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

Let’s go step by step to solve each problem. We’ll first identify the type of reaction (for the top section), then balance all equations carefully.

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Top Section: Types of Reactions THEN Balancing

We need to:
1. Identify the type of reaction.
2. Balance the equation.

Common types:
- Synthesis: A + B → AB
- Decomposition: AB → A + B
- Single Replacement: A + BC → AC + B
- Double Replacement: AB + CD → AD + CB
- Combustion: Hydrocarbon + O₂ → CO₂ + H₂O

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Problem 1:
___ NaBr + ___ H₃PO₄ → ___ Na₃PO₄ + ___ HBr

Type: Double replacement (ions swap partners)

Balance:
Left: Na=1, Br=1, H=3, P=1, O=4
Right: Na=3, Br=1, H=1, P=1, O=4

Need 3 Na on left → put 3 in front of NaBr
Now Br = 3 on left → need 3 HBr on right
H on right now = 3 → matches H₃PO₄

Balanced:
3 NaBr + 1 H₃PO₄ → 1 Na₃PO₄ + 3 HBr

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Problem 2:
___ Ca(OH)₂ + ___ Al₂(SO₄)₃ → ___ CaSO₄ + ___ Al(OH)₃

Type: Double replacement

Balance:
Left: Ca=1, O=2+12=14? Wait — better to count per element.

Ca(OH)₂ has Ca, 2O, 2H
Al₂(SO₄)₃ has 2Al, 3S, 12O

Right: CaSO₄ → Ca, S, 4O
Al(OH)₃ → Al, 3O, 3H

Try coefficients:

Put 3 Ca(OH)₂ → gives 3 Ca, 6 OH
Put 1 Al₂(SO₄)₃ → gives 2 Al, 3 SO₄

Then right: need 3 CaSO₄ (to match 3 Ca and 3 SO₄)
And 2 Al(OH)₃ (to match 2 Al)

Check H: Left: 3×2 = 6 H from Ca(OH)₂ → Right: 2×3 = 6 H from Al(OH)₃ ✔️
O: Left: 3×2 (from OH) + 12 (from sulfate) = 6+12=18
Right: 3×4 (from CaSO₄) + 2×3 (from Al(OH)₃) = 12+6=18 ✔️

Balanced:
3 Ca(OH)₂ + 1 Al₂(SO₄)₃ → 3 CaSO₄ + 2 Al(OH)₃

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Problem 3:
___ Mg + ___ Fe₂O₃ → ___ Fe + ___ MgO

Type: Single replacement (Mg replaces Fe)

Balance:
Left: Mg=1, Fe=2, O=3
Right: Fe=1, Mg=1, O=1

Need 2 Fe on right → put 2 Fe
Need 3 O on right → put 3 MgO → then Mg=3 on right → so 3 Mg on left

Balanced:
3 Mg + 1 Fe₂O₃ → 2 Fe + 3 MgO

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Problem 4:
___ C₂H₄ + ___ O₂ → ___ CO₂ + ___ H₂O

Type: Combustion (hydrocarbon burning)

Balance:
C₂H₄ has 2C, 4H
Right: CO₂ has 1C, H₂O has 2H

So: 2 CO₂ for carbon → 2C
2 H₂O for hydrogen → 4H

Now oxygen: Right: 2×2 + 2×1 = 4+2=6 O atoms → so 3 O₂ on left

Balanced:
1 C₂H₄ + 3 O₂ → 2 CO₂ + 2 H₂O

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Problem 5:
___ PbSO₄ → ___ PbSO₃ + ___ O₂

Type: Decomposition (one compound breaks into two)

Balance:
Left: Pb=1, S=1, O=4
Right: Pb=1, S=1, O=3+2=5? Wait — PbSO₃ has 3O, O₂ has 2O → total 5O? Not matching.

Actually, let’s check atoms:

PbSO₄ → PbSO₃ + ½ O₂? But we want whole numbers.

Multiply entire equation by 2:

2 PbSO₄ → 2 PbSO₃ + O₂

Check:
Left: Pb=2, S=2, O=8
Right: 2 PbSO₃ → Pb=2, S=2, O=6; plus O₂ → O=2 → total O=8 ✔️

Balanced:
2 PbSO₄ → 2 PbSO₃ + 1 O₂

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Problem 6:
___ NH₃ + ___ I₂ → ___ N₂I₆ + ___ H₂

Type: Synthesis? Or single replacement? Actually, it looks like a combination forming N₂I₆ and releasing H₂ — but this is unusual. Let’s just balance.

