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 step by step to balance each equation and identify the reaction type. We’ll do this carefully, one at a time.
---
a. Cu + O₂ → CuO
- Left: 1 Cu, 2 O
- Right: 1 Cu, 1 O
→ Need 2 CuO on right to match oxygen → now 2 Cu on right
→ So put 2 Cu on left
Balanced: 2Cu + O₂ → 2CuO
Type: Two elements combine → Formation
---
b. H₂O → H₂ + O₂
- Left: 2 H, 1 O
- Right: 2 H, 2 O
→ Need 2 H₂O on left → gives 4 H, 2 O
→ Then need 2 H₂ on right (to get 4 H)
Balanced: 2H₂O → 2H₂ + O₂
Type: One compound breaks into elements → Decomposition
---
c. Fe + H₂O → H₂ + Fe₃O₄
This is tricky. Let’s count atoms:
Right has Fe₃O₄ → 3 Fe, 4 O
Left: H₂O has 1 O per molecule → need 4 H₂O for 4 O → that gives 8 H
→ So need 4 H₂ on right (since each H₂ has 2 H → 4×2=8 H)
Now Fe: right has 3 Fe → so left needs 3 Fe
Balanced: 3Fe + 4H₂O → 4H₂ + Fe₃O₄
Type: Element replaces part of compound? Actually, it’s a redox reaction but not single replacement in simple terms → Other
*(Note: Some might call this “single replacement” if they think Fe replaces H, but since water becomes H₂ and Fe forms oxide, it’s more complex — we’ll say “other” to be safe.)*
---
d. AsCl₃ + H₂S → As₂S₃ + HCl
Left: As=1, Cl=3, H=2, S=1
Right: As=2, S=3, H=1, Cl=1
Need 2 As on left → 2AsCl₃ → now Cl=6
Need 3 S on left → 3H₂S → now H=6
Right: As₂S₃ is fine, HCl must be 6 to match H and Cl
Balanced: 2AsCl₃ + 3H₂S → As₂S₃ + 6HCl
Type: Ions swap partners → Double Replacement
---
e. CuSO₄•5H₂O → CuSO₄ + H₂O
This is dehydration. The dot means 5 water molecules are attached.
So: CuSO₄•5H₂O → CuSO₄ + 5H₂O
Already balanced!
Type: Compound breaks down → Decomposition
---
f. Fe₂O₃ + H₂ → Fe + H₂O
Left: Fe=2, O=3, H=2
Right: Fe=1, H=2, O=1
Need 2 Fe on right → 2Fe
Need 3 H₂O on right → gives 3 O and 6 H
→ So need 3 H₂ on left (gives 6 H)
Balanced: Fe₂O₃ + 3H₂ → 2Fe + 3H₂O
Type: H₂ takes oxygen from iron oxide → Single Replacement (or reduction, but in basic classification, often called single replacement)
---
g. CaCO₃ → CaO + CO₂
Already balanced!
Ca=1, C=1, O=3 on both sides.
Type: One compound breaks into two → Decomposition
---
h. Fe + S₈ → FeS
S₈ is 8 sulfur atoms. Each FeS has 1 S → need 8 FeS → so 8 Fe on left
Balanced: 8Fe + S₈ → 8FeS
Type: Two elements combine → Formation
---
i. H₂S + KOH → H₂O + K₂S
Left: H=2+1=3? Wait — H₂S has 2H, KOH has 1H → total 3H? No — better to balance properly.
Actually:
H₂S + 2KOH → 2H₂O + K₂S
Check:
Left: H=2+2=4, S=1, K=2, O=2
Right: H=4, O=2, K=2, S=1 → yes!
