Classify and Balance Chemical Equations Exercise Set with Keys - Free Printable
Educational worksheet: Classify and Balance Chemical Equations Exercise Set with Keys. Download and print for classroom or home learning activities.
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Step-by-step solution for: Classify and Balance Chemical Equations Exercise Set with Keys
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Show Answer Key & Explanations
Step-by-step solution for: Classify and Balance Chemical Equations Exercise Set with Keys
Let’s go through each reaction one by one. We’ll balance the atoms on both sides and then classify the type of reaction using the rules at the bottom.
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1. AlI₃ + HgCl₂ → AlCl₃ + HgI₂
- Left: Al=1, I=3, Hg=1, Cl=2
- Right: Al=1, Cl=3, Hg=1, I=2 → Not balanced
Try coefficients:
Put 2 in front of AlI₃ → now Al=2, I=6
Put 3 in front of HgCl₂ → Hg=3, Cl=6
Now right side: need 2 AlCl₃ (Al=2, Cl=6) and 3 HgI₂ (Hg=3, I=6)
✔ Balanced: 2 AlI₃ + 3 HgCl₂ → 2 AlCl₃ + 3 HgI₂
Classification: Two compounds swap partners → Double Replacement
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2. Cl₂ + KBr → KCl + Br₂
Left: Cl=2, K=1, Br=1
Right: K=1, Cl=1, Br=2 → Not balanced
Need 2 KBr to get 2 Br → then need 2 KCl to match K and Cl
So: Cl₂ + 2KBr → 2KCl + Br₂
✔ Balanced: 1 Cl₂ + 2 KBr → 2 KCl + 1 Br₂
Classification: One element replaces another in a compound → Single Replacement
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3. H₂ + N₂ → NH₃
Left: H=2, N=2
Right: N=1, H=3 → Not balanced
Find LCM for H: 2 and 3 → 6
So 3 H₂ (H=6), 2 NH₃ (H=6, N=2) → then N₂ is already 2 N
✔ Balanced: 3 H₂ + 1 N₂ → 2 NH₃
Classification: Two elements make one compound → Synthesis
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4. Al₂S₃ → Al + S
Left: Al=2, S=3
Right: Al=1, S=1 → Not balanced
Just put 2 in front of Al, 3 in front of S
✔ Balanced: 1 Al₂S₃ → 2 Al + 3 S
Classification: One compound breaks into elements → Decomposition
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5. C₃H₈ + O₂ → CO₂ + H₂O ← Combustion!
Rule: Balance C first, then H, then O
C₃H₈ has 3 C → so 3 CO₂
Has 8 H → so 4 H₂O (since 4×2=8 H)
Now count O on right: 3×2 = 6 from CO₂, 4×1 = 4 from H₂O → total 10 O atoms → need 5 O₂ on left
✔ Balanced: 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
Classification: Hydrocarbon + oxygen → Combustion
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6. Ba + Rb₂C₂O₄ → BaC₂O₄ + Rb
Left: Ba=1, Rb=2, C=2, O=4
Right: Ba=1, C=2, O=4, Rb=1 → Rb not balanced
Put 2 in front of Rb on right
✔ Balanced: 1 Ba + 1 Rb₂C₂O₄ → 1 BaC₂O₄ + 2 Rb
Classification: Element replaces another in compound → Single Replacement
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7. Cs + Ca₃(PO₄)₂ → Cs₃PO₄ + Ca
Left: Cs=1, Ca=3, P=2, O=8
Right: Cs=3, P=1, O=4, Ca=1 → Way off
First, fix PO₄: Ca₃(PO₄)₂ has 2 PO₄ groups → so need 2 Cs₃PO₄ on right → that gives Cs=6, P=2, O=8
Then Ca: left has 3 Ca → right needs 3 Ca
Cs: right has 6 Cs → left needs 6 Cs
✔ Balanced: 6 Cs + 1 Ca₃(PO₄)₂ → 2 Cs₃PO₄ + 3 Ca
Classification: Element replaces metal in compound → Single Replacement
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8. Fe(OH)₂ → FeO + H₂O
Left: Fe=1, O=2, H=2
Right: Fe=1, O=1+1=2, H=2 → Already balanced!
