Balancing Equations Worksheet for chemistry practice, featuring 15 equations to balance with blank spaces for coefficients.

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

We’ll write coefficients (numbers in front) to make that happen. We never change the subscripts — only the numbers in front!

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1) Al(NO₃)₃ + (NH₄)₃PO₄ → AlPO₄ + NHNO₃

Left:
Al = 1, N = 3 (from NO₃) + 3 (from NH₄) = 6? Wait — let’s break it down properly.

Actually, better to count per compound:

Al(NO₃)₃ has: 1 Al, 3 N, 9 O
(NH₄)₃PO₄ has: 3 N, 12 H, 1 P, 4 O
→ Total left: Al=1, N=6, H=12, P=1, O=13

Right:
AlPO₄: 1 Al, 1 P, 4 O
NH₄NO₃: 2 N, 4 H, 3 O → but we need more than one!

Try putting a 3 in front of NH₄NO₃:

Right: AlPO₄ + 3 NH₄NO₃
→ Al=1, P=1, O=4 + 9 = 13, N=6, H=12 → matches left!

So:
1 Al(NO₃)₃ + 1 (NH₄)₃PO₄ → 1 AlPO₄ + 3 NH₄NO₃

✔ Balanced.

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2) AgF + CaCl₂ → AgCl + CaF₂

Left: Ag=1, F=1, Ca=1, Cl=2
Right: Ag=1, Cl=1, Ca=1, F=2 → not balanced.

Need 2 AgF to get 2 F for CaF₂ → then 2 Ag on left → so 2 AgCl on right.

Then Cl: 2 on left (from CaCl₂), 2 on right (from 2 AgCl) → good.

Ca: 1 on each side.

So:
2 AgF + 1 CaCl₂ → 2 AgCl + 1 CaF₂

✔ Balanced.

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3) ZnBr₂ + Pb(NO₂)₂ → Zn(NO₂)₂ + PbBr₂

Check atoms:

Left: Zn=1, Br=2, Pb=1, N=2, O=4
Right: Zn=1, N=2, O=4, Pb=1, Br=2 → already balanced!

So:
1 ZnBr₂ + 1 Pb(NO₂)₂ → 1 Zn(NO₂)₂ + 1 PbBr₂

✔ Balanced.

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4) C₂H₄O₂ + O₂ → CO₂ + H₂O

This is combustion. Let’s balance carbon first.

C₂H₄O₂ has 2 C → so 2 CO₂ on right.

Has 4 H → so 2 H₂O on right (since each has 2 H).

Now oxygen:

Left: from C₂H₄O₂ → 2 O; from O₂ → ?
Right: 2 CO₂ → 4 O; 2 H₂O → 2 O → total 6 O

So left needs 6 O total → minus 2 from acid → need 4 O from O₂ → so 2 O₂ molecules.

Check:

Left: C₂H₄O₂ + 2 O₂ → C=2, H=4, O=2+4=6
Right: 2 CO₂ + 2 H₂O → C=2, H=4, O=4+2=6 → perfect.

So:
1 C₂H₄O₂ + 2 O₂ → 2 CO₂ + 2 H₂O

✔ Balanced.

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5) Ca + CuF₂ → CaF₂ + Cu

Left: Ca=1, Cu=1, F=2
Right: Ca=1, F=2, Cu=1 → already balanced!

So:
1 Ca + 1 CuF₂ → 1 CaF₂ + 1 Cu

✔ Balanced.

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6) H₂SO₄ + B(OH)₃ → B₂(SO₄)₃ + H₂O

Left: H=2+3=5? Wait — H₂SO₄ has 2 H, B(OH)₃ has 3 H → but we need to balance properly.

Better to think: B₂(SO₄)₃ has 2 B and 3 SO₄ → so need 3 H₂SO₄ and 2 B(OH)₃.

Try:

3 H₂SO₄ + 2 B(OH)₃ → 1 B₂(SO₄)₃ + ? H₂O

Left: H = 3×2 + 2×3 = 6 + 6 = 12 H
O = 3×4 + 2×3 = 12 + 6 = 18 O? Wait — let's count all.

Actually:

3 H₂SO₄: H=6, S=3, O=12
2 B(OH)₃: B=2, O=6, H=6 → total H=12, O=18, S=3, B=2

Right: B₂(SO₄)₃: B=2, S=3, O=12
Water: need to account for remaining H and O.

H left: 12 → so 6 H₂O (each has 2 H → 12 H)
O in water: 6 O → plus 12 O from sulfate → total 18 O → matches.

So:
3 H₂SO₄ + 2 B(OH)₃ → 1 B₂(SO₄)₃ + 6 H₂O

✔ Balanced.

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7) S₈ + O₂ → SO₂

S₈ has 8 S → so need 8 SO₂ on right.