Balanced: H₂S + 2KOH → 2H₂O + K₂S
Type: Acid-base neutralization → Double Replacement
---
j. NaCl → Na + Cl₂
Left: Na=1, Cl=1
Right: Na=1, Cl=2 → need 2NaCl on left → then 2Na on right
Balanced: 2NaCl → 2Na + Cl₂
Type: Breaks into elements → Decomposition
---
k. Al + H₂SO₄ → H₂ + Al₂(SO₄)₃
Right: Al=2, SO₄=3 → so need 3 H₂SO₄ on left → gives 6 H → so 3 H₂ on right
Al: need 2 on left
Balanced: 2Al + 3H₂SO₄ → 3H₂ + Al₂(SO₄)₃
Type: Metal displaces hydrogen → Single Replacement
---
l. H₃PO₄ + NHOH → H₂O + (NH₄)₃PO₄
Right: (NH₄)₃PO₄ → 3 NH₄, 1 PO₄
Left: H₃PO₄ has 1 PO₄ → good
NH₄OH has 1 NH₄ → need 3 NH₄OH
Then H: left = 3 (from acid) + 3 (from base) = 6 H → right: 3 H₂O has 6 H → perfect
Balanced: H₃PO₄ + 3NH₄OH → 3H₂O + (NH₄)₃PO₄
Type: Acid + base → salt + water → Double Replacement
---
m. C₃H₈ + O₂ → CO₂ + H₂O
Hydrocarbon combustion.
C₃H₈ → 3C → need 3CO₂
8H → need 4H₂O (since each has 2H)
Oxygen: right = 3×2 + 4×1 = 6+4=10 O → so need 5 O₂ on left
Balanced: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Type: Hydrocarbon + oxygen → Hydrocarbon Combustion
---
n. Al + O₂ → Al₂O₃
Left: Al=1, O=2
Right: Al=2, O=3 → LCM of O is 6 → so 3O₂ and 2Al₂O₃ → then Al=4 on right → so 4Al on left
Balanced: 4Al + 3O₂ → 2Al₂O₃
Type: Elements combine → Formation
---
o. CH₄ + O₂ → CO₂ + H₂O
Combustion again.
CH₄ → 1C → 1CO₂
4H → 2H₂O
O: right = 2 + 2 = 4 → so 2O₂ on left
Balanced: CH₄ + 2O₂ → CO₂ + 2H₂O
Type: Hydrocarbon Combustion
---
p. K₂SO₄ + BaCl₂ → KCl + BaSO₄
Swap ions: K with Cl, Ba with SO₄
Left: K=2, SO₄=1, Ba=1, Cl=2
Right: K=1, Cl=1, Ba=1, SO₄=1 → need 2KCl
Balanced: K₂SO₄ + BaCl₂ → 2KCl + BaSO₄
Type: Ions swap → Double Replacement
---
q. C₅H₁₂ + O₂ → CO₂ + H₂O
Pentane combustion.
C₅ → 5CO₂
H₁₂ → 6H₂O
O: right = 5×2 + 6×1 = 10+6=16 → so 8O₂ on left
Balanced: C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O
Type: Hydrocarbon Combustion
---
r. Ca(OH)₂ + NH₄Cl → NH₄OH + CaCl₂
Left: Ca=1, O=2, H=2+4=6? Wait — Ca(OH)₂ has 2OH → 2O, 2H; NH₄Cl has N, 4H, Cl
Better:
Ca(OH)₂ + 2NH₄Cl → 2NH₄OH + CaCl₂
Check:
Left: Ca=1, O=2, H=2+8=10, N=2, Cl=2
Right: NH₄OH ×2 → N=2, H=8+2=10? Wait — NH₄OH is NH₄⁺ and OH → so each has 5H? No:
NH₄OH = N, 5H, O? Actually, formula is often written as NH₄OH meaning ammonium hydroxide → which is NH₃(aq) + H₂O, but for balancing, treat as NH₄OH → N, 5H, O? That’s messy.
Standard way:
Ca(OH)₂ + 2NH₄Cl → CaCl₂ + 2NH₃ + 2H₂O
But the problem says NH₄OH — so let’s assume it’s written as such.
If we write:
Ca(OH)₂ + 2NH₄Cl → 2NH₄OH + CaCl₂
Atoms:
Left: Ca=1, O=2, H=2 (from OH) + 8 (from 2NH₄) = 10H, N=2, Cl=2
Right: 2NH₄OH → 2N, 10H, 2O; CaCl₂ → Ca, 2Cl → matches!
