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 provide 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

Then: 6 Li → 3 Li₂O

And Ga₂O₃ → 2 Ga

So:
1 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 seems problematic.

BaS₂? That’s barium disulfide — but typically barium forms BaS (not S₂). But assuming it's correct...

But look at sulfur: Na₂S has one S → BaS₂ has two S → so need 2 Na₂S

Then: 2 Na₂S → 4 Na → need 4 NaF → so need 4 F → BaF₂ only has 2 F → need 2 BaF₂

Try:
2 BaF₂ + 2 Na₂S → 2 BaS₂ + 4 NaF

But wait: BaS₂ has 2 S → 2 Na₂S gives 2 S → OK

But now:
- Ba: 2 = 2
- F: 4 = 4
- Na: 4 = 4
- S: 2 = 2

But wait: 2 Na₂S → 2 S → gives 2 S for 2 BaS₂ → each BaS₂ has 2 S → total 4 S needed → problem!

Ah! So if product is BaS₂, then each BaS₂ has 2 sulfur atoms, so for 1 BaS₂ → need 2 S → from 2 Na₂S

So:
1 BaF₂ + 2 Na₂S → 1 BaS₂ + 4 NaF

But F: left: 2 (from BaF₂), right: 4 (from 4 NaF) → not balanced

Need 2 BaF₂ → 4 F → 4 NaF

So:
2 BaF₂ + 2 Na₂S → 2 BaS₂ + 4 NaF?

No: 2 Na₂S → 2 S → but 2 BaS₂ → 4 S → too much

Wait: each BaS₂ needs 2 S → so for 2 BaS₂ → need 4 S → need 4 Na₂S

So:
2 BaF₂ + 4 Na₂S → 2 BaS₂ + 8 NaF

Now check:
- Ba: 2 = 2
- F: 4 = 8 → no! 2 BaF₂ → 4 F, but 8 NaF → 8 F → imbalance

So need 4 BaF₂ → 8 F → 8 NaF

Try:
4 BaF₂ + 4 Na₂S → 4 BaS₂ + 8 NaF

But S: 4 Na₂S → 4 S → 4 BaS₂ → 8 S → still mismatch

Wait: BaS₂ has two sulfur atoms per formula unit, so to make 1 BaS₂ → need 2 S → from 2 Na₂S

So for 1 BaS₂ → need 2 Na₂S

But also need 1 BaF₂ → 2 F → for 2 NaF

So:
1 BaF₂ + 2 Na₂S → 1 BaS₂ + 2 NaF

But Na: left: 4 (from 2 Na₂S), right: 2 (from 2 NaF) → not balanced

So need 4 NaF → 4 Na → need 2 Na₂S → but 2 Na₂S gives 4 Na → yes!

Wait: 2 Na₂S → 4 Na → so 4 NaF → 4 Na → good

But S: 2 Na₂S → 2 S → but BaS₂ needs 2 S → good

F: BaF₂ → 2 F → 4 NaF → 4 F → not enough

So need 2 BaF₂ → 4 F → 4 NaF → 4 Na → but 4 Na → need 2 Na₂S → 2 S → but BaS₂ needs 2 S → so only one BaS₂ possible

Wait: confusion.

Let’s define variables.

Let:
a BaF₂ + b Na₂S → c BaS₂ + d NaF

Atoms:
- Ba: a = c
- F: 2a = d
- Na: 2b = d
- S: b = 2c (because BaS₂ has 2 S)

From S: b = 2c
From Na: 2b = d
From F: 2a = d
From Ba: a = c

So:
a = c
b = 2c
d = 2b = 4c
d = 2a = 2c → contradiction unless c=0

Wait: d = 2b and d = 2a → so 2b = 2a → a = b
But b = 2c and a = c → so c = 2c → c = 0 → impossible

So no solution unless the compound is wrong.

Wait — perhaps BaS₂ is incorrect. Barium usually forms BaS, not BaS₂.

But if the problem says BaS₂, maybe it's a typo or unusual compound.

Alternatively, could it be BaS instead of BaS₂?

If it were:
BaF₂ + Na₂S → BaS + 2 NaF

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

✔ Perfectly balanced.

But the problem says BaS₂, which is not standard.

However, let’s suppose it’s BaS₂ — then we must have:

From earlier:
b = 2c (S atoms)
a = c (Ba)
d = 2a (F)
d = 2b (Na)

So:
d = 2a = 2c
d = 2b = 4c
So 2c = 4c → c = 0 → impossible.

