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

Let's solve each of these chemical equations by balancing them. I'll go through each one step-by-step, explaining the process and providing the balanced equation.

---

1. Ca(AlO₂)₂ + HCl → AlCl₃ + CaCl₂ + H₂O

Step 1: Identify atoms on both sides.

- Left: Ca, Al, O, H, Cl
- Right: Al, Cl, Ca, H, O

Step 2: Balance metals first.

- 1 Ca on both sides → OK.
- 2 Al on left (from Ca(AlO₂)₂), so need 2 AlCl₃ on right.

So:
> Ca(AlO₂)₂ + HCl → 2AlCl₃ + CaCl₂ + H₂O

Now Cl: 2×3 = 6 from AlCl₃ + 2 from CaCl₂ → total 8 Cl → need 8 HCl.

So:
> Ca(AlO₂)₂ + 8HCl → 2AlCl₃ + CaCl₂ + H₂O

Now H: 8 H on left → need 4 H₂O (since 4×2=8 H)

O: On left: 4 O from AlO₂ (each AlO₂ has 2 O, two AlO₂ → 4 O).
On right: 4 H₂O → 4 O → OK.

Check all:

| Atom | Left | Right |
|------|------|-------|
| Ca | 1 | 1 |
| Al | 2 | 2 |
| O | 4 | 4 |
| H | 8 | 8 |
| Cl | 8 | 6+2=8 |

✔ Balanced.

Answer:
> Ca(AlO₂)₂ + 8HCl → 2AlCl₃ + CaCl₂ + 4H₂O

---

2. CeO₂ + KI + HCl → KCl + CeCl₃ + H₂O + I₂

This is a redox reaction. Ce⁴⁺ → Ce³⁺, I⁻ → I₂.

Step 1: Assign oxidation states.

- Ce in CeO₂: +4 → CeCl₃: +3 → gains 1 e⁻ per Ce
- I⁻ → I₂: loses 1 e⁻ per I atom → 2I⁻ → I₂ + 2e⁻

So to balance electrons:
- 2 Ce⁴⁺ → 2 Ce³⁺ + 2e⁻
- 2I⁻ → I₂ + 2e⁻ → but we need 2I⁻ → I₂, so 2I⁻ → I₂

But KI provides I⁻, so need 2 KI → I₂

Also, CeO₂ → CeCl₃: need 3 Cl⁻, so 3 HCl per Ce.

But also, H⁺ needed to form water.

Try balancing:

Start with:
> CeO₂ + KI + HCl → KCl + CeCl₃ + H₂O + I₂

Let’s assume 2 CeO₂ → 2 CeCl₃ → need 6 Cl⁻ → 6 HCl

But also, 2 Ce⁴⁺ → 2 Ce³⁺ → gain 2e⁻ → need 2 I⁻ → 1 I₂ → need 2 KI

So far:
> 2CeO₂ + 2KI + 6HCl → 2KCl + 2CeCl₃ + H₂O + I₂

Now check H: 6 H → need 3 H₂O → 3 H₂O

O: 4 O from 2CeO₂ → 3 H₂O → 3 O → missing 1 O? Wait.

Wait: 2CeO₂ → 4 O atoms → must go to H₂O → 4 H₂O?

But H: 6 H → only enough for 3 H₂O.

Conflict.

Better approach: Use half-reactions.

Reduction: CeO₂ → CeCl₃

Ce⁴⁺ → Ce³⁺ + e⁻

But oxygen must be accounted for. In acidic medium:

CeO₂ + 2H⁺ + e⁻ → Ce³⁺ + H₂O

So CeO₂ + 2H⁺ + e⁻ → Ce³⁺ + H₂O

Oxidation: 2I⁻ → I₂ + 2e⁻

To balance electrons: multiply reduction by 2:

2CeO₂ + 4H⁺ + 2e⁻ → 2Ce³⁺ + 2H₂O

Add to oxidation:
2CeO₂ + 4H⁺ + 2I⁻ → 2Ce³⁺ + I₂ + 2H₂O

Now add spectator ions.

From above: 2CeO₂ + 2I⁻ + 4H⁺ → 2Ce³⁺ + I₂ + 2H₂O

Now include K⁺ and Cl⁻.

KI → K⁺ and I⁻ → so 2KI

HCl → H⁺ and Cl⁻ → 4 HCl

Products: KCl → need 2 K⁺ → 2 KCl

CeCl₃ → 2 CeCl₃ → needs 6 Cl⁻ → so 6 Cl⁻ → need 6 HCl

But we already used 4 HCl → conflict.