So balanced: Ca(OH)₂ + 2NH₄Cl → 2NH₄OH + CaCl₂
Type: Double replacement → Double Replacement
---
s. V₂O₅ + Ca → CaO + V
Left: V=2, O=5, Ca=1
Right: Ca=1, O=1, V=1 → need 5 CaO → so 5 Ca on left → then V=2 on right
Balanced: V₂O₅ + 5Ca → 5CaO + 2V
Type: Metal displaces another metal → Single Replacement
---
t. Na + ZnI₂ → NaI + Zn
Left: Na=1, Zn=1, I=2
Right: Na=1, I=1, Zn=1 → need 2NaI → so 2Na on left
Balanced: 2Na + ZnI₂ → 2NaI + Zn
Type: Single replacement → Single Replacement
---
u. C₇H₆O₃ + O₂ → CO₂ + H₂O
Benzoic acid combustion.
C₇ → 7CO₂
H₆ → 3H₂O
O: left = 3 (in compound) + ? from O₂
Right: 7×2 + 3×1 = 14+3=17 O → minus 3 from compound → need 14 O from O₂ → so 7O₂
Balanced: C₇H₆O₃ + 7O₂ → 7CO₂ + 3H₂O
Type: Hydrocarbon Combustion (even though it has oxygen, still combustion)
---
v. Ca + N₂ → Ca₃N₂
Left: Ca=1, N=2
Right: Ca=3, N=2 → need 3Ca on left
Balanced: 3Ca + N₂ → Ca₃N₂
Type: Elements combine → Formation
---
w. Fe₂O₃ + H₂ → Fe + H₂O
Same as f! Already did: Fe₂O₃ + 3H₂ → 2Fe + 3H₂O
Type: Single Replacement
---
x. C₁₅H₃₀ + O₂ → CO₂ + H₂O
Combustion.
C₁₅ → 15CO₂
H₃₀ → 15H₂O
O: right = 15×2 + 15×1 = 30+15=45 → so 22.5 O₂ → multiply all by 2 to eliminate fraction
Original: C₁₅H₃₀ + 22.5O₂ → 15CO₂ + 15H₂O
Multiply by 2: 2C₁₅H₃₀ + 45O₂ → 30CO₂ + 30H₂O
Type: Hydrocarbon Combustion
---
y. BN + F₂ → BF₃ + N₂
Left: B=1, N=1, F=2
Right: B=1, F=3, N=2 → need 2BN on left → then N=2 → good
F: right = 3 per BF₃ → need 2BF₃ → F=6 → so 3F₂ on left
Balanced: 2BN + 3F₂ → 2BF₃ + N₂
Type: Not standard categories → Other
---
z. C₁₂H₂₆ + O₂ → CO₂ + H₂O
Dodecane combustion.
C₁₂ → 12CO₂
H₂₆ → 13H₂O
O: right = 12×2 + 13×1 = 24+13=37 → so 18.5 O₂ → multiply by 2
2C₁₂H₂₆ + 37O₂ → 24CO₂ + 26H₂O
Type: Hydrocarbon Combustion
---
Now, let’s compile the final answers clearly.
