Balancing Chemical Equations worksheet with ten equations to practice adjusting coefficients for atom conservation.

Let's balance each of the chemical equations step by step. We'll adjust the coefficients (the numbers in front of each compound) so that the number of atoms of each element is the same on both sides of the equation.

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1. H₂ + O₂ → H₂O

- Left: H = 2, O = 2
- Right: H = 2, O = 1 → Not balanced

We need to balance oxygen. Put a 2 in front of H₂O:

→ H₂ + O₂ → 2H₂O

Now: Right has 4 H and 2 O
Left has 2 H and 2 O → H not balanced

So put 2 in front of H₂:

→ 2H₂ + O₂ → 2H₂O

Now:
Left: H = 4, O = 2
Right: H = 4, O = 2 ✔

✔ Balanced: 2H₂ + O₂ → 2H₂O

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2. Fe + S₂ → FeS₃

- Fe: 1 on both sides
- S: 2 on left, 3 on right → Not balanced

We need to balance sulfur. The least common multiple of 2 and 3 is 6.

So:
- 3S₂ gives 6 S
- 2FeS₃ gives 6 S

So:
→ 2Fe + 3S₂ → 2FeS₃

Check:
- Fe: 2 = 2 ✔
- S: 6 = 6 ✔

✔ Balanced: 2Fe + 3S₂ → 2FeS₃

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3. NH₃ + O₂ → N₂ + H₂O

- N: 1 left, 2 right → need 2 NH₃
- H: 3 per NH₃ → 6 H if 2 NH₃ → need 3 H₂O (since 3×2=6 H)
- O: 3 H₂O → 3 O → so need 3/2 O₂

But we avoid fractions. Multiply entire equation by 2:

Start over:
→ 4NH₃ + 3O₂ → 2N₂ + 6H₂O

Check:
- N: 4 = 4 ✔
- H: 12 = 12 ✔
- O: 6 = 6 ✔

✔ Balanced: 4NH₃ + 3O₂ → 2N₂ + 6H₂O

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4. Al + HCl → AlCl₃ + H₂

- Al: 1 = 1 ✔
- Cl: 1 left, 3 right → need 3 HCl
- H: 3 left → H₂ needs 3 H → need 3/2 H₂ → use 3 HCl and 3/2 H₂ → multiply by 2

→ 2Al + 6HCl → 2AlCl₃ + 3H₂

Check:
- Al: 2 = 2 ✔
- Cl: 6 = 6 ✔
- H: 6 = 6 ✔

✔ Balanced: 2Al + 6HCl → 2AlCl₃ + 3H₂

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5. C₄H₁₀ + O₂ → CO₂ + H₂O

This is combustion of butane.

- C: 4 → need 4 CO₂
- H: 10 → need 5 H₂O (since 5×2=10 H)
- O: Right side: 4×2 + 5×1 = 8 + 5 = 13 O → need 13/2 O₂ → multiply by 2

So:
→ 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

Check:
- C: 8 = 8 ✔
- H: 20 = 20 ✔
- O: 26 = 16 (from CO₂) + 10 (from H₂O) = 26 ✔

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

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6. Na + H₂O → NaOH + H₂

- Na: 1 = 1 ✔
- H: 2 left, 1 in NaOH + 2 in H₂? Wait — let’s count carefully

Left: H₂O → 2 H, 1 O
Right: NaOH → 1 H, 1 O, 1 Na; H₂ → 2 H

Total right: H = 1 + 2 = 3 → too many

Try:
→ 2Na + 2H₂O → 2NaOH + H₂

Now:
- Na: 2 = 2 ✔
- H: 4 left → 2 in 2NaOH + 2 in H₂ = 4 ✔
- O: 2 = 2 ✔

✔ Balanced: 2Na + 2H₂O → 2NaOH + H₂

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7. N₂ + H₂ → NH₃

- N: 2 left, 1 right → need 2 NH₃
- H: 6 right → need 3 H₂

→ N₂ + 3H₂ → 2NH₃

Check:
- N: 2 = 2 ✔
- H: 6 = 6 ✔

✔ Balanced: N₂ + 3H₂ → 2NH₃

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8. KClO₃ → KCl + O₂

Decomposition of potassium chlorate.

- K: 1 = 1 ✔
- Cl: 1 = 1 ✔
- O: 3 left, 2 right → need LCM of 3 and 2 = 6

So:
→ 2KClO₃ → 2KCl + 3O₂

Check:
- K: 2 = 2 ✔
- Cl: 2 = 2 ✔
- O: 6 = 6 ✔

✔ Balanced: 2KClO₃ → 2KCl + 3O₂

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9. CH₄ + O₂ → CO₂ + H₂O

Combustion of methane.

- C: 1 = 1 ✔
- H: 4 → need 2 H₂O (4 H)
- O: Right: 2 (CO₂) + 1 (H₂O) = 3 → need 3/2 O₂ → multiply by 2

→ 2CH₄ + 3O₂ → 2CO₂ + 4H₂O

Check:
- C: 2 = 2 ✔
- H: 8 = 8 ✔
- O: 6 = 4 + 2 = 6 ✔

✔ Balanced: 2CH₄ + 3O₂ → 2CO₂ + 4H₂O

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10. C₆H₁₂O₆ → C₂H₅OH + CO₂

This is fermentation of glucose.

- Left: C: 6, H: 12, O: 6
- Right: C₂H₅OH has 2 C, 6 H, 1 O; CO₂ has 1 C, 2 O

Suppose we have a C₂H₅OH and b CO₂

Then:
- C: 2a + b = 6
- H: 6a = 12 → a = 2
- Then: 2(2) + b = 6 → 4 + b = 6 → b = 2

So:
→ C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

Check:
- C: 6 = 4 + 2 = 6 ✔
- H: 12 = 2×6 = 12 ✔
- O: 6 = 2×1 (in ethanol) + 2×2 (in CO₂) = 2 + 4 = 6 ✔

✔ Balanced: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

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Final Balanced Equations:

1. 2H₂ + O₂ → 2H₂O
2. 2Fe + 3S₂ → 2FeS₃
3. 4NH₃ + 3O₂ → 2N₂ + 6H₂O
4. 2Al + 6HCl → 2AlCl₃ + 3H₂
5. 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
6. 2Na + 2H₂O → 2NaOH + H₂
7. N₂ + 3H₂ → 2NH₃
8. 2KClO₃ → 2KCl + 3O₂
9. 2CH₄ + 3O₂ → 2CO₂ + 4H₂O
10. C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

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✔ All equations are now balanced! Let me know if you'd like them formatted for printing or study.