Students can use this sheet to practice balancing atoms and classifying reaction mechanisms for twelve different chemical scenarios.

Here is the complete solution for balancing and identifying each of the 12 chemical equations. I’ll balance each equation step by step and classify the reaction type.

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(1) Al₂(SO₄)₃ + BaCl₂ → BaSO₄ + AlCl₃

Balanced Equation:
Al₂(SO₄)₃ + 3BaCl₂ → 3BaSO₄ + 2AlCl₃

✔ Check atoms:
- Al: 2 = 2
- S: 3 = 3
- O: 12 = 12
- Ba: 3 = 3
- Cl: 6 = 6

Type of reaction: Double Displacement (or Precipitation — since BaSO₄ is insoluble)

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(2) Al₂S₃ → Al + S

Balanced Equation:
Al₂S₃ → 2Al + 3S

✔ Check atoms:
- Al: 2 = 2
- S: 3 = 3

Type of reaction: Decomposition (one compound breaks into two elements)

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(3) NaOH + CuSO₄ → Na₃PO₄ + Cu(OH)₂

⚠️ Problem: The products are chemically incorrect. NaOH + CuSO₄ should produce Na₂SO₄ and Cu(OH)₂, not Na₃PO₄ (which contains phosphorus, not present in reactants).

This appears to be a typo. Likely intended:

Corrected Reaction:
NaOH + CuSO₄ → Na₂SO₄ + Cu(OH)₂

Balanced Equation:
2NaOH + CuSO₄ → Na₂SO₄ + Cu(OH)₂

✔ Check atoms:
- Na: 2 = 2
- O: 2+4=6 → 4+2=6
- H: 2 = 2
- Cu: 1 = 1
- S: 1 = 1

Type of reaction: Double Displacement (Precipitation — Cu(OH)₂ is insoluble)

> If we insist on the original (incorrect) products, it’s impossible to balance because P is not in reactants. So we assume typo.

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(4) Fe + H₂SO₄ → Fe₂SO₄ + H₂

⚠️ Problem: Iron(II) sulfate is FeSO₄, not Fe₂SO₄. Fe₂SO₄ doesn’t exist — correct formula is FeSO₄.

Corrected Reaction:
Fe + H₂SO₄ → FeSO₄ + H₂

Balanced Equation:
Fe + H₂SO₄ → FeSO₄ + H₂ ✔ Already balanced.

✔ Check atoms:
- Fe: 1 = 1
- H: 2 = 2
- S: 1 = 1
- O: 4 = 4

Type of reaction: Single Displacement (Fe displaces H from acid)

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(5) C₄H₁₂ + O₂ → CO₂ + H₂O

⚠️ Problem: C₄H₁₂ does not exist. Common hydrocarbon with 4 carbons is butane: C₄H₁₀.

Assuming typo → C₄H₁₀

Balanced Equation:
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

✔ Check atoms:
- C: 8 = 8
- H: 20 = 20
- O: 26 = 16 + 10 = 26

Type of reaction: Combustion (hydrocarbon + oxygen → CO₂ + H₂O)

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(6) H₂S + O₂ → SO₂ + H₂O

Balanced Equation:
2H₂S + 3O₂ → 2SO₂ + 2H₂O

✔ Check atoms:
- H: 4 = 4
- S: 2 = 2
- O: 6 = 4 + 2 = 6

Type of reaction: Combustion (sulfur compound burns to form SO₂ and H₂O)

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(7) C₅H₉O + O₂ → CO₂ + H₂O

⚠️ Problem: C₅H₉O is not a standard molecular formula (odd number of H atoms suggests radical or error). But we can still balance as given.

Let’s balance:

Assume: C₅H₉O + O₂ → 5CO₂ + 4.5H₂O → multiply by 2 to eliminate fraction.

Balanced Equation:
2C₅H₉O + 13O₂ → 10CO₂ + 9H₂O

✔ Check atoms:
- C: 10 = 10
- H: 18 = 18
- O: 2 + 26 = 28 → 20 + 9 = 29 ✘ Not balanced!

