Balancing Chemical Equations worksheet with ten practice problems for students to complete.

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

To balance oxygen, we need 2 H₂O on the right:
→ H₂ + O₂ → 2H₂O

Now H on right = 4, but left has only 2 H → Add 2 H₂ on left:
→ 2H₂ + O₂ → 2H₂O

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

✔ Balanced: 2H₂ + O₂ → 2H₂O

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2. Fe + S₂ → FeS₃

Note: S₂ is a molecule of sulfur (diatomic), but FeS₃ implies iron(III) sulfide.

But FeS₃ is not a common stable compound. However, assuming it’s correct as written:

- Left: Fe = 1, S = 2
- Right: Fe = 1, S = 3 → Not balanced

We need to balance S. LCM of 2 and 3 is 6.

So, use 3 S₂ → 6 S, and 2 FeS₃ → 6 S

So:
→ 2Fe + 3S₂ → 2FeS₃

Check:
Left: Fe = 2, S = 6
Right: Fe = 2, S = 6 ✔

✔ Balanced: 2Fe + 3S₂ → 2FeS₃

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3. NH₃ + O₂ → N₂ + H₂O

- Left: N = 1, H = 3, O = 2
- Right: N = 2, H = 2, O = 1 → Not balanced

Start with nitrogen: 2N on right, so need 2NH₃ on left:
→ 2NH₃ + O₂ → N₂ + H₂O

Now H: Left = 6, Right = 2 → Need 3H₂O
→ 2NH₃ + O₂ → N₂ + 3H₂O

Now O: Right = 3, Left = 2 → Need 3/2 O₂ → Multiply entire equation by 2 to eliminate fraction:

Multiply all coefficients by 2:
→ 4NH₃ + 3O₂ → 2N₂ + 6H₂O

Check:
Left: N = 4, H = 12, O = 6
Right: N = 4, H = 12, O = 6 ✔

✔ Balanced: 4NH₃ + 3O₂ → 2N₂ + 6H₂O

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4. Al + HCl → AlCl₃ + H₂

- Left: Al = 1, H = 1, Cl = 1
- Right: Al = 1, Cl = 3, H = 2 → Not balanced

AlCl₃ has 3 Cl → need 3 HCl
→ Al + 3HCl → AlCl₃ + H₂

Now H: Left = 3, Right = 2 → Need 3/2 H₂ → Multiply by 2:

→ 2Al + 6HCl → 2AlCl₃ + 3H₂

Check:
Left: Al = 2, H = 6, Cl = 6
Right: Al = 2, Cl = 6, H = 6 ✔

✔ Balanced: 2Al + 6HCl → 2AlCl₃ + 3H₂

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5. C₄H₁₀ + O₂ → CO₂ + H₂O

This is combustion of butane.

- Left: C = 4, H = 10, O = 2
- Right: C = 1, H = 2, O = 3 → Not balanced

Set up:
→ C₄H₁₀ + O₂ → 4CO₂ + 5H₂O (since 4C and 10H)

Now count O on right: 4×2 + 5×1 = 8 + 5 = 13 O atoms

So need 13/2 O₂ → Multiply entire equation by 2:

→ 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

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

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

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6. Na + H₂O → NaOH + H₂

- Left: Na = 1, H = 2, O = 1
- Right: Na = 1, O = 1, H = 1 (in NaOH) + 2 (in H₂) = 3 → Not balanced

Need more H on left → Try 2Na:

→ 2Na + H₂O → 2NaOH + H₂

Now check:
Left: Na = 2, H = 2, O = 1
Right: Na = 2, O = 2, H = 2 (in 2NaOH) + 2 (in H₂) = 4 → O and H mismatch

Wait: 2NaOH has 2O → need 2H₂O

Try:
→ 2Na + 2H₂O → 2NaOH + H₂

Now:
Left: Na = 2, H = 4, O = 2
Right: Na = 2, O = 2, H = 2 (in 2NaOH) + 2 (in H₂) = 4 ✔

✔ Balanced: 2Na + 2H₂O → 2NaOH + H₂

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7. N₂ + H₂ → NH₃

- Left: N = 2, H = 2
- Right: N = 1, H = 3 → Not balanced

Need 2NH₃ for N: → N₂ + H₂ → 2NH₃

Now H: Right = 6, Left = 2 → Need 3H₂

→ N₂ + 3H₂ → 2NH₃

Check: N = 2, H = 6 on both sides ✔

✔ Balanced: N₂ + 3H₂ → 2NH₃

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8. KClO₃ → KCl + O₂

Decomposition of potassium chlorate.

- Left: K = 1, Cl = 1, O = 3
- Right: K = 1, Cl = 1, O = 2 → Not balanced

O: 3 on left, 2 on right → LCM = 6

Use 2KClO₃ → 2KCl + 3O₂

Check:
Left: K = 2, Cl = 2, O = 6
Right: K = 2, Cl = 2, O = 6 ✔

✔ Balanced: 2KClO₃ → 2KCl + 3O₂

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9. CH₄ + O₂ → CO₂ + H₂O

Combustion of methane.

- Left: C = 1, H = 4, O = 2
- Right: C = 1, O = 2 (CO₂) + 1 (H₂O) = 3, H = 2 → Not balanced

Set:
→ CH₄ + O₂ → CO₂ + 2H₂O

Now H: 4 on both sides
O: Right = 2 (CO₂) + 2 (H₂O) = 4 → So need 2O₂

→ CH₄ + 2O₂ → CO₂ + 2H₂O

Check:
Left: C = 1, H = 4, O = 4
Right: C = 1, H = 4, O = 2 + 2 = 4 ✔

✔ Balanced: CH₄ + 2O₂ → CO₂ + 2H₂O

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10. C₆H₁₂O₆ → C₂H₅OH + CO₂

This is fermentation of glucose.

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

Assume products: ethanol (C₂H₅OH) and CO₂

Let’s suppose x C₂H₅OH and y CO₂

Then:
C: 2x + y = 6
H: 6x = 12 → x = 2
Then: 2(2) + y = 6 → 4 + y = 6 → y = 2

So: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

Check:
Left: C = 6, H = 12, O = 6
Right: C = 4 + 2 = 6, H = 12, O = 2 (ethanol) + 4 (CO₂) = 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. CH₄ + 2O₂ → CO₂ + 2H₂O
10. C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

All equations are now balanced! Let me know if you'd like them formatted neatly.