Balancing Chemical Equations Worksheet featuring 15 problems to practice balancing chemical reactions.

Let's go through each of these chemical equations and balance them step by step. Balancing chemical equations means making sure the number of atoms of each element is the same on both sides of the equation (conservation of mass).

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1) \_\_ Br₂ + \_\_ LiF → \_\_ LiBr + \_\_ F₂

Unbalanced:
Br₂ + LiF → LiBr + F₂

- Left: Br₂ (2 Br), Li, F
- Right: Li, Br, F₂ (2 F)

We need to balance Br and F.

Let’s try:
- 1 Br₂ → 2 Br → so we need 2 LiBr on right
- 2 LiBr → 2 Li → so we need 2 LiF on left
- 2 LiF → 2 F → so we need 1 F₂ on right

✔ Balanced:
1 Br₂ + 2 LiF → 2 LiBr + 1 F₂

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2) \_\_ H₃PO₄ + \_\_ Fe(OH)₃ → \_\_ H₂O + \_\_ Fe(PO₄)₂

Wait — Fe(PO₄)₂? That’s incorrect. Iron(III) phosphate is FePO₄, not Fe(PO₄)₂.

But let's check: Fe³⁺ and PO₄³⁻ → FePO₄ (1:1 ratio). So correct formula is FePO₄.

So corrected equation:
H₃PO₄ + Fe(OH)₃ → H₂O + FePO₄

Now balance:

Left: H₃PO₄ (3H, 1P, 4O), Fe(OH)₃ (1Fe, 3O, 3H)
Right: H₂O, FePO₄ (1Fe, 1P, 4O)

Try:
- 1 Fe(OH)₃ → 1 Fe → needs 1 FePO₄
- 1 FePO₄ → 1 P → needs 1 H₃PO₄
- Now H: Left = 3 (from H₃PO₄) + 3 (from Fe(OH)₃) = 6H
- Right: H₂O → 2H per molecule → need 3 H₂O

Check O:
- Left: 4 (H₃PO₄) + 3 (Fe(OH)₃) = 7 O
- Right: 3 H₂O → 3 O, FePO₄ → 4 O → total 7 O ✔

✔ Balanced:
1 H₃PO₄ + 1 Fe(OH)₃ → 3 H₂O + 1 FePO₄

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3) \_\_ C₂H₅OH + \_\_ O₂ → \_\_ CO₂ + \_\_ H₂O

Ethanol combustion.

C₂H₅OH + O₂ → CO₂ + H₂O

Balance:
- Carbon: 2 on left → 2 CO₂
- Hydrogen: 6H on left → 3 H₂O (since 2H per water)
- Oxygen: Count right side: 2×2 = 4 from CO₂ + 3×1 = 3 from H₂O → 7 O
- Left: 1 O in ethanol + 2 per O₂ → need 3 O₂ → 6 O + 1 O = 7 O

So:
C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O

✔ Balanced:
1 C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O

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4) \_\_ Ni(OH)₂ → \_\_ NiO + \_\_ H₂O

Decomposition.

Ni(OH)₂ → NiO + H₂O

Left: Ni, 2O, 2H
Right: NiO (Ni, O), H₂O (2H, O) → total 2O, 2H, Ni

So:
1 Ni(OH)₂ → 1 NiO + 1 H₂O

✔ Balanced.

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5) \_\_ K₂SO₄ + \_\_ Mn(OH)₂ → \_\_ KOH + \_\_ MnSO₄

Double displacement.

K₂SO₄ + Mn(OH)₂ → KOH + MnSO₄

Balance:
- Mn: 1 each side
- SO₄: 1 each side
- K: 2 on left → need 2 KOH on right
- OH: 2 on left (from Mn(OH)₂) → but 2 KOH has 2 OH → OK

So:
K₂SO₄ + Mn(OH)₂ → 2 KOH + MnSO₄

✔ Balanced:
1 K₂SO₄ + 1 Mn(OH)₂ → 2 KOH + 1 MnSO₄

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6) \_\_ NaOH + \_\_ H₂SO₄ → \_\_ H₂O + \_\_ Na₂SO₄

Acid-base neutralization.

NaOH + H₂SO₄ → H₂O + Na₂SO₄

Balance:
- Na: 2 on right → need 2 NaOH on left
- H: 2 NaOH → 2H, H₂SO₄ → 2H → total 4H → need 2 H₂O
- S: 1 each side
- O: check later

So:
2 NaOH + H₂SO₄ → 2 H₂O + Na₂SO₄

✔ Balanced:
2 NaOH + 1 H₂SO₄ → 2 H₂O + 1 Na₂SO₄

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7) \_\_ Li + \_\_ Pb(OH)₂ → \_\_ Pb + \_\_ LiOH

Single replacement.

Li + Pb(OH)₂ → Pb + LiOH

Balance:
- Pb: 1 each side
- OH: 2 on left → need 2 LiOH on right
- Li: 2 on right → need 2 Li on left

So:
2 Li + Pb(OH)₂ → Pb + 2 LiOH

✔ Balanced:
2 Li + 1 Pb(OH)₂ → 1 Pb + 2 LiOH

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8) \_\_ C₄H₈ + \_\_ O₂ → \_\_ CO₂ + \_\_ H₂O

Combustion of butene.

