Balancing Equations Worksheet: Practice balancing chemical equations with this educational worksheet.

Let's go through each chemical equation and balance them 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) Br₂ + LiF → LiBr + F₂

Unbalanced:
Br₂ + LiF → LiBr + F₂

Step-by-step:
- Br: 2 on left, 1 on right → need 2 LiBr
- F: 1 on left (from LiF), but 2 on right (in F₂) → so need 2 LiF
- Now: Br₂ + 2LiF → 2LiBr + F₂
- Check:
- Br: 2 = 2 ✔
- Li: 2 = 2 ✔
- F: 2 = 2 ✔

✔ Balanced:
Br₂ + 2LiF → 2LiBr + F₂

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2) H₃PO₄ + Fe(OH)₂ → H₂O + Fe₃(PO₄)₂

Unbalanced:
H₃PO₄ + Fe(OH)₂ → H₂O + Fe₃(PO₄)₂

Step-by-step:
- Fe: 1 on left, 3 on right → need 3 Fe(OH)₂
- PO₄: 1 on left, 2 on right → need 2 H₃PO₄
- Now: 2H₃PO₄ + 3Fe(OH)₂ → H₂O + Fe₃(PO₄)₂
- H: Left: 2×3 + 3×2 = 6 + 6 = 12 H
Right: only in H₂O → need 6 H₂O (since 6×2=12)
- O: Check later, but let’s write:
2H₃PO₄ + 3Fe(OH)₂ → 6H₂O + Fe₃(PO₄)₂
- Check atoms:
- P: 2 = 2 ✔
- Fe: 3 = 3 ✔
- H: Left: 2×3 (from H₃PO₄) + 3×2 (from Fe(OH)₂) = 6 + 6 = 12
Right: 6×2 = 12 ✔
- O: Left: 2×4 (H₃PO₄) + 3×2 (OH) = 8 + 6 = 14
Right: 6×1 (H₂O) + 8 (in PO₄) = 6 + 8 = 14 ✔

✔ Balanced:
2H₃PO₄ + 3Fe(OH)₂ → 6H₂O + Fe₃(PO₄)₂

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3) C₃H₇OH + O₂ → CO₂ + H₂O

This is combustion of propanol (C₃H₈O).

Unbalanced:
C₃H₇OH + O₂ → CO₂ + H₂O

Note: C₃H₇OH is C₃H₈O

Step-by-step:
- C: 3 → need 3 CO₂
- H: 8 → need 4 H₂O (since 4×2=8)
- O: Left: 1 (in alcohol) + 2×O₂
Right: 3×2 = 6 (CO₂) + 4×1 = 4 (H₂O) → total 10 O
- So: O₂ must supply 9 O atoms → need 9/2 O₂ → multiply all by 2 to eliminate fraction.

Try:
C₃H₈O + 9/2 O₂ → 3CO₂ + 4H₂O
Multiply by 2:
2C₃H₈O + 9O₂ → 6CO₂ + 8H₂O

Check:
- C: 6 = 6 ✔
- H: 16 = 16 ✔
- O: Left: 2×1 + 9×2 = 2 + 18 = 20
Right: 6×2 + 8×1 = 12 + 8 = 20 ✔

✔ Balanced:
2C₃H₈O + 9O₂ → 6CO₂ + 8H₂O

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4) Ni(OH)₃ → Ni₂O₃ + H₂O

Unbalanced:
Ni(OH)₃ → Ni₂O₃ + H₂O

Step-by-step:
- Ni: 1 on left, 2 on right → need 2 Ni(OH)₃
- O and H: 2 Ni(OH)₃ has 6 H and 6 O from OH, plus 2 O from Ni? Wait:

Ni(OH)₃ → Ni³⁺ and 3OH⁻
So 2Ni(OH)₃ → 2Ni³⁺ and 6OH⁻

Products: Ni₂O₃ and H₂O

Now:
2Ni(OH)₃ → Ni₂O₃ + 3H₂O

Check:
- Ni: 2 = 2 ✔
- O: Left: 2×3 = 6 (from OH), Right: 3 (in Ni₂O₃) + 3×1 = 3+3=6 ✔
- H: 2×3 = 6 → 3×2 = 6 ✔

✔ Balanced:
2Ni(OH)₃ → Ni₂O₃ + 3H₂O

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5) K₂SO₃ + Mn(OH)₂ → KOH + MnSO₃

Unbalanced:
K₂SO₃ + Mn(OH)₂ → KOH + MnSO₃

Looks like a double displacement.

