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 that the number of atoms of each element is the same on both sides (reactants and products), following the Law of Conservation of Mass.

We'll go through each one step by step.

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1)

NaCl + KOH → NaOH + KCl

- Left: Na, Cl, K, O, H
- Right: Na, O, H, K, Cl

Each element appears once on each side:
- Na: 1 = 1
- Cl: 1 = 1
- K: 1 = 1
- O: 1 = 1
- H: 1 = 1

✔ Already balanced!

Answer:
1 NaCl + 1 KOH → 1 NaOH + 1 KCl

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2)

MgO + Li₂SO₄ → MgSO₄ + Li₂O

Check atoms:

| Element | Reactants | Products |
|--------|----------|---------|
| Mg | 1 | 1 |
| O | 1 (MgO) + 4 (SO₄) = 5 | 4 (SO₄) + 1 (Li₂O) = 5 |
| Li | 2 | 2 |
| S | 1 | 1 |

All elements are balanced as written.

✔ Balanced!

Answer:
1 MgO + 1 Li₂SO₄ → 1 MgSO₄ + 1 Li₂O

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3)

H₂O → H₂ + O₂

This is decomposition of water.

Left: H=2, O=1
Right: H=2, O=2 → Oxygen not balanced.

To balance oxygen: need 2 H₂O → gives 2 O atoms → makes 1 O₂

So:
2 H₂O → 2 H₂ + O₂

Now check:
- H: 4 left → 4 right (2×2)
- O: 2 left → 2 right (in O₂)

✔ Balanced!

Answer:
2 H₂O → 2 H₂ + 1 O₂

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4)

RbF + Be(NO₃)₂ → RbNO₃ + BeF₂

Look at ions:
- Rb⁺ and F⁻
- Be²⁺ and NO₃⁻

BeF₂ requires 2 F⁻ → so need 2 RbF
Be(NO₃)₂ has 2 NO₃⁻ → so need 2 RbNO₃

Try:
2 RbF + Be(NO₃)₂ → 2 RbNO₃ + BeF₂

Check:
- Rb: 2 = 2
- F: 2 = 2
- Be: 1 = 1
- N: 2 = 2
- O: 6 = 6 (from 2 NO₃ groups)

✔ Balanced!

Answer:
2 RbF + 1 Be(NO₃)₂ → 2 RbNO₃ + 1 BeF₂

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5)

Ag + Cu(NO₃)₂ → AgNO₃ + Cu

This is a single displacement reaction.

Cu(NO₃)₂ provides 2 NO₃⁻ → so need 2 AgNO₃ → need 2 Ag atoms

So:
2 Ag + Cu(NO₃)₂ → 2 AgNO₃ + Cu

Check:
- Ag: 2 = 2
- Cu: 1 = 1
- N: 2 = 2
- O: 6 = 6

✔ Balanced!

Answer:
2 Ag + 1 Cu(NO₃)₂ → 2 AgNO₃ + 1 Cu

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6)

CO₂ + Cl₂ → CCl₄ + O₂

Left: C, O, Cl
Right: C, Cl, O

Carbon: 1 on both sides → OK
Oxygen: 2 on left → 2 on right (in O₂) → OK
Chlorine: 2 on left → 4 on right → need 2 Cl₂

Try:
CO₂ + 2 Cl₂ → CCl₄ + O₂

Now check:
- C: 1 = 1
- O: 2 = 2
- Cl: 4 = 4

✔ Balanced!

Answer:
1 CO₂ + 2 Cl₂ → 1 CCl₄ + 1 O₂

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7)

CuSO₄ + HCN → Cu(CN)₂ + H₂SO₄

Note: Cu(CN)₂ implies Cu²⁺ and 2 CN⁻ → so we need 2 HCN to supply 2 CN⁻

Also, H₂SO₄ needs 2 H⁺ → so 2 HCN give 2 H⁺

So:
1 CuSO₄ + 2 HCN → 1 Cu(CN)₂ + 1 H₂SO₄

Check:
- Cu: 1 = 1
- S: 1 = 1
- O: 4 = 4
- H: 2 = 2
- C: 2 = 2
- N: 2 = 2

✔ Balanced!

