Balancing Chemical Equations Worksheet - Free Printable

Let's solve each of these balanced chemical equations step by step. I'll go through them one at a time, explaining the balancing process and providing the correct coefficients.

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1) C₃H₈ + O₂ → CO₂ + H₂O

Unbalanced:
C₃H₈ + O₂ → CO₂ + H₂O

- Carbon: 3 on left → need 3 CO₂
- Hydrogen: 8 on left → need 4 H₂O (since each has 2 H)
- Oxygen: Right side: 3×2 = 6 from CO₂ + 4×1 = 4 from H₂O → total 10 O atoms → need 5 O₂ molecules

✔ Balanced:
1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

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2) Al + Fe₃N₂ → AlN + Fe

Unbalanced:
Al + Fe₃N₂ → AlN + Fe

- Fe₃N₂ has 3 Fe and 2 N
- So we need 2 AlN to balance N → 2 Al needed
- Then 3 Fe on right → so 3 Fe produced

✔ Balanced:
2 Al + Fe₃N₂ → 2 AlN + 3 Fe

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3) Na + Cl₂ → NaCl

Unbalanced:
Na + Cl₂ → NaCl

- Cl₂ has 2 Cl atoms → need 2 NaCl → need 2 Na

✔ Balanced:
2 Na + 1 Cl₂ → 2 NaCl

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4) H₂O → H₂O + O₂

Wait — this looks odd. Water decomposing?

Actually, it’s likely meant to be electrolysis of water:

H₂O → H₂ + O₂

But written as H₂O → H₂O + O₂? That doesn't make sense.

Wait — probably a typo. Likely:
H₂O → H₂ + O₂

But let's assume the equation is:
H₂O → H₂ + O₂

Balance:
- Left: 2 H, 1 O
- Right: 2 H in H₂, but O₂ has 2 O → need 2 H₂O

So:
2 H₂O → 2 H₂ + O₂

✔ Balanced:
2 H₂O → 2 H₂ + 1 O₂

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5) C₆H₁₂O₆ + O₂ → H₂O + CO₂

This is glucose combustion.

- C₆H₁₂O₆ → 6 C → 6 CO₂
- 12 H → 6 H₂O
- Now count O:
- Left: 6 (from glucose) + 2×x (from O₂)
- Right: 6×2 = 12 (CO₂) + 6×1 = 6 (H₂O) → 18 O
- So: 6 + 2x = 18 → x = 6

✔ Balanced:
1 C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O

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6) H₂O + CO₂ → C₇H₈ + O₂

Wait — this is not balanced chemically.
Left: H₂O + CO₂ → elements: H, O, C
Right: C₇H₈ (toluene) + O₂

But C₇H₈ has 7 C and 8 H → no way to get that from just H₂O and CO₂.

Possibly reversed? Maybe photosynthesis or something?

But actually, this seems like a synthesis of toluene — unlikely.

Wait — maybe it's supposed to be C₇H₈ → H₂O + CO₂? But that would be combustion.

But the arrow goes left to right: H₂O + CO₂ → C₇H₈ + O₂

That implies forming hydrocarbon from CO₂ and H₂O, which requires energy (like photosynthesis), but stoichiometrically:

Let’s try balancing:

Assume:
a H₂O + b CO₂ → c C₇H₈ + d O₂

Atoms:
- C: b = 7c
- H: 2a = 8c → a = 4c
- O: a + 2b = 2d

Substitute:
- a = 4c
- b = 7c
- O: 4c + 2(7c) = 4c + 14c = 18c = 2d → d = 9c

Let c = 1 → a=4, b=7, d=9

✔ Balanced:
4 H₂O + 7 CO₂ → 1 C₇H₈ + 9 O₂

But wait — this produces more O₂ than consumed? That can’t happen without energy input.

But mathematically it balances.

✔ So: 4 H₂O + 7 CO₂ → C₇H₈ + 9 O₂

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7) NaClO₃ → NaCl + O₂

Decomposition of sodium chlorate.

Unbalanced: NaClO₃ → NaCl + O₂

- Na: 1 both sides
- Cl: 1 both sides
- O: 3 on left → 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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8) (NH₄)₃PO₄ + Pb(NO₃)₄ → Pb₃(PO₄)₂ + NH₄NO₃

Double displacement.