Wait: H⁺ is 4 → need 4 HCl → gives 4 Cl⁻

But 2CeCl₃ needs 6 Cl⁻ → so need more Cl⁻.

So additional 2 HCl → total 6 HCl

But then H⁺ = 6 → so 6 H⁺ → gives 3 H₂O (6 H → 3 H₂O)

Now adjust:

From earlier:
2CeO₂ + 4H⁺ + 2I⁻ → 2Ce³⁺ + I₂ + 2H₂O

But now H⁺ = 6 → so let’s use:

2CeO₂ + 6H⁺ + 2I⁻ → 2Ce³⁺ + I₂ + 3H₂O

Now check O: 4 O → 3 H₂O → 3 O → missing 1 O? No: 2CeO₂ has 4 O → 3 H₂O has 3 O → not balanced.

Wait: 2CeO₂ → 4 O → need 4 H₂O → 8 H → so 8 H⁺

So:

2CeO₂ + 8H⁺ + 2I⁻ → 2Ce³⁺ + I₂ + 4H₂O

Electrons: 2Ce⁴⁺ → 2Ce³⁺ + 2e⁻

2I⁻ → I₂ + 2e⁻ → matches.

So 2CeO₂ + 8H⁺ + 2I⁻ → 2Ce³⁺ + I₂ + 4H₂O

Now add counterions.

H⁺ from HCl → 8 HCl

I⁻ from 2 KI

K⁺ → 2 KCl

Ce³⁺ → 2 CeCl₃ → needs 6 Cl⁻ → so 6 Cl⁻ from HCl

But HCl provides 8 Cl⁻ → 8 Cl⁻ total

Used: 6 in CeCl₃ → 2 in KCl → perfect.

So:

> 2CeO₂ + 2KI + 8HCl → 2KCl + 2CeCl₃ + I₂ + 4H₂O

Check:

| Atom | Left | Right |
|------|------|-------|
| Ce | 2 | 2 |
| O | 4 | 4 (in H₂O) |
| K | 2 | 2 |
| I | 2 | 2 |
| H | 8 | 8 (in 4 H₂O) |
| Cl | 8 | 2 (KCl) + 6 (CeCl₃) = 8 |

✔ Balanced.

Answer:
> 2CeO₂ + 2KI + 8HCl → 2KCl + 2CeCl₃ + I₂ + 4H₂O

---

3. Ho + H₂O → Ho(OH)₃ + H₂

Holmium metal reacts with water.

Ho → Ho³⁺ → Ho(OH)₃

H₂O → H₂ → H⁺ reduced

Balancing:

Ho → Ho(OH)₃ → needs 3 OH⁻ → but comes from water.

Also produces H₂.

Assume:
Ho + H₂O → Ho(OH)₃ + H₂

Left: Ho, 2H, 1O
Right: Ho, 3O, 3H, 2H → 5H, 3O → not balanced.

Try:

Ho + 3H₂O → Ho(OH)₃ + 3/2 H₂ → ×2

→ 2Ho + 6H₂O → 2Ho(OH)₃ + 3H₂

Check:

Left: 2Ho, 12H, 6O
Right: 2Ho, 6O, 6H (in OH) + 6H (in 3H₂) = 12H → OK

Yes.

Answer:
> 2Ho + 6H₂O → 2Ho(OH)₃ + 3H₂

---

4. IrCl₃ + NaOH → Ir₂O₃ + HCl + NaCl

IrCl₃ → Ir₂O₃ → oxidation state?

Ir in IrCl₃: +3
In Ir₂O₃: +3 → no change → not redox.

So likely hydrolysis.

Write:

2IrCl₃ + 6NaOH → Ir₂O₃ + 6NaCl + 3H₂O

But product has HCl? That doesn’t make sense.

Wait: products are Ir₂O₃, HCl, NaCl → HCl is acid, but NaOH is base → can't have both.

Perhaps it's:

IrCl₃ + NaOH → Ir(OH)₃ + NaCl → then Ir(OH)₃ → Ir₂O₃ + H₂O

But here it says Ir₂O₃ directly.

So perhaps:

2IrCl₃ + 6NaOH → Ir₂O₃ + 6NaCl + 3H₂O

But the problem lists HCl as product — that can't be.

Unless it's wrong.

