Practice balancing chemical equations with this worksheet featuring 13 reactions involving elements and compounds.

Let's balance each of the chemical equations step by step. The goal is to ensure that the number of atoms of each element is the same on both sides of the equation.

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1. Fe + O₂ → Fe₃O₄

- Right side: 3 Fe, 4 O
- Left side: 1 Fe, 2 O

To get 3 Fe on the left: use 3 Fe
To get 4 O: need 2 O₂ (since O₂ has 2 oxygen atoms)

So:
> 3Fe + 2O₂ → Fe₃O₄

✔ Balanced: 3 Fe, 4 O on both sides.

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2. Sr + O₂ → SrO

- Right: 1 Sr, 1 O
- Left: 1 Sr, 2 O

We need to balance oxygen. Use 2 SrO on right → 2 Sr, 2 O
Then need 2 Sr on left and 1 O₂

> 2Sr + O₂ → 2SrO

✔ Balanced: 2 Sr, 2 O on both sides.

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3. Sn + NaOH → Na₂SnO₂ + H₂

- Products: Na₂SnO₂ (2 Na, 1 Sn, 2 O), H₂ (2 H)
- Reactants: Sn, NaOH (1 Na, 1 O, 1 H)

We need:
- 2 Na → so 2 NaOH
- 2 O → from 2 NaOH
- 2 H → will make 1 H₂

But Sn is already balanced.

Try:
> Sn + 2NaOH → Na₂SnO₂ + H₂

Check:
- Left: Sn=1, Na=2, O=2, H=2
- Right: Na=2, Sn=1, O=2, H=2

✔ Balanced.

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4. K + Br₂ → KBr

- Right: K=1, Br=1
- Left: K=1, Br=2

Need 2 KBr → 2 K, 2 Br
So need 2 K on left

> 2K + Br₂ → 2KBr

✔ Balanced.

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5. C₈H₁₈ + O₂ → CO₂ + H₂O

This is combustion of octane.

- Left: C=8, H=18, O=?
- Right: CO₂ has 1 C, 2 O; H₂O has 2 H, 1 O

Balance C: 8 CO₂
Balance H: 18 H → 9 H₂O (since each has 2 H)

Now:
- Right: 8 CO₂ → 8 C, 16 O
9 H₂O → 9 O → total O = 16 + 9 = 25 O
- So need 25/2 = 12.5 O₂ → multiply whole equation by 2

Multiply all coefficients by 2:

> 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

Check:
- C: 16 = 16 ✔
- H: 36 = 36 ✔
- O: 50 on left (25×2), right: 16×2 + 18×1 = 32+18=50 ✔

✔ Balanced.

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6. Sb + I₂ → SbI₃

- Right: Sb=1, I=3
- Left: Sb=1, I=2

Need 3 I → so 3/2 I₂ → use fractions or multiply

Use:
> 2Sb + 3I₂ → 2SbI₃

Check:
- Sb: 2 = 2
- I: 6 = 6 ✔

✔ Balanced.

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7. COCl₂ + H₂O → HCl + CO₂

- Left: C=1, O=1 (from COCl₂) + 1 (from H₂O) = 2 O, Cl=2, H=2
- Right: HCl has H and Cl, CO₂ has C and O

We have:
- C: 1 → 1 CO₂
- Cl: 2 → 2 HCl
- H: 2 → 2 HCl → H=2
- O: COCl₂ has 1 O, H₂O has 1 O → 2 O
CO₂ has 2 O → matches

So:
> COCl₂ + H₂O → 2HCl + CO₂

Check:
- C: 1 = 1 ✔
- O: 1+1 = 2, CO₂ has 2 → ✔
- Cl: 2 = 2 ✔
- H: 2 = 2 ✔

✔ Balanced.