Answer:
1 CuSO₄ + 2 HCN → 1 Cu(CN)₂ + 1 H₂SO₄

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8)

Ga₂O₃ + Li → Li₂O + Ga

Ga₂O₃ has 2 Ga and 3 O → need 3 Li₂O to get 3 O → so 6 Li needed

Then 2 Ga produced

So:
Ga₂O₃ + 6 Li → 3 Li₂O + 2 Ga

Check:
- Ga: 2 = 2
- O: 3 = 3
- Li: 6 = 6

✔ Balanced!

Answer:
1 Ga₂O₃ + 6 Li → 3 Li₂O + 2 Ga

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9)

BaF₂ + Na₂S → BaS₂ + NaF

Wait — this might be incorrect. BaS₂ is not a common compound. But assuming it's correct...

But let’s analyze:
- BaF₂ has Ba²⁺, 2 F⁻
- Na₂S has 2 Na⁺, S²⁻
- BaS₂ would imply Ba²⁺ with 2 S²⁻ → total charge -4 → not possible

Wait! That can't be. BaS₂ doesn’t make sense chemically. Probably meant BaS, not BaS₂.

But if it's BaS₂, then Ba²⁺ + 2 S²⁻ → but that would require 2 S²⁻, so maybe from 2 Na₂S?

But even then, Na₂S provides only one S per molecule.

Assume it's a typo and should be BaS? But the problem says BaS₂.

Alternatively, perhaps it's BaS and NaF, but the product is written as BaS₂.

Let’s suppose the intended reaction is:

BaF₂ + Na₂S → BaS + 2 NaF

That makes more sense.

But the worksheet says: → BaS₂ + NaF

So let’s try balancing as written.

Suppose:
BaF₂ + Na₂S → BaS₂ + NaF

Left: Ba, 2F, 2Na, S
Right: Ba, 2S, Na, F

Not balanced.

To get 2 S in BaS₂, need 2 Na₂S → gives 4 Na

Then need 4 NaF → 4 F → but BaF₂ only has 2 F → not enough

So impossible.

Thus, likely a typo. The correct product should be BaS, not BaS₂.

But since the worksheet says BaS₂, perhaps it's a hypothetical or error.

Alternatively, could it be BaS and NaF, and BaS₂ is a mistake?

Let’s assume it’s BaS, then:

BaF₂ + Na₂S → BaS + 2 NaF

Check:
- Ba: 1 = 1
- F: 2 = 2
- Na: 2 = 2
- S: 1 = 1

✔ Balanced.

But the worksheet says BaS₂, so unless it's a special compound (like polysulfide), it's likely a typo.

But let’s proceed as written: BaS₂

To form BaS₂, need 2 S atoms → so need 2 Na₂S → 4 Na

Then need 4 NaF → 4 F → but BaF₂ only has 2 F → need 2 BaF₂

Try:
2 BaF₂ + 2 Na₂S → BaS₂ + 4 NaF → but now Ba: 2 vs 1 → no

So:
2 BaF₂ + 2 Na₂S → 2 BaS + 4 NaF → but again, BaS, not BaS₂

No way to get BaS₂ with only one S source.

Therefore, the equation as written is chemically implausible.

But for the sake of balancing, suppose we ignore chemistry and just balance atoms.

We want:
BaF₂ + Na₂S → BaS₂ + NaF

We need 2 S on right → need 2 Na₂S → 4 Na

Need 4 NaF → 4 F → need 2 BaF₂

Then Ba: 2 left → 1 right → so need 2 BaS₂

So:
2 BaF₂ + 2 Na₂S → 2 BaS₂ + 4 NaF

Now check:
- Ba: 2 = 2
- F: 4 = 4
- Na: 4 = 4
- S: 2 = 4 → no! 2 S left, 4 S right

No.

Try:
2 BaF₂ + 4 Na₂S → 2 BaS₂ + 8 NaF

Now:
- Ba: 2 = 2
- F: 4 = 8 → no

Can't match.

Actually, impossible to balance if BaS₂ is required.

Conclusion: The equation is likely miswritten. It should be:

BaF₂ + Na₂S → BaS + 2 NaF

Balanced.

But since the worksheet says BaS₂, and it's unbalanced and chemically unlikely, we’ll assume it’s a typo.

However, if we must balance as written, it's not possible with integer coefficients.

Wait — another idea: Maybe it's BaS and NaF, and BaS₂ is a typo.

Or perhaps BaS₂ is meant to be BaS.

Given that, I will assume it's BaS.