✔ Balanced: 1 Fe(OH)₂ → 1 FeO + 1 H₂O
Classification: One compound breaks into two → Decomposition
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9. K₂SO₄ + Mg(OH)₂ → KOH + MgSO₄
Left: K=2, S=1, O=4+2=6? Wait — let's list carefully:
K₂SO₄: K=2, S=1, O=4
Mg(OH)₂: Mg=1, O=2, H=2 → Total left: K=2, S=1, Mg=1, O=6, H=2
Right: KOH: K=1, O=1, H=1; MgSO₄: Mg=1, S=1, O=4 → Total right: K=1, Mg=1, S=1, O=5, H=1 → Not balanced
Need 2 KOH to match K=2 and H=2 → then O on right: 2 (from 2KOH) + 4 (from MgSO₄) = 6 → matches left
✔ Balanced: 1 K₂SO₄ + 1 Mg(OH)₂ → 2 KOH + 1 MgSO₄
Classification: Two compounds swap ions → Double Replacement
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10. KI + AgCN → AgI + KCN
Left: K=1, I=1, Ag=1, C=1, N=1
Right: same → already balanced!
✔ Balanced: 1 KI + 1 AgCN → 1 AgI + 1 KCN
Classification: Swap of ions → Double Replacement
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11. Mg + O₂ → MgO
Left: Mg=1, O=2
Right: Mg=1, O=1 → Not balanced
Need 2 MgO → then need 2 Mg on left
O₂ is already 2 O → perfect
✔ Balanced: 2 Mg + 1 O₂ → 2 MgO
Classification: Two elements form one compound → Synthesis
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12. Mg(OH)₂ + H₂SO₄ → MgSO₄ + H₂O
Left: Mg=1, O=2+4=6? Let’s break down:
Mg(OH)₂: Mg=1, O=2, H=2
H₂SO₄: H=2, S=1, O=4 → Total: Mg=1, S=1, O=6, H=4
Right: MgSO₄: Mg=1, S=1, O=4
H₂O: H=2, O=1 → if only 1 H₂O → H=2, O=5 → not enough
Need 2 H₂O → then H=4, O=2 → total O on right: 4 (MgSO₄) + 2 (2H₂O) = 6 → matches
✔ Balanced: 1 Mg(OH)₂ + 1 H₂SO₄ → 1 MgSO₄ + 2 H₂O
Classification: Acid + base → salt + water → Double Replacement (also called neutralization)
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13. Rb + Li₃P → Rb₃P + Li
Left: Rb=1, Li=3, P=1
Right: Rb=3, P=1, Li=1 → Not balanced
Need 3 Rb on left → 3 Rb
Need 3 Li on right → 3 Li
✔ Balanced: 3 Rb + 1 Li₃P → 1 Rb₃P + 3 Li
Classification: Element replaces another → Single Replacement
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14. C₄H₁₀ + O₂ → CO₂ + H₂O ← Combustion again
Balance C first: C₄ → 4 CO₂
H: 10 H → 5 H₂O (5×2=10)
O on right: 4×2=8 from CO₂, 5×1=5 from H₂O → total 13 O atoms → need 13/2 O₂ → so multiply entire equation by 2 to eliminate fraction
Original trial: C₄H₁₀ + ? O₂ → 4 CO₂ + 5 H₂O → O needed: 8+5=13 → so 13/2 O₂
Multiply all by 2:
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
✔ Balanced: 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
Classification: Hydrocarbon + oxygen → Combustion
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15. Rb₂CO₃ + LiBr → Li₂CO₃ + RbBr
Left: Rb=2, C=1, O=3, Li=1, Br=1
Right: Li=2, C=1, O=3, Rb=1, Br=1 → Not balanced
Need 2 LiBr on left → Li=2, Br=2
Then need 2 RbBr on right → Rb=2, Br=2
✔ Balanced: 1 Rb₂CO₃ + 2 LiBr → 1 Li₂CO₃ + 2 RbBr
