49 Balancing Chemical Equations Worksheets [with Answers] - Free Printable

Let's solve each of these chemical equations by balancing them. Balancing a chemical equation means ensuring the number of atoms of each element is equal on both sides (reactants and products), following the Law of Conservation of Mass.

We’ll go through each one step-by-step and provide the balanced coefficients.

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1) C₃H₈ + O₂ → CO₂ + H₂O

This is combustion of propane.

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

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

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2) Al + Fe₃N₂ → AlN + Fe

Single displacement.

- Fe₃N₂ has 3 Fe, 2 N
- Need 2 AlN to balance N → so 2 Al
- Then 3 Fe produced

✔ Balanced:
2 Al + 1 Fe₃N₂ → 2 AlN + 3 Fe

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3) Na + Cl₂ → NaCl

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

✔ Balanced:
2 Na + 1 Cl₂ → 2 NaCl

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4) H₂O₂ → H₂O + O₂

Decomposition of hydrogen peroxide.

- Left: 2 H, 2 O
- Right: H₂O has 2 H, 1 O; O₂ has 2 O → total 2 H, 3 O → unbalanced

Try 2 H₂O₂ → 2 H₂O + 1 O₂

Left: 4 H, 4 O
Right: 4 H, 2 O (from water) + 2 O (from O₂) = 4 O → balanced

✔ Balanced:
2 H₂O₂ → 2 H₂O + 1 O₂

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5) C₆H₁₂O₆ + O₂ → H₂O + CO₂

Combustion of glucose.

- C: 6 → 6 CO₂
- H: 12 → 6 H₂O
- O: Left: 6 (glucose) + 2×O₂ → Right: 6×1 (H₂O) + 6×2 = 12 O → total 18 O on right
- So O₂ must supply 12 O → 6 O₂

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

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6) H₂O + CO₂ → C₇H₈ + O₂

This is not a typical reaction — seems like photosynthesis or something artificial. But let’s balance it as written.

Wait: Left: H₂O + CO₂ → elements: H, O, C
Right: C₇H₈ (toluene) + O₂

C: 7 → need 7 CO₂
H: 8 → need 4 H₂O
But now O: left = 4×1 (from H₂O) + 7×2 = 4 + 14 = 18 O
Right: O₂ only → so 9 O₂

But we have C₇H₈ and O₂ on right. Let’s try:

Try:
4 H₂O + 7 CO₂ → 1 C₇H₈ + 9 O₂

Check:
- C: 7 = 7 ✔
- H: 8 = 8 ✔
- O: 4 + 14 = 18 → 9×2 = 18 ✔

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

> Note: This is not a real reaction under normal conditions, but mathematically balanced.

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7) NaClO₃ → NaCl + O₂

Decomposition.

- Na: 1 → 1 NaCl
- Cl: 1 → 1 NaCl
- O: 3 → O₂ needs even number → use 2 NaClO₃ → 2 NaCl + 3 O₂?

Try:
2 NaClO₃ → 2 NaCl + 3 O₂

Left: 2 Na, 2 Cl, 6 O
Right: 2 Na, 2 Cl, 6 O → ✔

✔ Balanced:
2 NaClO₃ → 2 NaCl + 3 O₂

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8) (NH₄)₃PO₄ + Pb(NO₃)₄ → Pb₃(PO₄)₄ + NH₄NO₃

Double displacement.

Note: Pb₃(PO₄)₄ is incorrect — should be Pb₃(PO₄)₂ (because Pb²⁺ and PO₄³⁻). But assuming the formula is correct as given, proceed.

Wait: Pb(NO₃)₄ implies Pb⁴⁺ → so Pb⁴⁺ and PO₄³⁻ → LCM of charges: Pb₃(PO₄)₄ makes sense.

So:
- Pb: 3 in product → need 3 Pb(NO₃)₄
- PO₄: 4 in product → need 4 (NH₄)₃PO₄
- Now NH₄: 4×3 = 12 → need 12 NH₄NO₃
- NO₃: from 3 Pb(NO₃)₄ → 12 NO₃ → matches 12 NH₄NO₃

✔ Balanced:
4 (NH₄)₃PO₄ + 3 Pb(NO₃)₄ → 1 Pb₃(PO₄)₄ + 12 NH₄NO₃

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9) BF₃ + Li₂SO₃ → B₂(SO₃)₃ + LiF

- B: 2 in product → need 2 BF₃
- SO₃: 3 in product → need 3 Li₂SO₃
- Li: 3×2 = 6 → need 6 LiF
- F: 2×3 = 6 → 6 LiF → good

✔ Balanced:
2 BF₃ + 3 Li₂SO₃ → 1 B₂(SO₃)₃ + 6 LiF

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10) C₇H₁₇ + O₂ → CO₂ + H₂O

Combustion of heptane.