Final Answer:
a. 2Cu(s) + O₂(g) → 2CuO(s) — Formation
b. 2H₂O(l) → 2H₂(g) + O₂(g) — Decomposition
c. 3Fe(s) + 4H₂O(g) → 4H₂(g) + Fe₃O₄(s) — Other
d. 2AsCl₃(aq) + 3H₂S(aq) → As₂S₃(s) + 6HCl(aq) — Double Replacement
e. CuSO₄•5H₂O(s) → CuSO₄(s) + 5H₂O(g) — Decomposition
f. Fe₂O₃(s) + 3H₂(g) → 2Fe(s) + 3H₂O(l) — Single Replacement
g. CaCO₃(s) → CaO(s) + CO₂(g) — Decomposition
h. 8Fe(s) + S₈(s) → 8FeS(s) — Formation
i. H₂S(aq) + 2KOH(aq) → 2H₂O(l) + K₂S(aq) — Double Replacement
j. 2NaCl(l) → 2Na(l) + Cl₂(g) — Decomposition
k. 2Al(s) + 3H₂SO₄(aq) → 3H₂(g) + Al₂(SO₄)₃(aq) — Single Replacement
l. H₃PO₄(aq) + 3NH₄OH(aq) → 3H₂O(l) + (NH₄)₃PO₄(aq) — Double Replacement
m. C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l) — Hydrocarbon Combustion
n. 4Al(s) + 3O₂(g) → 2Al₂O₃(s) — Formation
o. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) — Hydrocarbon Combustion
p. K₂SO₄(aq) + BaCl₂(aq) → 2KCl(aq) + BaSO₄(s) — Double Replacement
q. C₅H₁₂(l) + 8O₂(g) → 5CO₂(g) + 6H₂O(g) — Hydrocarbon Combustion
r. Ca(OH)₂(aq) + 2NH₄Cl(aq) → 2NH₄OH(aq) + CaCl₂(aq) — Double Replacement
s. V₂O₅(s) + 5Ca(s) → 5CaO(s) + 2V(s) — Single Replacement
t. 2Na(s) + ZnI₂(aq) → 2NaI(aq) + Zn(s) — Single Replacement
u. C₇H₆O₃(l) + 7O₂(g) → 7CO₂(g) + 3H₂O(l) — Hydrocarbon Combustion
v. 3Ca(s) + N₂(g) → Ca₃N₂(s) — Formation
w. Fe₂O₃(s) + 3H₂(g) → 2Fe(s) + 3H₂O(l) — Single Replacement
x. 2C₁₅H₀(l) + 45O₂(g) → 30CO₂(g) + 30H₂O(g) — Hydrocarbon Combustion
y. 2BN(s) + 3F₂(g) → 2BF₃(s) + N₂(g) — Other
z. 2C₁₂H₂₆(l) + 37O₂(g) → 24CO₂(g) + 26H₂O(g) — Hydrocarbon Combustion
---
a. Cu + O₂ → CuO
- Left: 1 Cu, 2 O
- Right: 1 Cu, 1 O
→ Need 2 CuO on right to match oxygen → now 2 Cu on right
→ So put 2 Cu on left
Balanced: 2Cu + O₂ → 2CuO
Type: Two elements combine → Formation
---
b. H₂O → H₂ + O₂
- Left: 2 H, 1 O
- Right: 2 H, 2 O
→ Need 2 H₂O on left → gives 4 H, 2 O
→ Then need 2 H₂ on right (to get 4 H)
Balanced: 2H₂O → 2H₂ + O₂
Type: One compound breaks into elements → Decomposition
---
c. Fe + H₂O → H₂ + Fe₃O₄
This is tricky. Let’s count atoms:
Right has Fe₃O₄ → 3 Fe, 4 O
Left: H₂O has 1 O per molecule → need 4 H₂O for 4 O → that gives 8 H
→ So need 4 H₂ on right (since each H₂ has 2 H → 4×2=8 H)
Now Fe: right has 3 Fe → so left needs 3 Fe
Balanced: 3Fe + 4H₂O → 4H₂ + Fe₃O₄
Type: Element replaces part of compound? Actually, it’s a redox reaction but not single replacement in simple terms → Other
*(Note: Some might call this “single replacement” if they think Fe replaces H, but since water becomes H₂ and Fe forms oxide, it’s more complex — we’ll say “other” to be safe.)*
---
d. AsCl₃ + H₂S → As₂S₃ + HCl
Left: As=1, Cl=3, H=2, S=1
Right: As=2, S=3, H=1, Cl=1
Need 2 As on left → 2AsCl₃ → now Cl=6
Need 3 S on left → 3H₂S → now H=6
Right: As₂S₃ is fine, HCl must be 6 to match H and Cl
Balanced: 2AsCl₃ + 3H₂S → As₂S₃ + 6HCl
Type: Ions swap partners → Double Replacement
---
e. CuSO₄•5H₂O → CuSO₄ + H₂O
This is dehydration. The dot means 5 water molecules are attached.