Classification: Swap of ions → Double Replacement
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16. Sr + AgNO₃ → Sr(NO₃)₂ + Ag
Left: Sr=1, Ag=1, N=1, O=3
Right: Sr=1, N=2, O=6, Ag=1 → Not balanced
Sr(NO₃)₂ has 2 NO₃ → so need 2 AgNO₃ on left → then Ag=2 on left → need 2 Ag on right
✔ Balanced: 1 Sr + 2 AgNO₃ → 1 Sr(NO₃)₂ + 2 Ag
Classification: Element replaces metal in compound → Single Replacement
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Final Answer:
1. 2 AlI₃ + 3 HgCl₂ → 2 AlCl₃ + 3 HgI₂ — Double Replacement
2. 1 Cl₂ + 2 KBr → 2 KCl + 1 Br₂ — Single Replacement
3. 3 H₂ + 1 N₂ → 2 NH₃ — Synthesis
4. 1 Al₂S₃ → 2 Al + 3 S — Decomposition
5. 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O — Combustion
6. 1 Ba + 1 Rb₂C₂O₄ → 1 BaC₂O₄ + 2 Rb — Single Replacement
7. 6 Cs + 1 Ca₃(PO₄)₂ → 2 Cs₃PO₄ + 3 Ca — Single Replacement
8. 1 Fe(OH)₂ → 1 FeO + 1 H₂O — Decomposition
9. 1 K₂SO₄ + 1 Mg(OH)₂ → 2 KOH + 1 MgSO₄ — Double Replacement
10. 1 KI + 1 AgCN → 1 AgI + 1 KCN — Double Replacement
11. 2 Mg + 1 O₂ → 2 MgO — Synthesis
12. 1 Mg(OH)₂ + 1 H₂SO₄ → 1 MgSO₄ + 2 H₂O — Double Replacement
13. 3 Rb + 1 Li₃P → 1 Rb₃P + 3 Li — Single Replacement
14. 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O — Combustion
15. 1 Rb₂CO₃ + 2 LiBr → 1 Li₂CO₃ + 2 RbBr — Double Replacement
16. 1 Sr + 2 AgNO₃ → 1 Sr(NO₃)₂ + 2 Ag — Single Replacement
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1. AlI₃ + HgCl₂ → AlCl₃ + HgI₂
- Left: Al=1, I=3, Hg=1, Cl=2
- Right: Al=1, Cl=3, Hg=1, I=2 → Not balanced
Try coefficients:
Put 2 in front of AlI₃ → now Al=2, I=6
Put 3 in front of HgCl₂ → Hg=3, Cl=6
Now right side: need 2 AlCl₃ (Al=2, Cl=6) and 3 HgI₂ (Hg=3, I=6)
✔ Balanced: 2 AlI₃ + 3 HgCl₂ → 2 AlCl₃ + 3 HgI₂
Classification: Two compounds swap partners → Double Replacement
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2. Cl₂ + KBr → KCl + Br₂
Left: Cl=2, K=1, Br=1
Right: K=1, Cl=1, Br=2 → Not balanced
Need 2 KBr to get 2 Br → then need 2 KCl to match K and Cl
So: Cl₂ + 2KBr → 2KCl + Br₂
✔ Balanced: 1 Cl₂ + 2 KBr → 2 KCl + 1 Br₂
Classification: One element replaces another in a compound → Single Replacement
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3. H₂ + N₂ → NH₃
Left: H=2, N=2
Right: N=1, H=3 → Not balanced
Find LCM for H: 2 and 3 → 6
So 3 H₂ (H=6), 2 NH₃ (H=6, N=2) → then N₂ is already 2 N
✔ Balanced: 3 H₂ + 1 N₂ → 2 NH₃
Classification: Two elements make one compound → Synthesis
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4. Al₂S₃ → Al + S
Left: Al=2, S=3
Right: Al=1, S=1 → Not balanced
Just put 2 in front of Al, 3 in front of S
✔ Balanced: 1 Al₂S₃ → 2 Al + 3 S
Classification: One compound breaks into elements → Decomposition
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5. C₃H₈ + O₂ → CO₂ + H₂O ← Combustion!