- K: 2 on left → need 2 KOH
- Mn: 1 = 1 ✔
- SO₃: 1 = 1 ✔
- OH: Left: 2 (from Mn(OH)₂), Right: 2 (from 2KOH) → balanced?

Try:
K₂SO₃ + Mn(OH)₂ → 2KOH + MnSO₃

Check:
- K: 2 = 2 ✔
- S: 1 = 1 ✔
- O: Left: 3 + 2 = 5? Wait:
- K₂SO₃: 3 O
- Mn(OH)₂: 2 O + 2 H → but O count: 2 from OH
Total O: 3 + 2 = 5
Right: 2KOH → 2 O, MnSO₃ → 3 O → total 5 ✔
- H: 2 = 2 ✔

✔ Balanced:
K₂SO₃ + Mn(OH)₂ → 2KOH + MnSO₃

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6) NaOH + H₂SO₄ → H₂O + Na₂SO₄

Acid-base neutralization.

Unbalanced:
NaOH + H₂SO₄ → H₂O + Na₂SO₄

- Na: 1 → 2 → need 2 NaOH
- H: 2NaOH has 2H, H₂SO₄ has 2H → total 4H
- Water: 4H → 2 H₂O
- SO₄: 1 = 1 ✔

Try:
2NaOH + H₂SO₄ → 2H₂O + Na₂SO₄

Check:
- Na: 2 = 2 ✔
- O: Left: 2×1 + 4 = 6; Right: 2×1 + 4 = 6 ✔
- H: 2 + 2 = 4 → 2×2 = 4 ✔
- S: 1 = 1 ✔

✔ Balanced:
2NaOH + H₂SO₄ → 2H₂O + Na₂SO₄

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7) Li + Pb(OH)₂ → Pb + LiOH

Single displacement.

Unbalanced:
Li + Pb(OH)₂ → Pb + LiOH

- Pb: 1 = 1 ✔
- OH: 2 on left → need 2 LiOH
- Li: 2 on right → need 2 Li on left

So:
2Li + Pb(OH)₂ → Pb + 2LiOH

Check:
- Li: 2 = 2 ✔
- Pb: 1 = 1 ✔
- O: 2 = 2 ✔
- H: 2 = 2 ✔

✔ Balanced:
2Li + Pb(OH)₂ → Pb + 2LiOH

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8) C₄H₈ + O₂ → CO₂ + H₂O

Combustion of butene.

Unbalanced:
C₄H₈ + O₂ → CO₂ + H₂O

- C: 4 → need 4 CO₂
- H: 8 → need 4 H₂O
- O: Right: 4×2 + 4×1 = 8 + 4 = 12
Left: O₂ → need 6 O₂

So:
C₄H₈ + 6O₂ → 4CO₂ + 4H₂O

Check:
- C: 4 = 4 ✔
- H: 8 = 8 ✔
- O: 12 = 8 + 4 = 12 ✔

✔ Balanced:
C₄H₈ + 6O₂ → 4CO₂ + 4H₂O

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9) Ga(OH)₃ + KF → KOH + GaF₃

Double displacement.

Unbalanced:
Ga(OH)₃ + KF → KOH + GaF₃

- Ga: 1 = 1 ✔
- F: 1 → 3 → need 3 KF
- K: 3 → need 3 KOH
- OH: 3 → 3 → OK

So:
Ga(OH)₃ + 3KF → 3KOH + GaF₃

Check:
- Ga: 1 = 1 ✔
- F: 3 = 3 ✔
- K: 3 = 3 ✔
- O: 3 = 3 ✔
- H: 3 = 3 ✔

✔ Balanced:
Ga(OH)₃ + 3KF → 3KOH + GaF₃

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10) V + ZnBr₂ → VBr₃ + Zn

Single displacement.