So: CuSO₄•5H₂O → CuSO₄ + 5H₂O
Already balanced!
Type: Compound breaks down → Decomposition
---
f. Fe₂O₃ + H₂ → Fe + H₂O
Left: Fe=2, O=3, H=2
Right: Fe=1, H=2, O=1
Need 2 Fe on right → 2Fe
Need 3 H₂O on right → gives 3 O and 6 H
→ So need 3 H₂ on left (gives 6 H)
Balanced: Fe₂O₃ + 3H₂ → 2Fe + 3H₂O
Type: H₂ takes oxygen from iron oxide → Single Replacement (or reduction, but in basic classification, often called single replacement)
---
g. CaCO₃ → CaO + CO₂
Already balanced!
Ca=1, C=1, O=3 on both sides.
Type: One compound breaks into two → Decomposition
---
h. Fe + S₈ → FeS
S₈ is 8 sulfur atoms. Each FeS has 1 S → need 8 FeS → so 8 Fe on left
Balanced: 8Fe + S₈ → 8FeS
Type: Two elements combine → Formation
---
i. H₂S + KOH → H₂O + K₂S
Left: H=2+1=3? Wait — H₂S has 2H, KOH has 1H → total 3H? No — better to balance properly.
Actually:
H₂S + 2KOH → 2H₂O + K₂S
Check:
Left: H=2+2=4, S=1, K=2, O=2
Right: H=4, O=2, K=2, S=1 → yes!
Balanced: H₂S + 2KOH → 2H₂O + K₂S
Type: Acid-base neutralization → Double Replacement
---
j. NaCl → Na + Cl₂
Left: Na=1, Cl=1
Right: Na=1, Cl=2 → need 2NaCl on left → then 2Na on right
Balanced: 2NaCl → 2Na + Cl₂
Type: Breaks into elements → Decomposition
---
k. Al + H₂SO₄ → H₂ + Al₂(SO₄)₃
Right: Al=2, SO₄=3 → so need 3 H₂SO₄ on left → gives 6 H → so 3 H₂ on right
Al: need 2 on left
Balanced: 2Al + 3H₂SO₄ → 3H₂ + Al₂(SO₄)₃
Type: Metal displaces hydrogen → Single Replacement
---
l. H₃PO₄ + NHOH → H₂O + (NH₄)₃PO₄
Right: (NH₄)₃PO₄ → 3 NH₄, 1 PO₄
Left: H₃PO₄ has 1 PO₄ → good
NH₄OH has 1 NH₄ → need 3 NH₄OH
Then H: left = 3 (from acid) + 3 (from base) = 6 H → right: 3 H₂O has 6 H → perfect
Balanced: H₃PO₄ + 3NH₄OH → 3H₂O + (NH₄)₃PO₄
Type: Acid + base → salt + water → Double Replacement
---
m. C₃H₈ + O₂ → CO₂ + H₂O
Hydrocarbon combustion.
C₃H₈ → 3C → need 3CO₂
8H → need 4H₂O (since each has 2H)
Oxygen: right = 3×2 + 4×1 = 6+4=10 O → so need 5 O₂ on left
Balanced: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Type: Hydrocarbon + oxygen → Hydrocarbon Combustion
---
n. Al + O₂ → Al₂O₃
Left: Al=1, O=2
Right: Al=2, O=3 → LCM of O is 6 → so 3O₂ and 2Al₂O₃ → then Al=4 on right → so 4Al on left
Balanced: 4Al + 3O₂ → 2Al₂O₃
Type: Elements combine → Formation
---
o. CH₄ + O₂ → CO₂ + H₂O
Combustion again.
CH₄ → 1C → 1CO₂
4H → 2H₂O
O: right = 2 + 2 = 4 → so 2O₂ on left
Balanced: CH₄ + 2O₂ → CO₂ + 2H₂O
Type: Hydrocarbon Combustion
---
p. K₂SO₄ + BaCl₂ → KCl + BaSO₄
Swap ions: K with Cl, Ba with SO₄
Left: K=2, SO₄=1, Ba=1, Cl=2
Right: K=1, Cl=1, Ba=1, SO₄=1 → need 2KCl
Balanced: K₂SO₄ + BaCl₂ → 2KCl + BaSO₄
Type: Ions swap → Double Replacement
---
q. C₅H₁₂ + O₂ → CO₂ + H₂O
Pentane combustion.