Each SO₂ has 2 O → 8 × 2 = 16 O → so 8 O₂ molecules (since each O₂ has 2 O).

So:
1 S₈ + 8 O₂ → 8 SO₂

✔ Balanced.

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8) H₂O₂ → O₂ + H₂O

Left: H=2, O=2
Right: O₂ has 2 O, H₂O has 2 H and 1 O → total O=3, H=2 → not balanced.

Try 2 H₂O₂ → 4 H, 4 O

Right: if 1 O₂ (2 O) and 2 H₂O (4 H, 2 O) → total O=4, H=4 → perfect.

So:
2 H₂O₂ → 1 O₂ + 2 H₂O

✔ Balanced.

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9) K + F₂ → KF

Left: K=1, F=2
Right: K=1, F=1 → need 2 KF → then 2 K on left.

So:
2 K + 1 F₂ → 2 KF

✔ Balanced.

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10) AgNO₃ + Ga → Ag + Ga(NO₃)₃

Ga(NO₃)₃ has 3 NO₃ → so need 3 AgNO₃ on left.

Then 3 Ag on right.

Ga: 1 on each side.

So:
3 AgNO₃ + 1 Ga → 3 Ag + 1 Ga(NO₃)₃

✔ Balanced.

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11) N₂ + H₂ → NH₃

N₂ has 2 N → so 2 NH₃ on right → which has 6 H → so 3 H₂ on left.

So:
1 N₂ + 3 H₂ → 2 NH₃

✔ Balanced.

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12) NaCl + H₂SO₄ → NaHSO₄ + HCl

Left: Na=1, Cl=1, H=2, S=1, O=4
Right: Na=1, H=1+1=2, S=1, O=4, Cl=1 → already balanced!

So:
1 NaCl + 1 H₂SO₄ → 1 NaHSO₄ + 1 HCl

✔ Balanced.

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13) KMnO₄ → K₂MnO₄ + MnO₂ + O₂

Left: K=1, Mn=1, O=4
Right: K₂MnO₄ has K=2, Mn=1, O=4; MnO₂ has Mn=1, O=2; O₂ has O=2 → total K=2, Mn=2, O=8

So need 2 KMnO₄ on left → K=2, Mn=2, O=8

Right: 1 K₂MnO₄ (K=2, Mn=1, O=4) + 1 MnO₂ (Mn=1, O=2) + ? O₂

O so far: 4 + 2 = 6 → need 2 more O → so 1 O₂.

Total O: 4+2+2=8 → yes.

Mn: 1+1=2 → yes.

K: 2 → yes.

So:
2 KMnO₄ → 1 K₂MnO₄ + 1 MnO₂ + 1 O₂

✔ Balanced.

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14) Fe + AgNO₃ → Fe(NO₃)₂ + Ag

Fe(NO₃)₂ has 2 NO₃ → so need 2 AgNO₃ on left.

Then 2 Ag on right.

Fe: 1 on each side.

So:
1 Fe + 2 AgNO₃ → 1 Fe(NO₃)₂ + 2 Ag

✔ Balanced.

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15) Li₂SO₄ + K₃PO₄ → Li₃PO₄ + K₂SO₄

Left: Li=2, S=1, O=4, K=3, P=1
Right: Li=3, P=1, O=4, K=2, S=1 → not balanced.

Need common multiples.

Li: 2 vs 3 → LCM 6 → so 3 Li₂SO₄ and 2 Li₃PO₄
K: 3 vs 2 → LCM 6 → so 2 K₃PO₄ and 3 K₂SO₄

Try:

3 Li₂SO₄ + 2 K₃PO₄ → 2 Li₃PO₄ + 3 K₂SO₄

Check:

Left: Li=6, S=3, O=12 + 8? Wait — better:

3 Li₂SO₄: Li=6, S=3, O=12
2 K₃PO₄: K=6, P=2, O=8 → total O=20? No — wait, PO₄ has 4 O → 2×4=8 O → total O=12+8=20

Right: 2 Li₃PO₄: Li=6, P=2, O=8
3 K₂SO₄: K=6, S=3, O=12 → total O=8+12=20 → yes.

All match.

So:
3 Li₂SO₄ + 2 K₃PO₄ → 2 Li₃PO₄ + 3 K₂SO₄

✔ Balanced.

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Final Answer:

1) 1, 1, 1, 3
2) 2, 1, 2, 1
3) 1, 1, 1, 1
4) 1, 2, 2, 2
5) 1, 1, 1, 1
6) 3, 2, 1, 6
7) 1, 8, 8
8) 2, 1, 2
9) 2, 1, 2
10) 3, 1, 3, 1
11) 1, 3, 2
12) 1, 1, 1, 1
13) 2, 1, 1, 1
14) 1, 2, 1, 2
15) 3, 2, 2, 3