3, 1, 6
6. 2, 1, 1, 2
7. 1, 1, 1
8. 4, 1, 2
9. 16, 3, 8
10. 6, 1, 2
11. 1, 1, 1
12. 10, 2, 6, 1
13. 2, 15, 12, 6
14. 1, 3, 2
15. 1, 1, 1
16. 1, 5, 3, 4
17. 3, 1, 3, 1
18. 3, 1, 1
19. 2, 3, 1, 6
20. 16, 1, 8
21. 1, 2, 1, 1
22. 1, 3, 1, 3
23. 1, 12, 3, 4
24. 1, 2, 1, 1
25. 1, 1, 1, 1
Final Answer:
1. 1, 3, 1, 3
2. 1, 1, 1, 1
3. 10, 2, 6, 1
4. 2, 5, 2
5. 2, 3, 1, 6
6. 2, 1, 1, 2
7. 1, 1, 1
8. 4, 1, 2
9. 16, 3, 8
10. 6, 1, 2
11. 1, 1, 1
12. 10, 2, 6, 1
13. 2, 15, 12, 6
14. 1, 3, 2
15. 1, 1, 1
16. 1, 5, 3, 4
17. 3, 1, 3, 1
18. 3, 1, 1
19. 2, 3, 1, 6
20. 16, 1, 8
21. 1, 2, 1, 1
22. 1, 3, 1, 3
23. 1, 12, 3, 4
24. 1, 2, 1, 1
25. 1, 1, 1, 1
Parent Tip: Review the logic above to help your child master the concept of chemistry balancing equations worksheet.
![49 Balancing Chemical Equations Worksheets [with Answers]](/images2/chemistry-balancing-equations-worksheet/chemistry-balancing-equations-worksheet-0.jpg)