Rule: Balance C first, then H, then O
C₃H₈ has 3 C → so 3 CO₂
Has 8 H → so 4 H₂O (since 4×2=8 H)
Now count O on right: 3×2 = 6 from CO₂, 4×1 = 4 from H₂O → total 10 O atoms → need 5 O₂ on left
✔ Balanced: 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
Classification: Hydrocarbon + oxygen → Combustion
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6. Ba + Rb₂C₂O₄ → BaC₂O₄ + Rb
Left: Ba=1, Rb=2, C=2, O=4
Right: Ba=1, C=2, O=4, Rb=1 → Rb not balanced
Put 2 in front of Rb on right
✔ Balanced: 1 Ba + 1 Rb₂C₂O₄ → 1 BaC₂O₄ + 2 Rb
Classification: Element replaces another in compound → Single Replacement
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7. Cs + Ca₃(PO₄)₂ → Cs₃PO₄ + Ca
Left: Cs=1, Ca=3, P=2, O=8
Right: Cs=3, P=1, O=4, Ca=1 → Way off
First, fix PO₄: Ca₃(PO₄)₂ has 2 PO₄ groups → so need 2 Cs₃PO₄ on right → that gives Cs=6, P=2, O=8
Then Ca: left has 3 Ca → right needs 3 Ca
Cs: right has 6 Cs → left needs 6 Cs
✔ Balanced: 6 Cs + 1 Ca₃(PO₄)₂ → 2 Cs₃PO₄ + 3 Ca
Classification: Element replaces metal in compound → Single Replacement
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8. Fe(OH)₂ → FeO + H₂O
Left: Fe=1, O=2, H=2
Right: Fe=1, O=1+1=2, H=2 → Already balanced!
✔ Balanced: 1 Fe(OH)₂ → 1 FeO + 1 H₂O
Classification: One compound breaks into two → Decomposition
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9. K₂SO₄ + Mg(OH)₂ → KOH + MgSO₄
Left: K=2, S=1, O=4+2=6? Wait — let's list carefully:
K₂SO₄: K=2, S=1, O=4
Mg(OH)₂: Mg=1, O=2, H=2 → Total left: K=2, S=1, Mg=1, O=6, H=2
Right: KOH: K=1, O=1, H=1; MgSO₄: Mg=1, S=1, O=4 → Total right: K=1, Mg=1, S=1, O=5, H=1 → Not balanced
Need 2 KOH to match K=2 and H=2 → then O on right: 2 (from 2KOH) + 4 (from MgSO₄) = 6 → matches left
✔ Balanced: 1 K₂SO₄ + 1 Mg(OH)₂ → 2 KOH + 1 MgSO₄
Classification: Two compounds swap ions → Double Replacement
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10. KI + AgCN → AgI + KCN
Left: K=1, I=1, Ag=1, C=1, N=1
Right: same → already balanced!