- Pb₃(PO₄)₂ has 3 Pb and 2 PO₄
- So need 3 Pb(NO₃)₄ and 2 (NH₄)₃PO₄

Now:
- Left: 2 × (NH₄)₃PO₄ → 6 NH₄⁺
- 3 × Pb(NO₃)₄ → 12 NO₃⁻
- Right: Pb₃(PO₄)₂ + ? NH₄NO₃

We have 6 NH₄⁺ and 12 NO₃⁻ → so 6 NH₄NO₃

✔ Balanced:
2 (NH₄)₃PO₄ + 3 Pb(NO₃)₄ → Pb₃(PO₄)₂ + 6 NH₄NO₃

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9) BF₃ + Li₂SO₃ → B₂(SO₃)₃ + LiF

- B₂(SO₃)₃ has 2 B and 3 SO₃
- So need 2 BF₃ and 3 Li₂SO₃
- Then Li: 3×2 = 6 Li → need 6 LiF

✔ Balanced:
2 BF₃ + 3 Li₂SO₃ → B₂(SO₃)₃ + 6 LiF

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10) C₇H₁₇ + O₂ → CO₂ + H₂O

Combustion of heptane.

- C₇H₁₇ → 7 C → 7 CO₂
- 17 H → 17/2 H₂O → so 17 H₂O
- O: Right: 7×2 = 14 + 17×1 = 17 → total 31 O atoms
- Left: O₂ → 31/2 = 15.5 → so multiply all by 2

Multiply:
- 2 C₇H₁₇ → 14 CO₂ + 17 H₂O
- O: 14×2 = 28 + 17 = 45 O → need 45/2 = 22.5 O₂ → ×2 again

Better:
Start over:

C₇H₁₇ + O₂ → 7 CO₂ + 8.5 H₂O → ×2

2 C₇H₁₇ + 33 O₂ → 14 CO₂ + 17 H₂O

Check O: 33×2 = 66 O
Right: 14×2 = 28 + 17×1 = 17 → 45 → not matching!

Wait — C₇H₁₇ is heptyl radical, but usually heptane is C₇H₁₆

Possibly typo — should be C₇H₁₆

Try C₇H₁₆:

C₇H₁₆ + O₂ → 7 CO₂ + 8 H₂O

O: Right: 14 + 8 = 22 → need 11 O₂

✔ Balanced:
1 C₇H₁₆ + 11 O₂ → 7 CO₂ + 8 H₂O

But given is C₇H₁₇ — maybe it's an alkyl group?

Alternatively, accept C₇H₁₇ as is.

C₇H₁₇ → 7 C → 7 CO₂
17 H → 17/2 H₂O → 8.5 H₂O

O: Right: 7×2 + 8.5×1 = 14 + 8.5 = 22.5 → need 11.25 O₂ → ×4

So:
4 C₇H₁₇ + 45 O₂ → 28 CO₂ + 34 H₂O

Check:
- C: 4×7 = 28 → OK
- H: 4×17 = 68 → 34 H₂O → 68 H → OK
- O: 45×2 = 90 → right: 28×2 = 56 + 34×1 = 34 → 90 → OK

✔ Balanced:
4 C₇H₁₇ + 45 O₂ → 28 CO₂ + 34 H₂O

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11) CaCO₃ + H₃PO₄ → Ca₃(PO₄)₂ + H₂CO₃

Double displacement.

Ca₃(PO₄)₂ needs 3 Ca and 2 PO₄

So:
- 3 CaCO₃
- 2 H₃PO₄
- Products: Ca₃(PO₄)₂ + 3 H₂CO₃

Check H: 2×3 = 6 H → 3 H₂CO₃ → 6 H → OK

✔ Balanced:
3 CaCO₃ + 2 H₃PO₄ → Ca₃(PO₄)₂ + 3 H₂CO₃

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12) Ag₂S → Ag + S₈

Decomposition.