C₅ → 5CO₂
H₁₂ → 6H₂O
O: right = 5×2 + 6×1 = 10+6=16 → so 8O₂ on left
Balanced: C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O
Type: Hydrocarbon Combustion
---
r. Ca(OH)₂ + NH₄Cl → NH₄OH + CaCl₂
Left: Ca=1, O=2, H=2+4=6? Wait — Ca(OH)₂ has 2OH → 2O, 2H; NH₄Cl has N, 4H, Cl
Better:
Ca(OH)₂ + 2NH₄Cl → 2NH₄OH + CaCl₂
Check:
Left: Ca=1, O=2, H=2+8=10, N=2, Cl=2
Right: NH₄OH ×2 → N=2, H=8+2=10? Wait — NH₄OH is NH₄⁺ and OH → so each has 5H? No:
NH₄OH = N, 5H, O? Actually, formula is often written as NH₄OH meaning ammonium hydroxide → which is NH₃(aq) + H₂O, but for balancing, treat as NH₄OH → N, 5H, O? That’s messy.
Standard way:
Ca(OH)₂ + 2NH₄Cl → CaCl₂ + 2NH₃ + 2H₂O
But the problem says NH₄OH — so let’s assume it’s written as such.
If we write:
Ca(OH)₂ + 2NH₄Cl → 2NH₄OH + CaCl₂
Atoms:
Left: Ca=1, O=2, H=2 (from OH) + 8 (from 2NH₄) = 10H, N=2, Cl=2
Right: 2NH₄OH → 2N, 10H, 2O; CaCl₂ → Ca, 2Cl → matches!
So balanced: Ca(OH)₂ + 2NH₄Cl → 2NH₄OH + CaCl₂
Type: Double replacement → Double Replacement
---
s. V₂O₅ + Ca → CaO + V
Left: V=2, O=5, Ca=1
Right: Ca=1, O=1, V=1 → need 5 CaO → so 5 Ca on left → then V=2 on right
Balanced: V₂O₅ + 5Ca → 5CaO + 2V
Type: Metal displaces another metal → Single Replacement
---
t. Na + ZnI₂ → NaI + Zn
Left: Na=1, Zn=1, I=2
Right: Na=1, I=1, Zn=1 → need 2NaI → so 2Na on left
Balanced: 2Na + ZnI₂ → 2NaI + Zn
Type: Single replacement → Single Replacement
---
u. C₇H₆O₃ + O₂ → CO₂ + H₂O
Benzoic acid combustion.
C₇ → 7CO₂
H₆ → 3H₂O
O: left = 3 (in compound) + ? from O₂
Right: 7×2 + 3×1 = 14+3=17 O → minus 3 from compound → need 14 O from O₂ → so 7O₂
Balanced: C₇H₆O₃ + 7O₂ → 7CO₂ + 3H₂O
Type: Hydrocarbon Combustion (even though it has oxygen, still combustion)
---
v. Ca + N₂ → Ca₃N₂
Left: Ca=1, N=2
Right: Ca=3, N=2 → need 3Ca on left
Balanced: 3Ca + N₂ → Ca₃N₂
Type: Elements combine → Formation
---
w. Fe₂O₃ + H₂ → Fe + H₂O
Same as f! Already did: Fe₂O₃ + 3H₂ → 2Fe + 3H₂O
Type: Single Replacement
---
x. C₁₅H₃₀ + O₂ → CO₂ + H₂O
Combustion.