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8. CS₂ + O₂ → CO₂ + SO₂

- Left: C=1, S=2, O=?
- Right: CO₂ (C=1, O=2), SO₂ (S=1, O=2)

We need 2 SO₂ for 2 S → 2 SO₂ → 2 S, 4 O
Plus CO₂ → 1 C, 2 O → total O = 6

So need 3 O₂ (6 O)

> CS₂ + 3O₂ → CO₂ + 2SO₂

Check:
- C: 1 = 1 ✔
- S: 2 = 2 ✔
- O: 6 = 2 + 4 = 6 ✔

✔ Balanced.

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9. H₂SO₄ + NaCN → HCN + Na₂SO₄

- Left: H₂SO₄ (H=2, S=1, O=4), NaCN (Na=1, C=1, N=1)
- Right: HCN (H=1, C=1, N=1), Na₂SO₄ (Na=2, S=1, O=4)

We need:
- Na: 2 on right → need 2 NaCN
- Then: 2 NaCN → 2 C, 2 N, 2 Na
- HCN: need 2 HCN → 2 H, 2 C, 2 N

But H₂SO₄ has only 2 H → can supply 2 H for 2 HCN

So:
> H₂SO₄ + 2NaCN → 2HCN + Na₂SO₄

Check:
- H: 2 = 2 ✔
- S: 1 = 1 ✔
- O: 4 = 4 ✔
- Na: 2 = 2 ✔
- C: 2 = 2 ✔
- N: 2 = 2 ✔

✔ Balanced.

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10. KClO₃ → KCl + O₂

Decomposition of potassium chlorate.

- Left: K=1, Cl=1, O=3
- Right: KCl (K=1, Cl=1), O₂ (O=2)

Need to balance O: 3 O on left → but O₂ comes in pairs

So need 3/2 O₂ → multiply entire equation by 2:

> 2KClO₃ → 2KCl + 3O₂

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

✔ Balanced.

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11. H₂ + F₂ → HF

- Right: H=1, F=1
- Left: H=2, F=2

So need 2 HF → 2 H, 2 F

> H₂ + F₂ → 2HF

✔ Balanced.

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12. BaCl₂ + KIO₃ → Ba(IO₃)₂ + KCl

- Left: Ba=1, Cl=2, K=1, I=1, O=3
- Right: Ba(IO₃)₂ → Ba=1, I=2, O=6; KCl → K=1, Cl=1

Need:
- I: 2 on right → need 2 KIO₃
- Then: 2 KIO₃ → 2 K, 2 I, 6 O
- KCl: need 2 KCl → 2 K, 2 Cl
- But BaCl₂ has only 2 Cl → so need 1 BaCl₂

So:
> BaCl₂ + 2KIO₃ → Ba(IO₃)₂ + 2KCl

Check:
- Ba: 1 = 1 ✔
- Cl: 2 = 2 ✔
- K: 2 = 2 ✔
- I: 2 = 2 ✔
- O: 6 = 6 ✔

✔ Balanced.

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13. Mg + HCl → MgCl₂ + H₂

- Left: Mg=1, H=1, Cl=1
- Right: Mg=1, Cl=2, H=2

Need 2 HCl to provide 2 Cl and 2 H

> Mg + 2HCl → MgCl₂ + H₂

Check:
- Mg: 1 = 1 ✔
- H: 2 = 2 ✔
- Cl: 2 = 2 ✔

✔ Balanced.

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✔ Final Balanced Equations:

1. 3Fe + 2O₂ → Fe₃O₄
2. 2Sr + O₂ → 2SrO
3. Sn + 2NaOH → Na₂SnO₂ + H₂
4. 2K + Br₂ → 2KBr
5. 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O
6. 2Sb + 3I₂ → 2SbI₃
7. COCl₂ + H₂O → 2HCl + CO₂
8. CS₂ + 3O₂ → CO₂ + 2SO₂
9. H₂SO₄ + 2NaCN → 2HCN + Na₂SO₄
10. 2KClO₃ → 2KCl + 3O₂
11. H₂ + F₂ → 2HF
12. BaCl₂ + 2KIO₃ → Ba(IO₃)₂ + 2KCl
13. Mg + 2HCl → MgCl₂ + H₂

All equations are now balanced! Let me know if you'd like explanations for any specific one.