Wait — let's recalculate:

Left: O from 2C₅H₉O = 2 O; from 13O₂ = 26 O → total 28 O atoms
Right: 10CO₂ = 20 O; 9H₂O = 9 O → total 29 O → mismatch.

Try again:

Set up algebraically:

a C₅H₉O + b O₂ → c CO₂ + d H₂O

C: 5a = c
H: 9a = 2d → d = 9a/2
O: a + 2b = 2c + d

Substitute c = 5a, d = 9a/2:

a + 2b = 2(5a) + 9a/2
a + 2b = 10a + 4.5a = 14.5a
→ 2b = 13.5a → b = 6.75a

Multiply all by 4 to eliminate decimals:

a=4, b=27, c=20, d=18

Balanced Equation:
4C₅H₉O + 27O₂ → 20CO₂ + 18H₂O

✔ Check atoms:
- C: 20 = 20
- H: 36 = 36
- O: 4 + 54 = 58 → 40 + 18 = 58 ✔️

Type of reaction: Combustion

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(8) Al + NiBr₂ → AlBr₃ + Ni

Balanced Equation:
2Al + 3NiBr₂ → 2AlBr₃ + 3Ni

✔ Check atoms:
- Al: 2 = 2
- Ni: 3 = 3
- Br: 6 = 6

Type of reaction: Single Displacement (Al displaces Ni — aluminum is more reactive)

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(9) Al + O₂ → Al₂O₃

Balanced Equation:
4Al + 3O₂ → 2Al₂O₃

✔ Check atoms:
- Al: 4 = 4
- O: 6 = 6

Type of reaction: Synthesis (Combination) — two elements form a compound

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(10) H₂O₂ → H₂O + O₂

Balanced Equation:
2H₂O₂ → 2H₂O + O₂

✔ Check atoms:
- H: 4 = 4
- O: 4 = 4

Type of reaction: Decomposition (hydrogen peroxide breaks down)

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(11) K + Cl₂ → KCl

Balanced Equation:
2K + Cl₂ → 2KCl

✔ Check atoms:
- K: 2 = 2
- Cl: 2 = 2

Type of reaction: Synthesis (Combination) — metal + nonmetal → ionic compound

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(12) Na + MgCl₂ → NaCl + Mg

Balanced Equation:
2Na + MgCl₂ → 2NaCl + Mg

✔ Check atoms:
- Na: 2 = 2
- Mg: 1 = 1
- Cl: 2 = 2

Type of reaction: Single Displacement — BUT WAIT: Sodium is more reactive than magnesium? Actually, NO — sodium is *more* reactive than magnesium, so this reaction should occur.

✔ Reactivity series: Na > Mg → so yes, Na can displace Mg.

Type of reaction: Single Displacement

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## ✔ FINAL ANSWERS SUMMARY:

| # | Balanced Equation | Type of Reaction |
|---|---------------------------------------------|---------------------------|
| 1 | Al₂(SO₄)₃ + 3BaCl₂ → 3BaSO₄ + 2AlCl₃ | Double Displacement |
| 2 | Al₂S₃ → 2Al + 3S | Decomposition |
| 3 | 2NaOH + CuSO₄ → Na₂SO₄ + Cu(OH)₂ | Double Displacement |
| 4 | Fe + H₂SO₄ → FeSO₄ + H₂ | Single Displacement |
| 5 | 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O | Combustion |
| 6 | 2H₂S + 3O₂ → 2SO₂ + 2H₂O | Combustion |
| 7 | 4C₅H₉O + 27O₂ → 20CO₂ + 18H₂O | Combustion |
| 8 | 2Al + 3NiBr₂ → 2AlBr₃ + 3Ni | Single Displacement |
| 9 | 4Al + 3O₂ → 2Al₂O₃ | Synthesis |
|10 | 2H₂O₂ → 2H₂O + O₂ | Decomposition |
|11 | 2K + Cl₂ → 2KCl | Synthesis |
|12 | 2Na + MgCl₂ → 2NaCl + Mg | Single Displacement |

> Note: Equations (3), (4), and (5) had likely typos in the original worksheet. I corrected them based on chemical plausibility and common reactions.

Let me know if you’d like explanations for any specific reaction type or balancing technique!