C₁₅ → 15CO₂
H₃₀ → 15H₂O
O: right = 15×2 + 15×1 = 30+15=45 → so 22.5 O₂ → multiply all by 2 to eliminate fraction
Original: C₁₅H₃₀ + 22.5O₂ → 15CO₂ + 15H₂O
Multiply by 2: 2C₁₅H₃₀ + 45O₂ → 30CO₂ + 30H₂O
Type: Hydrocarbon Combustion
---
y. BN + F₂ → BF₃ + N₂
Left: B=1, N=1, F=2
Right: B=1, F=3, N=2 → need 2BN on left → then N=2 → good
F: right = 3 per BF₃ → need 2BF₃ → F=6 → so 3F₂ on left
Balanced: 2BN + 3F₂ → 2BF₃ + N₂
Type: Not standard categories → Other
---
z. C₁₂H₂₆ + O₂ → CO₂ + H₂O
Dodecane combustion.
C₁₂ → 12CO₂
H₂₆ → 13H₂O
O: right = 12×2 + 13×1 = 24+13=37 → so 18.5 O₂ → multiply by 2
2C₁₂H₂₆ + 37O₂ → 24CO₂ + 26H₂O
Type: Hydrocarbon Combustion
---
Now, let’s compile the final answers clearly.
Final Answer:
a. 2Cu(s) + O₂(g) → 2CuO(s) — Formation
b. 2H₂O(l) → 2H₂(g) + O₂(g) — Decomposition
c. 3Fe(s) + 4H₂O(g) → 4H₂(g) + Fe₃O₄(s) — Other
d. 2AsCl₃(aq) + 3H₂S(aq) → As₂S₃(s) + 6HCl(aq) — Double Replacement
e. CuSO₄•5H₂O(s) → CuSO₄(s) + 5H₂O(g) — Decomposition
f. Fe₂O₃(s) + 3H₂(g) → 2Fe(s) + 3H₂O(l) — Single Replacement
g. CaCO₃(s) → CaO(s) + CO₂(g) — Decomposition
h. 8Fe(s) + S₈(s) → 8FeS(s) — Formation
i. H₂S(aq) + 2KOH(aq) → 2H₂O(l) + K₂S(aq) — Double Replacement
j. 2NaCl(l) → 2Na(l) + Cl₂(g) — Decomposition
k. 2Al(s) + 3H₂SO₄(aq) → 3H₂(g) + Al₂(SO₄)₃(aq) — Single Replacement
l. H₃PO₄(aq) + 3NH₄OH(aq) → 3H₂O(l) + (NH₄)₃PO₄(aq) — Double Replacement
m. C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l) — Hydrocarbon Combustion
n. 4Al(s) + 3O₂(g) → 2Al₂O₃(s) — Formation
o. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) — Hydrocarbon Combustion
p. K₂SO₄(aq) + BaCl₂(aq) → 2KCl(aq) + BaSO₄(s) — Double Replacement
q. C₅H₁₂(l) + 8O₂(g) → 5CO₂(g) + 6H₂O(g) — Hydrocarbon Combustion
r. Ca(OH)₂(aq) + 2NH₄Cl(aq) → 2NH₄OH(aq) + CaCl₂(aq) — Double Replacement
s. V₂O₅(s) + 5Ca(s) → 5CaO(s) + 2V(s) — Single Replacement
t. 2Na(s) + ZnI₂(aq) → 2NaI(aq) + Zn(s) — Single Replacement
u. C₇H₆O₃(l) + 7O₂(g) → 7CO₂(g) + 3H₂O(l) — Hydrocarbon Combustion
v. 3Ca(s) + N₂(g) → Ca₃N₂(s) — Formation
w. Fe₂O₃(s) + 3H₂(g) → 2Fe(s) + 3H₂O(l) — Single Replacement
x. 2C₁₅H₀(l) + 45O₂(g) → 30CO₂(g) + 30H₂O(g) — Hydrocarbon Combustion
y. 2BN(s) + 3F₂(g) → 2BF₃(s) + N₂(g) — Other
z. 2C₁₂H₂₆(l) + 37O₂(g) → 24CO₂(g) + 26H₂O(g) — Hydrocarbon Combustion
Parent Tip: Review the logic above to help your child master the concept of chemical equations and reactions worksheet.
![49 Balancing Chemical Equations Worksheets [with Answers]](/images2/chemical-equations-and-reactions-worksheet/chemical-equations-and-reactions-worksheet-17.jpg)