Ag₂S → 2 Ag + 1/8 S₈ → ×8

8 Ag₂S → 16 Ag + S₈

✔ Balanced:
8 Ag₂S → 16 Ag + 1 S₈

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13) KBr + Fe(OH)₃ → KOH + FeBr₃

Single replacement? But Br⁻ and OH⁻?

Fe(OH)₃ is insoluble — may not react.

But assuming double displacement:

Fe(OH)₃ + 3 KBr → 3 KOH + FeBr₃

Check:
- Fe: 1
- O and H: 3 OH → 3 KOH → OK
- K: 3
- Br: 3

✔ Balanced:
3 KBr + Fe(OH)₃ → 3 KOH + FeBr₃

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14) KNO₃ + H₂CO₃ → K₂CO₃ + HNO₃

Acid-base reaction.

K₂CO₃ needs 2 K → so 2 KNO₃

Then H₂CO₃ provides H⁺ and CO₃²⁻

So:
2 KNO₃ + H₂CO₃ → K₂CO₃ + 2 HNO₃

✔ Balanced:
2 KNO₃ + 1 H₂CO₃ → 1 K₂CO₃ + 2 HNO₃

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15) Pb(OH)₄ + Cu₂O → PbO₂ + CuOH

Redox? Or decomposition?

Pb(OH)₄ → PbO₂ + 2 H₂O

But here: Pb(OH)₄ + Cu₂O → PbO₂ + CuOH

Cu₂O → 2 Cu⁺ → possibly oxidized to Cu²⁺?

But CuOH is unstable — might be Cu(OH)₂

Assume it's CuOH.

Try balancing:

Pb(OH)₄ → PbO₂ + 2 H₂O

But here: Pb(OH)₄ + Cu₂O → PbO₂ + 2 CuOH

Check atoms:

Left: Pb, 4 O, 4 H, 2 Cu, 1 O → total O: 5, H: 4

Right: PbO₂ + 2 CuOH → Pb, 2 O + 2 O = 4 O, 2 H → only 2 H → not balanced

Try: Pb(OH)₄ + Cu₂O → PbO₂ + 2 CuOH + H₂O

Now H: 4 → 2 + 2 = 4 → OK

O: Left: 4 + 1 = 5; Right: 2 + 2 + 1 = 5 → OK

So:

✔ Balanced:
1 Pb(OH)₄ + 1 Cu₂O → 1 PbO₂ + 2 CuOH + 1 H₂O

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16) Cr(NO₂)₂ + (NH₄)₂SO₄ → CrSO₄ + NH₄NO₂

Double displacement.

CrSO₄ → Cr²⁺ and SO₄²⁻

So need 1 Cr(NO₂)₂ and 1 (NH₄)₂SO₄ → CrSO₄ + 2 NH₄NO₂

Yes:

✔ Balanced:
1 Cr(NO₂)₂ + 1 (NH₄)₂SO₄ → 1 CrSO₄ + 2 NH₄NO₂

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17) KOH + Co₃(PO₄)₂ → K₃PO₄ + Co(OH)₂

Co₃(PO₄)₂ → 3 Co²⁺, 2 PO₄³⁻

So need 3 Co(OH)₂ and 2 K₃PO₄

K: 2×3 = 6 → need 6 KOH

So:

6 KOH + Co₃(PO₄)₂ → 2 K₃PO₄ + 3 Co(OH)₂

Check:
- K: 6 → 6 → OK
- O and H: 6 H₂O → 3 Co(OH)₂ → 6 OH → 6 H → OK
- PO₄: 2 → 2 → OK
- Co: 3 → 3 → OK

✔ Balanced:
6 KOH + 1 Co₃(PO₄)₂ → 2 K₃PO₄ + 3 Co(OH)₂

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18) Sn(NO₃)₄ + Pt₃N₄ → Sn₃N₄ + Pt(NO₃)₄

Looks like a double displacement.

