Balancing Chemical Equations Practice Problems 19

Let's go through each of these chemical equations and balance them step by step. Balancing chemical equations means making sure the number of atoms of each element is the same on both sides of the equation.

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1. C₃H₈ + O₂ → CO₂ + H₂O

Step-by-step:
- Carbon: 3 on left → need 3 CO₂ on right
- Hydrogen: 8 on left → need 4 H₂O (since each has 2 H)
- Now oxygen: Right side = 3×2 (from CO₂) + 4×1 (from H₂O) = 6 + 4 = 10 O atoms
- So, need 5 O₂ molecules (since each has 2 O atoms)

Balanced:
> C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

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2. Al₂(SO₄)₃ + NaOH → Na₂SO₄ + Al(OH)₃

Step-by-step:
- Al: 2 on left → need 2 Al(OH)₃ on right
- SO₄: 3 on left → need 3 Na₂SO₄ on right
- Na: 3×2 = 6 on right → need 6 NaOH on left
- OH: 6 from NaOH → matches 2×3 = 6 in 2 Al(OH)₃

Balanced:
> Al₂(SO₄)₃ + 6NaOH → 3Na₂SO₄ + 2Al(OH)₃

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3. Al₂O₃ + Fe → Fe₃O₄ + Al

This is a redox reaction (thermite-like). Let's balance:

- Al: 2 on left → need 2 Al on right
- Fe: 3 on right → need 3 Fe on left
- O: 3 on left, 4 on right → not balanced yet

Try balancing with least common multiple for O:
- Left: Al₂O₃ → 3 O
- Right: Fe₃O₄ → 4 O

LCM of 3 and 4 is 12 → use 4 Al₂O₃ (12 O) and 3 Fe₃O₄ (12 O)

Now:
- Al: 4×2 = 8 → need 8 Al on right
- Fe: 3×3 = 9 → need 9 Fe on left

So:
> 4Al₂O₃ + 9Fe → 3Fe₃O₄ + 8Al

Check:
- Al: 8 = 8 ✔
- O: 12 = 12 ✔
- Fe: 9 = 9 ✔

Balanced:
> 4Al₂O₃ + 9Fe → 3Fe₃O₄ + 8Al

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4. FeSO₄ + AgCl → Ag₂SO₄ + FeCl₂

Step-by-step:
- SO₄: 1 on each side → OK
- Fe: 1 on each side → OK
- Ag: 1 on left, 2 on right → need 2 AgCl on left
- Cl: 2 on left → need FeCl₂ on right → OK

So:
> FeSO₄ + 2AgCl → Ag₂SO₄ + FeCl₂

Check:
- Fe: 1 = 1 ✔
- S: 1 = 1 ✔
- O: 4 = 4 ✔
- Ag: 2 = 2 ✔
- Cl: 2 = 2 ✔

Balanced:
> FeSO₄ + 2AgCl → Ag₂SO₄ + FeCl₂

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5. NH₄NO₃ → N₂O + H₂O

Decomposition:
- N: 2 on left → need 1 N₂O (has 2 N)
- H: 4 on left → need 2 H₂O (each has 2 H)
- O: 3 on left → N₂O has 1 O, 2 H₂O has 2 O → total 3 O ✔

Balanced:
> NH₄NO₃ → N₂O + 2H₂O

Wait — check atoms:
- Left: N=2, H=4, O=3
- Right: N₂O → N=2, O=1; 2H₂O → H=4, O=2 → total O=3 ✔

Yes!

Balanced:
> NH₄NO₃ → N₂O + 2H₂O

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6. KOH + HBr → KBr + H₂O

Simple acid-base neutralization.

- K: 1 = 1 ✔
- O: 1 = 1 ✔
- H: 1+1=2 on left → H₂O has 2 H ✔
- Br: 1 = 1 ✔

Already balanced!