- C: 7 → 7 CO₂
- H: 17 → need 8.5 H₂O → multiply whole equation by 2

Start with:
C₇H₁₇ + O₂ → 7 CO₂ + 8.5 H₂O → ×2

→ 2 C₇H₁₇ + O₂ → 14 CO₂ + 17 H₂O

Now O: right = 14×2 + 17 = 28 + 17 = 45 O → need 45/2 = 22.5 O₂ → ×2 again

→ 4 C₇H₁₇ + 45 O₂ → 28 CO₂ + 34 H₂O

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

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

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11) CaCO₃ + H₃PO₄ → Ca₃(PO₄)₂ + H₂CO₃

Acid-base reaction.

- Ca₃(PO₄)₂ → 3 Ca → need 3 CaCO₃
- PO₄: 2 → need 2 H₃PO₄
- H: 2×3 = 6 H → H₂CO₃ has 2 H → need 3 H₂CO₃
- C: 3 → 3 H₂CO₃ → good

✔ Balanced:
3 CaCO₃ + 2 H₃PO₄ → 1 Ca₃(PO₄)₂ + 3 H₂CO₃

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12) Ag₂S → Ag + S₈

Decomposition.

- S₈ has 8 S → need 8 Ag₂S → 16 Ag
- Ag: 16 → 16 Ag

✔ Balanced:
8 Ag₂S → 16 Ag + 1 S₈

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13) KBr + Fe(OH)₃ → KOH + FeBr₃

Double displacement.

Fe(OH)₃ → Fe³⁺, OH⁻
FeBr₃ → Fe³⁺, Br⁻

So:
- Fe: 1 → 1 FeBr₃
- Br: 3 → need 3 KBr
- K: 3 → 3 KOH
- OH: 3 → 3 KOH

✔ Balanced:
3 KBr + 1 Fe(OH)₃ → 3 KOH + 1 FeBr₃

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14) KNO₃ + H₂CO₃ → K₂CO₃ + HNO₃

Acid-base.

K₂CO₃ → 2 K → need 2 KNO₃
H₂CO₃ → 2 H → 2 HNO₃

So:
2 KNO₃ + 1 H₂CO₃ → 1 K₂CO₃ + 2 HNO₃

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

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15) Pb(OH)₄ + Cu₂O → PbO₂ + CuOH

Redox? Let’s check.

Pb(OH)₄ → PbO₂ + 2 H₂O? But here we have Cu₂O → 2 CuOH

Assume:
- Pb: 1 → 1 PbO₂
- Cu: 2 → 2 CuOH
- O and H: balance

Try:
1 Pb(OH)₄ + 1 Cu₂O → 1 PbO₂ + 2 CuOH

Left: Pb, 4 O, 4 H, 2 Cu, 1 O → total O: 5, H: 4
Right: PbO₂ → Pb, 2 O; 2 CuOH → 2 Cu, 2 O, 2 H → total: Pb, 2 Cu, 4 O, 2 H → mismatch

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

So combine:
Pb(OH)₄ + Cu₂O → PbO₂ + 2 CuOH

Left: Pb, 4 O, 4 H, 2 Cu, 1 O → total O: 5, H: 4
Right: PbO₂ (2 O), 2 CuOH (2 O, 2 H) → total O: 4, H: 2 → no

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

Now check:
Left: Pb, 4 O, 4 H, 2 Cu, 1 O → total O: 5, H: 4
Right: PbO₂ (2 O), 2 CuOH (2 O, 2 H), H₂O (1 O, 2 H) → total O: 5, H: 4 → ✔

But this adds extra H₂O. Is that allowed? Only if it's part of reaction.

But original equation doesn't include H₂O.

Alternatively, maybe the reaction is not feasible. But let's assume stoichiometry.

Wait — perhaps:
Pb(OH)₄ → PbO₂ + 2 H₂O
Cu₂O + H₂O → 2 CuOH
So overall: Pb(OH)₄ + Cu₂O → PbO₂ + 2 CuOH + H₂O

But since H₂O isn’t in original, maybe not.

Alternatively, suppose we want no extra species.

Try balancing as written:

Assume:
a Pb(OH)₄ + b Cu₂O → c PbO₂ + d CuOH

Pb: a = c
Cu: 2b = d
O: 4a + b = 2c + d
H: 4a = d

From H: d = 4a
From Cu: 2b = 4a → b = 2a
From O: 4a + 2a = 2a + 4a → 6a = 6a ✔

So choose a = 1 → b = 2, c = 1, d = 4

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

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

Double displacement.