✔ Balanced: 1 KI + 1 AgCN → 1 AgI + 1 KCN
Classification: Swap of ions → Double Replacement
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11. Mg + O₂ → MgO
Left: Mg=1, O=2
Right: Mg=1, O=1 → Not balanced
Need 2 MgO → then need 2 Mg on left
O₂ is already 2 O → perfect
✔ Balanced: 2 Mg + 1 O₂ → 2 MgO
Classification: Two elements form one compound → Synthesis
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12. Mg(OH)₂ + H₂SO₄ → MgSO₄ + H₂O
Left: Mg=1, O=2+4=6? Let’s break down:
Mg(OH)₂: Mg=1, O=2, H=2
H₂SO₄: H=2, S=1, O=4 → Total: Mg=1, S=1, O=6, H=4
Right: MgSO₄: Mg=1, S=1, O=4
H₂O: H=2, O=1 → if only 1 H₂O → H=2, O=5 → not enough
Need 2 H₂O → then H=4, O=2 → total O on right: 4 (MgSO₄) + 2 (2H₂O) = 6 → matches
✔ Balanced: 1 Mg(OH)₂ + 1 H₂SO₄ → 1 MgSO₄ + 2 H₂O
Classification: Acid + base → salt + water → Double Replacement (also called neutralization)
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13. Rb + Li₃P → Rb₃P + Li
Left: Rb=1, Li=3, P=1
Right: Rb=3, P=1, Li=1 → Not balanced
Need 3 Rb on left → 3 Rb
Need 3 Li on right → 3 Li
✔ Balanced: 3 Rb + 1 Li₃P → 1 Rb₃P + 3 Li
Classification: Element replaces another → Single Replacement
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14. C₄H₁₀ + O₂ → CO₂ + H₂O ← Combustion again
Balance C first: C₄ → 4 CO₂
H: 10 H → 5 H₂O (5×2=10)
O on right: 4×2=8 from CO₂, 5×1=5 from H₂O → total 13 O atoms → need 13/2 O₂ → so multiply entire equation by 2 to eliminate fraction
Original trial: C₄H₁₀ + ? O₂ → 4 CO₂ + 5 H₂O → O needed: 8+5=13 → so 13/2 O₂
Multiply all by 2:
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
✔ Balanced: 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
Classification: Hydrocarbon + oxygen → Combustion
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15. Rb₂CO₃ + LiBr → Li₂CO₃ + RbBr
Left: Rb=2, C=1, O=3, Li=1, Br=1
Right: Li=2, C=1, O=3, Rb=1, Br=1 → Not balanced
Need 2 LiBr on left → Li=2, Br=2
Then need 2 RbBr on right → Rb=2, Br=2
✔ Balanced: 1 Rb₂CO₃ + 2 LiBr → 1 Li₂CO₃ + 2 RbBr
Classification: Swap of ions → Double Replacement
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16. Sr + AgNO₃ → Sr(NO₃)₂ + Ag
Left: Sr=1, Ag=1, N=1, O=3
Right: Sr=1, N=2, O=6, Ag=1 → Not balanced
Sr(NO₃)₂ has 2 NO₃ → so need 2 AgNO₃ on left → then Ag=2 on left → need 2 Ag on right
✔ Balanced: 1 Sr + 2 AgNO₃ → 1 Sr(NO₃)₂ + 2 Ag
Classification: Element replaces metal in compound → Single Replacement
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Final Answer:
1. 2 AlI₃ + 3 HgCl₂ → 2 AlCl₃ + 3 HgI₂ — Double Replacement
2. 1 Cl₂ + 2 KBr → 2 KCl + 1 Br₂ — Single Replacement
3. 3 H₂ + 1 N₂ → 2 NH₃ — Synthesis
4. 1 Al₂S₃ → 2 Al + 3 S — Decomposition
5. 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O — Combustion
6. 1 Ba + 1 Rb₂C₂O₄ → 1 BaC₂O₄ + 2 Rb — Single Replacement
7. 6 Cs + 1 Ca₃(PO₄)₂ → 2 Cs₃PO₄ + 3 Ca — Single Replacement
8. 1 Fe(OH)₂ → 1 FeO + 1 H₂O — Decomposition
9. 1 K₂SO₄ + 1 Mg(OH)₂ → 2 KOH + 1 MgSO₄ — Double Replacement
10. 1 KI + 1 AgCN → 1 AgI + 1 KCN — Double Replacement
11. 2 Mg + 1 O₂ → 2 MgO — Synthesis
12. 1 Mg(OH)₂ + 1 H₂SO₄ → 1 MgSO₄ + 2 H₂O — Double Replacement
13. 3 Rb + 1 Li₃P → 1 Rb₃P + 3 Li — Single Replacement
14. 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O — Combustion
15. 1 Rb₂CO₃ + 2 LiBr → 1 Li₂CO₃ + 2 RbBr — Double Replacement
16. 1 Sr + 2 AgNO₃ → 1 Sr(NO₃)₂ + 2 Ag — Single Replacement
Parent Tip: Review the logic above to help your child master the concept of classification of chemical reaction worksheet.