C₄H₈ + O₂ → CO₂ + H₂O

Balance:
- C: 4 → 4 CO₂
- H: 8 → 4 H₂O
- O: right: 4×2 = 8 from CO₂ + 4×1 = 4 from H₂O → 12 O
- Left: O₂ → need 6 O₂ (12 O)

So:
C₄H₈ + 6 O₂ → 4 CO₂ + 4 H₂O

✔ Balanced:
1 C₄H₈ + 6 O₂ → 4 CO₂ + 4 H₂O

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9) \_\_ Ga(OH)₃ + \_\_ KF → \_\_ KOH + \_\_ GaF₃

Double displacement.

Ga(OH)₃ + KF → KOH + GaF₃

Balance:
- Ga: 1 each side
- F: 3 on right → need 3 KF on left
- K: 3 → 3 KOH on right
- OH: 3 on left → 3 KOH → 3 OH → OK

So:
Ga(OH)₃ + 3 KF → 3 KOH + GaF₃

✔ Balanced:
1 Ga(OH)₃ + 3 KF → 3 KOH + 1 GaF₃

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10) \_\_ V + \_\_ ZnBr₂ → \_\_ VBr₃ + \_\_ Zn

Single replacement.

V + ZnBr₂ → VBr₃ + Zn

Balance:
- V: 1 each side
- Br: 2 on left, 3 on right → LCM = 6
- So: 3 ZnBr₂ → 6 Br → 2 VBr₃ → 6 Br
- Then V: 2 on right → need 2 V on left
- Zn: 3 on left → need 3 Zn on right

So:
2 V + 3 ZnBr₂ → 2 VBr₃ + 3 Zn

✔ Balanced:
2 V + 3 ZnBr₂ → 2 VBr₃ + 3 Zn

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11) \_\_ As₂O₅ + \_\_ H₂O → \_\_ H₃AsO₄

This is a hydration reaction.

As₂O₅ + H₂O → H₃AsO₄

Balance:
- As: 2 on left → 2 H₃AsO₄ on right
- O: left: 5 + 1 = 6; right: 2×4 = 8 → too many
- Wait: H₃AsO₄ has 4 O, so 2 H₃AsO₄ → 8 O
- As₂O₅ has 5 O → need 3 more → need 3 H₂O?

Try:
As₂O₅ + 3 H₂O → 2 H₃AsO₄

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

✔ Balanced:
1 As₂O₅ + 3 H₂O → 2 H₃AsO₄

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12) \_\_ NH₃ + \_\_ O₂ → \_\_ NO + \_\_ H₂O

Ammonia oxidation.

NH₃ + O₂ → NO + H₂O

Balance:
- N: 1 each side
- H: 3 on left → need 3/2 H₂O → multiply by 2
- Try doubling:
2 NH₃ + O₂ → 2 NO + 3 H₂O? Check H: 6 on left, 6 on right ✔
O: right: 2×1 + 3×1 = 5 O → left: O₂ → need 5/2 O₂ → multiply whole equation by 2

So:
4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

Check:
- N: 4 = 4
- H: 12 = 12
- O: 10 on left; 4 + 6 = 10 on right ✔

✔ Balanced:
4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

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13) \_\_ C₄H₈ + \_\_ O₂ → \_\_ CO₂ + \_\_ H₂O

Wait — this is the same as #8? Yes, same compound.

Already balanced above:
1 C₄H₈ + 6 O₂ → 4 CO₂ + 4 H₂O

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14) \_\_ NaClO₃ → \_\_ NaCl + \_\_ O₂

Decomposition of sodium chlorate.

NaClO₃ → NaCl + O₂

Balance:
- Na: 1 each side
- Cl: 1 each side
- O: 3 on left → O₂ → need 3/2 O₂ → multiply by 2

So:
2 NaClO₃ → 2 NaCl + 3 O₂

✔ Balanced:
2 NaClO₃ → 2 NaCl + 3 O₂

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15) \_\_ Ca + \_\_ O₂ → \_\_ CaO

Calcium oxide formation.

Ca + O₂ → CaO

Balance:
- O: 2 on left → need 2 CaO on right
- Ca: 2 → need 2 Ca on left

So:
2 Ca + O₂ → 2 CaO

✔ Balanced:
2 Ca + 1 O₂ → 2 CaO

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✔ Final Balanced Equations:

1) 1 Br₂ + 2 LiF → 2 LiBr + 1 F₂
2) 1 H₃PO₄ + 1 Fe(OH)₃ → 3 H₂O + 1 FePO₄
3) 1 C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O
4) 1 Ni(OH)₂ → 1 NiO + 1 H₂O
5) 1 K₂SO₄ + 1 Mn(OH)₂ → 2 KOH + 1 MnSO₄
6) 2 NaOH + 1 H₂SO₄ → 2 H₂O + 1 Na₂SO₄
7) 2 Li + 1 Pb(OH)₂ → 1 Pb + 2 LiOH
8) 1 C₄H₈ + 6 O₂ → 4 CO₂ + 4 H₂O
9) 1 Ga(OH)₃ + 3 KF → 3 KOH + 1 GaF₃
10) 2 V + 3 ZnBr₂ → 2 VBr₃ + 3 Zn
11) 1 As₂O₅ + 3 H₂O → 2 H₃AsO₄
12) 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O
13) 1 C₄H₈ + 6 O₂ → 4 CO₂ + 4 H₂O
14) 2 NaClO₃ → 2 NaCl + 3 O₂
15) 2 Ca + 1 O₂ → 2 CaO

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Let me know if you'd like these written with coefficients filled in the blanks!