Cr²⁺, SO₄²⁻ → CrSO₄
NO₂⁻ and NH₄⁺ → NH₄NO₂

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

Check:
- Cr: 1 ✔
- S: 1 ✔
- N: left: 2 (from Cr(NO₂)₂) + 2 (from (NH₄)₂SO₄) = 4 → right: 2 NH₄NO₂ → 2 N → no

Wait: Cr(NO₂)₂ → 2 NO₂⁻
(NH₄)₂SO₄ → 2 NH₄⁺
So ions: Cr²⁺, 2 NO₂⁻, 2 NH₄⁺, SO₄²⁻

Products: CrSO₄ (Cr²⁺, SO₄²⁻), and 2 NH₄NO₂ (2 NH₄⁺, 2 NO₂⁻)

Yes! So:

✔ 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

Double displacement.

Co₃(PO₄)₂ → 3 Co²⁺, 2 PO₄³⁻
Need 3 Co(OH)₂ → 3 Co²⁺, 6 OH⁻
And 2 K₃PO₄ → 6 K⁺, 2 PO₄³⁻

So:
- PO₄: 2 → 2 K₃PO₄
- K: 6 → need 6 KOH
- OH: 6 → 3 Co(OH)₂ → 6 OH⁻

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

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18) Sn(NO₂)₄ + Pt₃N₄ → Sn₃N₄ + Pt(NO₂)₄

Looks like exchange.

Sn₃N₄ → 3 Sn, 4 N
Pt₃N₄ → 3 Pt, 4 N

So:
- Sn: 3 → need 3 Sn(NO₂)₄
- Pt: 3 → need 1 Pt₃N₄
- N: 4 from Pt₃N₄, 3×4 = 12 from Sn(NO₂)₄ → total 16 N
- Products: Sn₃N₄ → 4 N, Pt(NO₂)₄ → 4 N → only 8 N → too low

Wait: Pt(NO₂)₄ → 1 Pt, 4 NO₂ → 4 N
Sn₃N₄ → 4 N

Total N: 4 + 4 = 8 → but left: 3×4 = 12 from Sn(NO₂)₄, plus 4 from Pt₃N₄ → 16 N → imbalance

Try:

Let’s say:
a Sn(NO₂)₄ + b Pt₃N₄ → c Sn₃N₄ + d Pt(NO₂)₄

Sn: a = 3c
Pt: 3b = d
N: 4a + 4b = 4c + 4d → divide by 4: a + b = c + d
O: 8a = 8d → a = d

From a = d and 3b = d → a = 3b
From a = 3c → 3b = 3c → b = c
From a + b = c + d → 3b + b = b + 3b → 4b = 4b ✔

So pick b = 1 → c = 1, a = 3, d = 3

✔ Balanced:
3 Sn(NO₂)₄ + 1 Pt₃N₄ → 1 Sn₃N₄ + 3 Pt(NO₂)₄

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19) B₂Br₆ + HNO₃ → B(NO₃)₃ + HBr

Boron tribromide reacts with nitric acid.

- B: 2 → 2 B(NO₃)₃
- Br: 6 → 6 HBr
- H: 6 → 6 HBr → need 6 HNO₃
- NO₃: 6 → 6 HNO₃

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

So:
3 ZnS + 2 AlP → Zn₃P₂ + Al₂S₃

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

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## ✔ Final Answers (Balanced Equations):

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 H₂O₂ → 2 H₂O + 1 O₂
5) 1 C₆H₁₂O₆ + 6 O₂ → 6 H₂O + 6 CO₂
6) 4 H₂O + 7 CO₂ → 1 C₇H₈ + 9 O₂
7) 2 NaClO₃ → 2 NaCl + 3 O₂
8) 4 (NH₄)₃PO₄ + 3 Pb(NO₃)₄ → 1 Pb₃(PO₄)₄ + 12 NH₄NO₃
9) 2 BF₃ + 3 Li₂SO₃ → 1 B₂(SO₃)₃ + 6 LiF
10) 4 C₇H₁₇ + 45 O₂ → 28 CO₂ + 34 H₂O
11) 3 CaCO₃ + 2 H₃PO₄ → 1 Ca₃(PO₄)₂ + 3 H₂CO₃
12) 8 Ag₂S → 16 Ag + 1 S₈
13) 3 KBr + 1 Fe(OH)₃ → 3 KOH + 1 FeBr₃
14) 2 KNO₃ + 1 H₂CO₃ → 1 K₂CO₃ + 2 HNO₃
15) 1 Pb(OH)₄ + 2 Cu₂O → 1 PbO₂ + 4 CuOH
16) 1 Cr(NO₂)₂ + 1 (NH₄)₂SO₄ → 1 CrSO₄ + 2 NH₄NO₂
17) 6 KOH + 1 Co₃(PO₄)₂ → 2 K₃PO₄ + 3 Co(OH)₂
18) 3 Sn(NO₂)₄ + 1 Pt₃N₄ → 1 Sn₃N₄ + 3 Pt(NO₂)₄
19) 1 B₂Br₆ + 6 HNO₃ → 2 B(NO₃)₃ + 6 HBr
20) 3 ZnS + 2 AlP → 1 Zn₃P₂ + 1 Al₂S₃

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