Balancing Chemical Equations Practice Worksheet | PDF - Free Printable

Let’s go through each equation one by one and balance them step by step. Balancing means making sure the number of atoms of each element is the same on both sides of the arrow.

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1. Fe + O₂ → Fe₃O₄

Left: Fe = 1, O = 2
Right: Fe = 3, O = 4

We need 3 Fe on left → put 3 in front of Fe
We need 4 O on left → O₂ has 2 oxygen atoms, so we need 2 O₂ (which gives 4 O)

→ 3Fe + 2O₂ → Fe₃O₄

Check: Left: Fe=3, O=4; Right: Fe=3, O=4 ✔

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2. Sr + O₂ → SrO

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

Need 2 O on right → put 2 in front of SrO → now Sr=2 on right
So put 2 in front of Sr on left

→ 2Sr + O₂ → 2SrO

Check: Left: Sr=2, O=2; Right: Sr=2, O=2 ✔

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3. Sn + NaOH → Na₂SnO₂ + H₂

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

Start with Na: need 2 Na on left → put 2 in front of NaOH → now Na=2, O=2, H=2
Now check H: left has 2H, right has 2H in H₂ → good
O: left has 2O, right has 2O in Na₂SnO₂ → good
Sn: 1 on each side → good

→ Sn + 2NaOH → Na₂SnO₂ + H₂

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

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4. K + Br₂ → KBr

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

Need 2 Br on right → put 2 in front of KBr → now K=2 on right
So put 2 in front of K on left

→ 2K + Br₂ → 2KBr

Check: Left: K=2, Br=2; Right: K=2, Br=2 ✔

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5. C₈H₁₈ + O₂ → CO₂ + H₂O

This is combustion. Start with C and H first.

Left: C=8, H=18, O=?
Right: C=1 (in CO₂), H=2 (in H₂O), O=?

Put 8 in front of CO₂ → C balanced
Put 9 in front of H₂O → H: 9×2=18 → balanced

Now count O on right:
CO₂: 8 × 2 = 16 O
H₂O: 9 × 1 = 9 O
Total O = 25 → so O₂ must provide 25 O → that’s 25/2 = 12.5 molecules → not whole!

Multiply entire equation by 2 to eliminate fraction:

Original after balancing C and H:
C₈H₁₈ + ?O₂ → 8CO₂ + 9H₂O

O needed: 8×2 + 9×1 = 16+9=25 → so O₂ coefficient = 25/2

Multiply all by 2:

→ 2C₈H₁ + 25O₂ → 16CO₂ + 18H₂O

Check:
Left: C=16, H=36, O=50
Right: C=16, H=36, O=32 (from CO₂) + 18 (from H₂O) = 50 ✔

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6. Sb + I₂ → SbI₃

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

Find LCM of 2 and 3 → 6

So need 6 I on both sides → put 3 in front of I₂ (gives 6 I)
Put 2 in front of SbI₃ (gives 6 I and 2 Sb)
Then put 2 in front of Sb on left

→ 2Sb + 3I₂ → 2SbI₃

Check: Left: Sb=2, I=6; Right: Sb=2, I=6 ✔

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7. COCl₂ + H₂O → HCl + CO₂

Left: C=1, O=1 (from COCl₂) +1 (from H₂O)=2? Wait — let's list properly:

COCl₂: C=1, O=1, Cl=2
H₂O: H=2, O=1
Total left: C=1, O=2, Cl=2, H=2

Right: HCl: H=1, Cl=1; CO₂: C=1, O=2

So if we put 2 in front of HCl → H=2, Cl=2 → matches left

→ COCl₂ + H₂O → 2HCl + CO₂

Check: Left: C=1, O=2, Cl=2, H=2; Right: H=2, Cl=2, C=1, O=2 ✔

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8. CS₂ + O₂ → CO₂ + SO₂

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

Need 2 S on right → put 2 in front of SO₂ → now S=2, O from SO₂=4, plus CO₂ has 2 O → total O=6

So O₂ must give 6 O → 3 O₂

→ CS₂ + 3O₂ → CO₂ + 2SO₂

Check: Left: C=1, S=2, O=6; Right: C=1, S=2, O=2+4=6 ✔

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9. H₂SO₄ + NaCN → HCN + Na₂SO₄

Left: H=2, S=1, O=4, Na=1, C=1, N=1
Right: H=1, C=1, N=1, Na=2, S=1, O=4

Na: need 2 on left → put 2 in front of NaCN → now Na=2, C=2, N=2
But right has only 1 HCN → so put 2 in front of HCN → now H=2 on right (good), C=2, N=2

Left H: from H₂SO₄ → 2H → matches right 2H in 2HCN

→ H₂SO₄ + 2NaCN → 2HCN + Na₂SO₄

Check: Left: H=2, S=1, O=4, Na=2, C=2, N=2
Right: H=2, C=2, N=2, Na=2, S=1, O=4 ✔

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10. KClO₃ → KCl + O₂

Left: K=1, Cl=1, O=3
Right: K=1, Cl=1, O=2

Need to balance O. LCM of 3 and 2 is 6.

So put 2 in front of KClO₃ → O=6
Put 3 in front of O₂ → O=6
Then K and Cl: 2 on left → put 2 in front of KCl

→ 2KClO₃ → 2KCl + 3O₂

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

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11. H₂ + F₂ → HF

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

Put 2 in front of HF → H=2, F=2

→ H₂ + F₂ → 2HF

Check: Left: H=2, F=2; Right: H=2, F=2 ✔

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12. BaCl₂ + KIO₃ → Ba(IO₃)₂ + KCl

Left: Ba=1, Cl=2, K=1, I=1, O=3
Right: Ba=1, I=2, O=6, K=1, Cl=1

I and O are off. Right has 2 I and 6 O → so need 2 KIO₃ on left → now K=2, I=2, O=6

Then KCl: need 2 K on right → put 2 in front of KCl → Cl=2 → matches left BaCl₂

→ BaCl₂ + 2KIO₃ → Ba(IO₃)₂ + 2KCl

Check: Left: Ba=1, Cl=2, K=2, I=2, O=6
Right: Ba=1, I=2, O=6, K=2, Cl=2 ✔

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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 Cl and 2 H on left → put 2 in front of HCl

→ Mg + 2HCl → MgCl₂ + H₂

Check: Left: Mg=1, H=2, Cl=2; Right: Mg=1, Cl=2, H=2 ✔

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Final Answer:
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