Sn₃N₄ → 3 Sn, 4 N

Pt₃N₄ → 3 Pt, 4 N

So need 3 Sn(NO₃)₄ and 1 Pt₃N₄ → Sn₃N₄ + 3 Pt(NO₃)₄

Check:
- Sn: 3 → 3
- N: 3×4 = 12 from Sn(NO₃)₄ + 4 from Pt₃N₄ → total N: 16?
Wait — Pt₃N₄ has 4 N atoms → 4 N
Sn₃N₄ has 4 N → so N must come from both

But NO₃⁻ has N → so:

Left:
- 3 Sn(NO₃)₄ → 3 Sn, 12 NO₃⁻ → 12 N
- Pt₃N₄ → 3 Pt, 4 N

Total N: 12 + 4 = 16

Right:
- Sn₃N₄ → 4 N
- 3 Pt(NO₃)₄ → 3×4 = 12 N → total 16 → OK

Other atoms:
- Sn: 3 → 3
- Pt: 3 → 3
- O: 3×12 = 36 from NO₃ → 3×12 = 36 → OK

✔ Balanced:
3 Sn(NO₃)₄ + 1 Pt₃N₄ → 1 Sn₃N₄ + 3 Pt(NO₃)₄

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19) B₂Br₆ + HNO₃ → B(NO₃)₃ + HBr

B₂Br₆ → 2 B, 6 Br

Need 2 B(NO₃)₃ and 6 HBr

H: 6 H → need 6 HNO₃

So:
B₂Br₆ + 6 HNO₃ → 2 B(NO₃)₃ + 6 HBr

Check:
- B: 2 → 2
- Br: 6 → 6
- H: 6 → 6
- N: 6 → 6
- O: 6×3 = 18 → 2×3×3 = 18 → OK

✔ Balanced:
1 B₂Br₆ + 6 HNO₃ → 2 B(NO₃)₃ + 6 HBr

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20) ZnS + AlP → Zn₃P₂ + Al₂S₃

Double displacement.

Zn₃P₂ → 3 Zn, 2 P
Al₂S₃ → 2 Al, 3 S

So need:
- 3 ZnS → 3 Zn, 3 S
- 2 AlP → 2 Al, 2 P

But we need 2 P and 3 S → so:

3 ZnS + 2 AlP → Zn₃P₂ + Al₂S₃

Check:
- Zn: 3 → 3
- S: 3 → 3
- Al: 2 → 2
- P: 2 → 2

✔ Balanced:
3 ZnS + 2 AlP → 1 Zn₃P₂ + 1 Al₂S₃

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## ✔ Final Answers:

1) 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
2) 2 Al + 1 Fe₃N₂ → 2 AlN + 3 Fe
3) 2 Na + 1 Cl₂ → 2 NaCl
4) 2 H₂O → 2 H₂ + 1 O₂
5) 1 C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O
6) 4 H₂O + 7 CO₂ → 1 C₇H₈ + 9 O₂
7) 2 NaClO₃ → 2 NaCl + 3 O₂
8) 2 (NH₄)₃PO₄ + 3 Pb(NO₃)₄ → Pb₃(PO₄)₂ + 6 NH₄NO₃
9) 2 BF₃ + 3 Li₂SO₃ → B₂(SO₃)₃ + 6 LiF
10) 4 C₇H₁₇ + 45 O₂ → 28 CO₂ + 34 H₂O
11) 3 CaCO₃ + 2 H₃PO₄ → Ca₃(PO₄)₂ + 3 H₂CO₃
12) 8 Ag₂S → 16 Ag + 1 S₈
13) 3 KBr + Fe(OH)₃ → 3 KOH + FeBr₃
14) 2 KNO₃ + H₂CO₃ → K₂CO₃ + 2 HNO₃
15) 1 Pb(OH)₄ + 1 Cu₂O → 1 PbO₂ + 2 CuOH + 1 H₂O
16) 1 Cr(NO₂)₂ + 1 (NH₄)₂SO₄ → 1 CrSO₄ + 2 NH₄NO₂
17) 6 KOH + Co₃(PO₄)₂ → 2 K₃PO₄ + 3 Co(OH)₂
18) 3 Sn(NO₃)₄ + 1 Pt₃N₄ → 1 Sn₃N₄ + 3 Pt(NO₃)₄
19) 1 B₂Br₆ + 6 HNO₃ → 2 B(NO₃)₃ + 6 HBr
20) 3 ZnS + 2 AlP → 1 Zn₃P₂ + 1 Al₂S₃

Let me know if you'd like these formatted for printing or with explanations!