Balanced:
> KOH + HBr → KBr + H₂O

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7. P₄O₁₀ + H₂O → H₃PO₄

- P: 4 on left → need 4 H₃PO₄ on right
- O: 10 + 1 = 11 on left? Wait — H₂O adds 1 O, but we have 4 H₃PO₄ → 4×4 = 16 O?
Wait — let's count carefully.

P₄O₁₀ has 4 P and 10 O
Each H₃PO₄ has 1 P, 4 O, 3 H → so 4 H₃PO₄ has 4 P, 16 O, 12 H

But left side: P₄O₁₀ has 10 O, H₂O has 1 O → total 11 O → not enough.

We need more water.

Let’s suppose:
- 4 H₃PO₄ needs 4 P → OK
- Needs 16 O → P₄O₁₀ gives 10 O → need 6 more O → so need 6 H₂O (6 O and 12 H)
- H: 6 H₂O gives 12 H → 4 H₃PO₄ needs 12 H → perfect

So:
> P₄O₁₀ + 6H₂O → 4H₃PO₄

Check:
- P: 4 = 4 ✔
- O: 10 + 6 = 16 → 4×4 = 16 ✔
- H: 12 = 12 ✔

Balanced:
> P₄O₁₀ + 6H₂O → 4H₃PO₄

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8. Al + H₂SO₄ → Al₂(SO₄)₃ + H₂

- Al: 2 on right → need 2 Al on left
- SO₄: 3 on right → need 3 H₂SO₄ on left
- H: 3×2 = 6 on left → need 3 H₂ on right

So:
> 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

Check:
- Al: 2 = 2 ✔
- S: 3 = 3 ✔
- O: 12 = 12 ✔
- H: 6 = 6 ✔

Balanced:
> 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

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9. Be₂C + H₂O → Be(OH)₂ + CH₄

- Be: 2 on left → need 2 Be(OH)₂ on right
- C: 1 on left → need 1 CH₄ on right
- Now H: right side → 2 Be(OH)₂ has 4 H, CH₄ has 4 H → total 8 H
- O: 2 Be(OH)₂ has 4 O → need 4 H₂O on left (provides 4 O and 8 H)

So:
> Be₂C + 4H₂O → 2Be(OH)₂ + CH₄

Check:
- Be: 2 = 2 ✔
- C: 1 = 1 ✔
- O: 4 = 4 ✔
- H: 8 = 8 ✔

Balanced:
> Be₂C + 4H₂O → 2Be(OH)₂ + CH₄

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10. NH₃ + CuO → Cu + N₂ + H₂O

- N: 2 on right (N₂) → need 2 NH₃ on left
- H: 2×3 = 6 H → need 3 H₂O on right (6 H)
- O: 3 H₂O → 3 O → need 3 CuO on left
- Cu: 3 on left → need 3 Cu on right

So:
> 2NH₃ + 3CuO → 3Cu + N₂ + 3H₂O

Check:
- N: 2 = 2 ✔
- H: 6 = 6 ✔
- Cu: 3 = 3 ✔
- O: 3 = 3 ✔

Balanced:
> 2NH₃ + 3CuO → 3Cu + N₂ + 3H₂O

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✔ Final Balanced Equations:

1. C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
2. Al₂(SO₄)₃ + 6NaOH → 3Na₂SO₄ + 2Al(OH)₃
3. 4Al₂O₃ + 9Fe → 3Fe₃O₄ + 8Al
4. FeSO₄ + 2AgCl → Ag₂SO₄ + FeCl₂
5. NH₄NO₃ → N₂O + 2H₂O
6. KOH + HBr → KBr + H₂O
7. P₄O₁₀ + 6H₂O → 4H₃PO₄
8. 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂
9. Be₂C + 4H₂O → 2Be(OH)₂ + CH₄
10. 2NH₃ + 3CuO → 3Cu + N₂ + 3H₂O

Let me know if you'd like explanations for any specific one!