So:

1 BaF₂ + 1 Na₂S → 1 BaS + 2 NaF

✔ Balanced.

But the worksheet says BaS₂, so if you're forced to use that, it's invalid.

For now, I'll go with BaS.

But let’s see if there's any possibility.

Alternative: Could it be BaS₂ formed from two sulfides? Unlikely.

Perhaps it's a double displacement with excess sulfur? Not standard.

So best guess: typo, should be BaS.

But let's suppose the answer expects us to balance as written.

Let me try:

We want:
BaF₂ + Na₂S → BaS₂ + NaF

To get 2 S atoms on right → need 2 Na₂S → 4 Na

Need 4 NaF → 4 F → need 2 BaF₂

Then Ba: 2 left → 1 right → so need 2 BaS₂

So:
2 BaF₂ + 2 Na₂S → 2 BaS₂ + 4 NaF

Now:
- Ba: 2 = 2
- F: 4 = 4
- Na: 4 = 4
- S: 2 = 4 → no!

Still not balanced.

To get 4 S on right, need 4 Na₂S → 8 Na → 8 NaF → 8 F → 4 BaF₂

Then:
4 BaF₂ + 4 Na₂S → 4 BaS₂ + 8 NaF

Now:
- Ba: 4 = 4
- F: 8 = 8
- Na: 8 = 8
- S: 4 = 8 → no

Wait — 4 Na₂S has 4 S, but BaS₂ has 2 S per molecule → 4 BaS₂ has 8 S → mismatch.

So to have 8 S on right, need 8 Na₂S → 16 Na → 16 NaF → 16 F → 8 BaF₂

Then:
8 BaF₂ + 8 Na₂S → 8 BaS₂ + 16 NaF

Now:
- Ba: 8 = 8
- F: 16 = 16
- Na: 16 = 16
- S: 8 = 16 → still not balanced

Wait — 8 Na₂S has 8 S atoms, but 8 BaS₂ has 16 S atoms → impossible.

So no solution.

Therefore, the equation cannot be balanced as written.

Likely typo: should be BaS, not BaS₂

So correct version:
1 BaF₂ + 1 Na₂S → 1 BaS + 2 NaF

✔

But since the worksheet says BaS₂, and it's clearly wrong, I'll note that.

But perhaps the student is expected to write:

1 BaF₂ + 1 Na₂S → 1 BaS₂ + 2 NaF — but that's not balanced.

So I think it's a typo.

Final Answer (assuming typo):
1 BaF₂ + 1 Na₂S → 1 BaS + 2 NaF

But since the problem says BaS₂, and it's unbalanced, I suspect an error.

Alternatively, maybe it's BaS and NaF, and BaS₂ is a mistake.

I'll proceed with the corrected version.

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10)

NH₃ → N₂ + H₂

Decomposition of ammonia.

Left: N=1, H=3
Right: N=2, H=2 → not balanced

Need 2 NH₃ → 2 N and 6 H → so N₂ and 3 H₂

So:
2 NH₃ → 1 N₂ + 3 H₂

Check:
- N: 2 = 2
- H: 6 = 6

✔ Balanced!

Answer:
2 NH₃ → 1 N₂ + 3 H₂

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✔ Final Answers:

1) 1 NaCl + 1 KOH → 1 NaOH + 1 KCl
2) 1 MgO + 1 Li₂SO₄ → 1 MgSO₄ + 1 Li₂O
3) 2 H₂O → 2 H₂ + 1 O₂
4) 2 RbF + 1 Be(NO₃)₂ → 2 RbNO₃ + 1 BeF₂
5) 2 Ag + 1 Cu(NO₃)₂ → 2 AgNO₃ + 1 Cu
6) 1 CO₂ + 2 Cl₂ → 1 CCl₄ + 1 O₂
7) 1 CuSO₄ + 2 HCN → 1 Cu(CN)₂ + 1 H₂SO₄
8) 1 Ga₂O₃ + 6 Li → 3 Li₂O + 2 Ga
9) 1 BaF₂ + 1 Na₂S → 1 BaS + 2 NaF *(Note: BaS₂ likely a typo)*
10) 2 NH₃ → 1 N₂ + 3 H₂

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If you believe #9 is indeed BaS₂, please double-check the original problem — it may be a typo. Otherwise, the equation cannot be balanced as written.