Note: N₂I₆ is diiodine hexaiodide? Probably meant as N₂I₆ (like hydrazine derivative). Anyway, balance atoms.

Left: N=1, H=3, I=2
Right: N=2, I=6, H=2

So need 2 NH₃ → gives 2N, 6H
Need 3 I₂ → gives 6I
Right: 1 N₂I₆ → 2N, 6I
And 3 H₂ → 6H

Balanced:
2 NH₃ + 3 I₂ → 1 N₂I₆ + 3 H₂

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Problem 7:
___ H₂O + ___ SO₃ → ___ H₂SO₄

Type: Synthesis (two compounds make one)

Balance:
Left: H=2, O=1+3=4, S=1
Right: H=2, S=1, O=4 → already balanced!

Balanced:
1 H₂O + 1 SO₃ → 1 H₂SO₄

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Problem 8:
___ H₂SO₄ + ___ NH₄OH → ___ H₂O + ___ (NH₄)₂SO₄

Type: Double replacement (acid-base neutralization)

Balance:
Left: H₂SO₄ → 2H, S, 4O
NH₄OH → N, 5H, O? Wait — NH₄OH is NH₄⁺ and OH⁻ → so N, 5H, O? Actually, formula is often written as NH₄OH meaning ammonia water, but atomically: N, 5H, O? No — NH₄ has 4H, OH has 1H and 1O → total N, 5H, O? That can’t be right.

Actually, standard way: NH₄OH contributes NH₄⁺ and OH⁻ → so when reacting with H₂SO₄, it forms (NH₄)₂SO₄ and H₂O.

So: H₂SO₄ + 2 NH₄OH → (NH₄)₂SO₄ + 2 H₂O

Check atoms:

Left: H₂SO₄ → H=2, S=1, O=4
2 NH₄OH → N=2, H=2×(4+1)=10? Wait — NH₄OH is actually considered as having N, 5H, O? But that’s not accurate. Better to think ionically.

Standard balanced equation:

H₂SO₄ + 2 NH₃ → (NH₄)₂SO₄ — but here it’s NH₄OH.

Actually, NH₄OH dissociates to NH₄⁺ and OH⁻, so:

H₂SO₄ + 2 NH₄OH → (NH₄)₂SO₄ + 2 H₂O

Atoms:

Left:
H₂SO₄: H=2, S=1, O=4
2 NH₄OH: N=2, H=2×5=10? Wait — if NH₄OH is written as such, it implies N, 5H, O per molecule? But that would be incorrect because NH₄ is 4H, OH is 1H and 1O — so yes, 5H total per NH₄OH.

But in reality, when you write NH₄OH, it's equivalent to NH₃(aq) + H₂O, but for balancing, we treat it as given.

So:

Left: H₂SO₄ + 2 NH₄OH →
H: 2 + 2×5 = 12? Too many.

Wait — I think there’s confusion. Standard practice: NH₄OH is treated as providing NH₄⁺ and OH, so the reaction is:

H₂SO₄ + 2 NH₄OH → (NH₄)₂SO₄ + 2 H₂O

Now count atoms properly:

Assume NH₄OH means: N, 5H, O? But that leads to imbalance.

Better approach: Think of it as acid-base:

H₂SO₄ has 2H⁺
Each NH₄OH provides 1 OH⁻ → so need 2 NH₄OH to neutralize.

Products: (NH₄)₂SO₄ and 2 H₂O

So:

H₂SO₄ + 2 NH₄OH → (NH₄)₂SO₄ + 2 H₂O

Now count:

Left:
H₂SO₄: H=2, S=1, O=4
2 NH₄OH: If we consider NH₄OH as N, 5H, O — then 2N, 10H, 2O → total H=2+10=12, O=4+2=6, S=1, N=2

Right:
(NH₄)₂SO₄: 2N, 8H, S, 4O
2 H₂O: 4H, 2O → total H=8+4=12, O=4+2=6, S=1, N=2 ✔️

Yes! So it balances.