Thus, this equation cannot be balanced as written.

But perhaps it's a typo and should be BaS, not BaS₂

Assume typo: BaS instead of BaS₂

Then:
BaF₂ + Na₂S → BaS + 2 NaF

Balanced.

✔ Yes.

So likely intended:
1 BaF₂ + 1 Na₂S → 1 BaS + 2 NaF

But since the worksheet says BaS₂, let’s recheck.

Alternatively, maybe it’s BaS₂ and Na₂S₂? No.

Another possibility: double replacement with sulfur as S₂²⁻? But unlikely.

Perhaps the product is BaS and NaF, and BaS₂ is a mistake.

Given the inconsistency, and since BaS₂ doesn't make sense stoichiometrically here, I suspect a typo.

But let’s assume it's BaS.

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

But since the question says BaS₂, and it can’t be balanced, perhaps the intended answer is:

Wait — maybe it's BaS₂ and Na₂S₂? But not given.

Alternatively, maybe it's BaF₂ + Na₂S → BaS + 2 NaF, and BaS₂ is a typo.

I’ll proceed with that assumption.

But let’s try again: maybe BaS₂ is correct, and we need 2 S per Ba, so need 2 Na₂S per BaF₂?

Try:
1 BaF₂ + 2 Na₂S → 1 BaS₂ + 4 NaF

Now:
- Ba: 1 = 1
- F: 2 = 4 → no
- Na: 4 = 4 → OK
- S: 2 = 2 → OK

F: left: 2, right: 4 → not balanced

So need 2 BaF₂ → 4 F → 4 NaF → 4 Na → need 2 Na₂S → 2 S → but BaS₂ needs 2 S → so one BaS₂

So:
2 BaF₂ + 2 Na₂S → 1 BaS₂ + 4 NaF

Now:
- Ba: 2 = 1 → no

So need 2 BaS₂ → 4 S → need 4 Na₂S → 8 Na → 8 NaF → 8 F → need 4 BaF₂

So:
4 BaF₂ + 4 Na₂S → 2 BaS₂ + 8 NaF

Now:
- Ba: 4 = 2 → no

Wait: 4 BaF₂ → 4 Ba → so need 4 BaS₂

So:
4 BaF₂ + 4 Na₂S → 4 BaS₂ + 8 NaF

Now:
- Ba: 4 = 4
- F: 8 = 8
- Na: 8 = 8
- S: 4 = 8 → no! 4 Na₂S → 4 S, but 4 BaS₂ → 8 S → need 8 S → 8 Na₂S

So:
4 BaF₂ + 8 Na₂S → 4 BaS₂ + 16 NaF

Now:
- Ba: 4 = 4
- F: 8 = 16 → no

F: 4 BaF₂ → 8 F → 16 NaF → 16 F → need 8 BaF₂

So:
8 BaF₂ + 8 Na₂S → 8 BaS₂ + 16 NaF

Now:
- Ba: 8 = 8
- F: 16 = 16
- Na: 16 = 16
- S: 8 = 16 → no

Still not working.

The issue is: each BaS₂ requires 2 S, so for n BaS₂ → need 2n S → from n Na₂S → only n S → so need 2n Na₂S

But Na₂S provides 2 Na per molecule → 2n Na₂S → 4n Na → need 4n NaF → 4n F → need 2n BaF₂ → 2n Ba → but need n BaS₂ → n Ba → contradiction.

So no consistent solution.

Therefore, the equation as written cannot be balanced due to stoichiometry.

Conclusion: Likely a typo. Should be BaS, not BaS₂

So:
1 BaF₂ + 1 Na₂S → 1 BaS + 2 NaF

✔ Balanced.

Answer (corrected):
1 BaF₂ + 1 Na₂S → 1 BaS + 2 NaF

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10)

NH₃ → N₂ + H₂

Decomposition of ammonia.

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

Need even number of N atoms → use 2 NH₃ → 2 N → 1 N₂

2 NH₃ → N₂ + 3 H₂

Check:
- N: 2 = 2
- H: 6 = 6 (3 H₂ molecules)

✔ 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 *(assuming typo in BaS₂)*
10) 2 NH₃ → 1 N₂ + 3 H₂

> Note: Problem #9 likely contains a typo. BaS₂ is not chemically reasonable here; it should probably be BaS. If BaS₂ is intentional, the equation cannot be balanced. But based on standard chemistry, BaS is correct.

Let me know if you'd like explanations for any specific steps!