Unbalanced:
V + ZnBr₂ → VBr₃ + Zn

- V: 1 = 1 ✔
- Br: 2 → 3 → need LCM of 2 and 3 → 6
- So: 3ZnBr₂ → 6Br
- 2VBr₃ → 6Br
- So: 2V + 3ZnBr₂ → 2VBr₃ + 3Zn

Check:
- V: 2 = 2 ✔
- Zn: 3 = 3 ✔
- Br: 6 = 6 ✔

✔ Balanced:
2V + 3ZnBr₂ → 2VBr₃ + 3Zn

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11) As₂O₅ + H₂O → H₃AsO₄

Unbalanced:
As₂O₅ + H₂O → H₃AsO₄

- As: 2 → need 2 H₃AsO₄
- O: Left: 5 + 1 = 6; Right: 2×4 = 8 → too many
Wait: H₃AsO₄ has 4 O per molecule → 2×4 = 8 O
Left: As₂O₅ has 5 O, H₂O has 1 → total 6 → not enough

But we need to form 2 H₃AsO₄ → needs 8 O

So: As₂O₅ + 3H₂O → 2H₃AsO₄

Check:
- As: 2 = 2 ✔
- O: 5 + 3 = 8; Right: 2×4 = 8 ✔
- H: 3×2 = 6; Right: 2×3 = 6 ✔

✔ Balanced:
As₂O₅ + 3H₂O → 2H₃AsO₄

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12) NH₃ + O₂ → NO + H₂O

Combustion of ammonia.

Unbalanced:
NH₃ + O₂ → NO + H₂O

- N: 1 = 1 ✔
- H: 3 → need 3/2 H₂O → multiply by 2

Try:
4NH₃ + 5O₂ → 4NO + 6H₂O

Why?
- N: 4 = 4 ✔
- H: 12 → 6×2 = 12 ✔
- O: 5×2 = 10; Right: 4×1 + 6×1 = 4 + 6 = 10 ✔

✔ Balanced:
4NH₃ + 5O₂ → 4NO + 6H₂O

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13) C₃H₈ + O₂ → CO₂ + H₂O

Combustion of propane.

Unbalanced:
C₃H₈ + O₂ → CO₂ + H₂O

- C: 3 → 3 CO₂
- H: 8 → 4 H₂O
- O: Right: 3×2 + 4×1 = 6 + 4 = 10 → need 5 O₂

So:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Check:
- C: 3 = 3 ✔
- H: 8 = 8 ✔
- O: 10 = 6 + 4 = 10 ✔

✔ Balanced:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

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14) NaClO₃ → NaCl + O₂

Decomposition.

Unbalanced:
NaClO₃ → NaCl + O₂

- Na: 1 = 1 ✔
- Cl: 1 = 1 ✔
- O: 3 → 2 → need LCM 6

So: 2NaClO₃ → 2NaCl + 3O₂

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

✔ Balanced:
2NaClO₃ → 2NaCl + 3O₂

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15) Ca + O₂ → CaO

Unbalanced:
Ca + O₂ → CaO

- O: 2 → need 2 CaO
- Ca: 2 → need 2 Ca

So:
2Ca + O₂ → 2CaO

Check:
- Ca: 2 = 2 ✔
- O: 2 = 2 ✔

✔ Balanced:
2Ca + O₂ → 2CaO

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✔ Final Answers:

1) Br₂ + 2LiF → 2LiBr + F₂
2) 2H₃PO₄ + 3Fe(OH)₂ → 6H₂O + Fe₃(PO₄)₂
3) 2C₃H₈O + 9O₂ → 6CO₂ + 8H₂O
4) 2Ni(OH)₃ → Ni₂O₃ + 3H₂O
5) K₂SO₃ + Mn(OH)₂ → 2KOH + MnSO₃
6) 2NaOH + H₂SO₄ → 2H₂O + Na₂SO₄
7) 2Li + Pb(OH)₂ → Pb + 2LiOH
8) C₄H₈ + 6O₂ → 4CO₂ + 4H₂O
9) Ga(OH)₃ + 3KF → 3KOH + GaF₃
10) 2V + 3ZnBr₂ → 2VBr₃ + 3Zn
11) As₂O₅ + 3H₂O → 2H₃AsO₄
12) 4NH₃ + 5O₂ → 4NO + 6H₂O
13) C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
14) 2NaClO₃ → 2NaCl + 3O₂
15) 2Ca + O₂ → 2CaO

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