Balanced:
1 H₂SO₄ + 2 NH₄OH → 2 H₂O + 1 (NH₄)₂SO₄

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Now, second part: Balancing Equations Practice Worksheet

Just balance these — no need to classify.

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Practice Problem 1:
___ NaNO₃ + ___ PbO → ___ Pb(NO₃)₂ + ___ Na₂O

Balance:
Left: Na=1, N=1, O=3+1=4, Pb=1
Right: Pb=1, N=2, O=6+1=7? Pb(NO₃)₂ has Pb, 2N, 6O; Na₂O has 2Na, 1O

So need 2 NaNO₃ on left → gives 2Na, 2N, 6O
Plus PbO → Pb, 1O → total O=7, Na=2, N=2, Pb=1

Right: Pb(NO₃)₂ → Pb, 2N, 6O
Na₂O → 2Na, 1O → total O=7, Na=2, N=2, Pb=1 ✔️

Balanced:
2 NaNO₃ + 1 PbO → 1 Pb(NO₃)₂ + 1 Na₂O

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Practice Problem 2:
___ AgI + ___ Fe₂(CO₃)₃ → ___ FeI₃ + ___ Ag₂CO₃

Balance:
Left: Ag=1, I=1, Fe=2, C=3, O=9
Right: Fe=1, I=3, Ag=2, C=1, O=3

Need to balance Fe: left has 2 Fe → so 2 FeI₃ on right → then I=6 → so 6 AgI on left → Ag=6 → so 3 Ag₂CO₃ on right → then C=3, O=9 → matches Fe₂(CO₃)₃

Check:

Left: 6 AgI → Ag=6, I=6
Fe₂(CO₃)₃ → Fe=2, C=3, O=9

Right: 2 FeI₃ → Fe=2, I=6
3 Ag₂CO₃ → Ag=6, C=3, O=9 ✔️

Balanced:
6 AgI + 1 Fe₂(CO₃)₃ → 2 FeI₃ + 3 Ag₂CO₃

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Practice Problem 3:
___ C₂H₄O₂ + ___ O₂ → ___ CO₂ + ___ H₂O

This is combustion of acetic acid or similar.

C₂H₄O₂ has 2C, 4H, 2O

Right: CO₂ and H₂O

Set: 2 CO₂ for carbon → 2C
2 H₂O for hydrogen → 4H

Oxygen: Right: 2×2 + 2×1 = 6 O
Left: C₂H₄O₂ has 2O, so O₂ must provide 4O → so 2 O₂

Balanced:
1 C₂H₄O₂ + 2 O₂ → 2 CO₂ + 2 H₂O

Check:
Left: C=2, H=4, O=2+4=6
Right: C=2, H=4, O=4+2=6 ✔️

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Practice Problem 4:
___ ZnSO₄ + ___ Li₂CO₃ → ___ ZnCO₃ + ___ Li₂SO₄

Already balanced? Check:

Left: Zn=1, S=1, O=4+3=7, Li=2, C=1
Right: Zn=1, C=1, O=3+4=7, Li=2, S=1 ✔️

Balanced:
1 ZnSO₄ + 1 Li₂CO₃ → 1 ZnCO₃ + 1 Li₂SO₄

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Practice Problem 5:
___ V₂O₅ + ___ CaS → ___ CaO + ___ V₂S₅

Balance:
Left: V=2, O=5, Ca=1, S=1
Right: Ca=1, O=1, V=2, S=5

Need 5 S on right → so 5 CaS on left → then Ca=5 → so 5 CaO on right
V₂O₅ has 5 O → matches 5 CaO

Balanced:
1 V₂O₅ + 5 CaS → 5 CaO + 1 V₂S₅

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Practice Problem 6:
___ Mn(NO₂)₂ + ___ BeCl₂ → ___ Be(NO₂)₂ + ___ MnCl₂

Looks like double replacement. Already balanced?

Left: Mn=1, N=2, O=4, Be=1, Cl=2
Right: Be=1, N=2, O=4, Mn=1, Cl=2 ✔️

Balanced:
1 Mn(NO₂)₂ + 1 BeCl₂ → 1 Be(NO₂)₂ + 1 MnCl₂

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Practice Problem 7:
___ AgBr + ___ GaPO₄ → ___ Ag₃PO₄ + ___ GaBr₃

Balance:
Left: Ag=1, Br=1, Ga=1, P=1, O=4
Right: Ag=3, P=1, O=4, Ga=1, Br=3

Need 3 AgBr on left → Ag=3, Br=3
GaPO₄ unchanged → Ga=1, P=1, O=4
Right: Ag₃PO₄ → Ag=3, P=1, O=4
GaBr₃ → Ga=1, Br=3 ✔️

Balanced:
3 AgBr + 1 GaPO₄ → 1 Ag₃PO₄ + 1 GaBr₃

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Practice Problem 8:
___ H₂SO₄ + ___ B(OH)₃ → ___ B₂(SO₄)₃ + ___ H₂O

Balance:
Left: H₂SO₄ → H=2, S=1, O=4
B(OH)₃ → B=1, O=3, H=3

Right: B₂(SO₄)₃ → B=2, S=3, O=12
H₂O → H=2, O=1

Need 2 B on right → so 2 B(OH)₃ on left → B=2, O=6, H=6
Need 3 S on right → so 3 H₂SO₄ on left → H=6, S=3, O=12

Total left: H=6+6=12, O=12+6=18, S=3, B=2

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

H on right: from H₂O → each has 2H → need 6 H₂O to get 12H → then O from water=6

Total O on right: 12 (from sulfate) + 6 (from water) = 18 ✔️

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

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Practice Problem 9:
___ S₈ + ___ O₂ → ___ SO₂

Balance:
S₈ has 8 S → so 8 SO₂ on right → then O=16 → so 8 O₂ on left

Balanced:
1 S₈ + 8 O₂ → 8 SO₂

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Practice Problem 10:
___ Fe + ___ AgNO₃ → ___ Fe(NO₃)₂ + ___ Ag

Single replacement.

Fe goes from 0 to +2, Ag from +1 to 0.

So: Fe + 2 AgNO₃ → Fe(NO₃)₂ + 2 Ag

Check:
Left: Fe=1, Ag=2, N=2, O=6
Right: Fe=1, N=2, O=6, Ag=2 ✔️

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

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Final Answer:

Types of Reactions THEN Balancing:

1) Type: Double replacement
Balanced: 3 NaBr + 1 H₃PO₄ → 1 Na₃PO₄ + 3 HBr

2) Type: Double replacement
Balanced: 3 Ca(OH)₂ + 1 Al₂(SO₄)₃ → 3 CaSO₄ + 2 Al(OH)₃

3) Type: Single replacement
Balanced: 3 Mg + 1 Fe₂O₃ → 2 Fe + 3 MgO

4) Type: Combustion
Balanced: 1 C₂H₄ + 3 O₂ → 2 CO₂ + 2 H₂O

5) Type: Decomposition
Balanced: 2 PbSO₄ → 2 PbSO₃ + 1 O₂

6) Type: Synthesis (or combination)
Balanced: 2 NH₃ + 3 I₂ → 1 N₂I₆ + 3 H₂

7) Type: Synthesis
Balanced: 1 H₂O + 1 SO₃ → 1 H₂SO₄

8) Type: Double replacement (neutralization)
Balanced: 1 H₂SO₄ + 2 NH₄OH → 2 H₂O + 1 (NH₄)₂SO₄

Balancing Equations Practice Worksheet:

1) 2 NaNO₃ + 1 PbO → 1 Pb(NO₃)₂ + 1 Na₂O
2) 6 AgI + 1 Fe₂(CO₃)₃ → 2 FeI₃ + 3 Ag₂CO₃
3) 1 C₂H₄O₂ + 2 O₂ → 2 CO₂ + 2 H₂O
4) 1 ZnSO₄ + 1 Li₂CO₃ → 1 ZnCO₃ + 1 Li₂SO₄
5) 1 V₂O₅ + 5 CaS → 5 CaO + 1 V₂S₅
6) 1 Mn(NO₂)₂ + 1 BeCl₂ → 1 Be(NO₂)₂ + 1 MnCl₂
7) 3 AgBr + 1 GaPO₄ → 1 Ag₃PO₄ + 1 GaBr₃
8) 3 H₂SO₄ + 2 B(OH)₃ → 1 B₂(SO₄)₃ + 6 H₂O
9) 1 S₈ + 8 O₂ → 8 SO₂
10) 1 Fe + 2 AgNO₃ → 1 Fe(NO₃)₂ + 2 Ag