Wait: maybe it's:

IrCl₃ + NaOH → Ir₂O₃ + NaCl + HCl → but HCl and NaOH would react.

So probably typo — or it's meant to be:

IrCl₃ + NaOH → Ir(OH)₃ + NaCl → but Ir(OH)₃ dehydrates.

But given products: Ir₂O₃, HCl, NaCl → impossible unless HCl is produced.

Alternative: maybe it's a disproportionation?

But Ir is +3 in both.

No.

Wait — perhaps it's:

2IrCl₃ + 6NaOH → Ir₂O₃ + 6NaCl + 3H₂O

Then HCl is not needed.

But the problem says "HCl" as product — so maybe error.

Alternatively, could be:

IrCl₃ + NaOH → Ir₂O₃ + NaCl + HCl — but that implies HCl is formed from NaOH — impossible.

So likely a mistake in problem — should be H₂O instead of HCl.

But assuming it's correct, perhaps it's:

IrCl₃ + NaOH → Ir₂O₃ + NaCl + HCl

But that would require:

2IrCl₃ + 6NaOH → Ir₂O₃ + 6NaCl + 6HCl → but then H⁺ and OH⁻ cancel.

No.

Best guess: HCl is a typo, should be H₂O

So:

> 2IrCl₃ + 6NaOH → Ir₂O₃ + 6NaCl + 3H₂O

But since HCl is listed, perhaps:

Wait — maybe it's written as:

IrCl₃ + NaOH → Ir₂O₃ + NaCl + HCl

But that can't happen.

Alternatively, if we write:

2IrCl₃ + 6NaOH → Ir₂O₃ + 6NaCl + 3H₂O

Then HCl is not present.

So I think HCl is a typo — should be H₂O.

So:

Answer:
> 2IrCl₃ + 6NaOH → Ir₂O₃ + 6NaCl + 3H₂O

---

5. MoO₃ + Zn + H₂SO₄ → Mo₂O₃ + ZnSO₄ + H₂O

MoO₃ → Mo₂O₃ → Mo changes from +6 to +3 → reduction

Zn → Zn²⁺ → oxidation

So redox.

MoO₃ → Mo₂O₃: 2 Mo from +6 to +3 → each gains 3e⁻ → total 6e⁻ gained

Zn → Zn²⁺ + 2e⁻ → so need 3 Zn → 6e⁻

So:

2MoO₃ + 3Zn → Mo₂O₃ + 3Zn²⁺

Now sulfate: ZnSO₄ → need SO₄²⁻ → from H₂SO₄

So 3 ZnSO₄ → 3 H₂SO₄

Now H₂SO₄ provides H⁺ and SO₄²⁻

But H⁺ must go to H₂O

H₂O: from H⁺ and O

Left: 2MoO₃ → 6 O

Right: Mo₂O₃ → 3 O → so 3 O left → need 3 H₂O → 6 H → from 3 H₂SO₄ → 6 H → OK

So:

2MoO₃ + 3Zn + 3H₂SO₄ → Mo₂O₃ + 3ZnSO₄ + 3H₂O

Check:

| Atom | Left | Right |
|------|------|-------|
| Mo | 2 | 2 |
| O | 6 + 12 = 18? Wait: MoO₃: 2×3=6 O; H₂SO₄: 3×4=12 O → total 18 O
Right: Mo₂O₃: 3 O; 3ZnSO₄: 3×4=12 O; 3H₂O: 3 O → total 3+12+3=18 → OK
H: 6 → 6 → OK
S: 3 → 3 → OK
Zn: 3 → 3 → OK

✔ Balanced.

Answer:
> 2MoO₃ + 3Zn + 3H₂SO₄ → Mo₂O₃ + 3ZnSO₄ + 3H₂O

---

6. Cu + H₂SO₄ → CuSO₄ + SO₂ + H₂O

Cu → Cu²⁺ → oxidation

S in H₂SO₄: +6 → SO₂: +4 → reduction

So redox.

Cu → Cu²⁺ + 2e⁻

S⁶⁺ + 2e⁻ → S⁴⁺ → so 1 S reduced per Cu

So:

Cu + H₂SO₄ → CuSO₄ + SO₂ + H₂O

But H₂SO₄ provides SO₄²⁻ for CuSO₄, and another S for SO₂.

So need 2 H₂SO₄: one for CuSO₄, one for SO₂

Try:

Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O

Check:

Left: Cu, 4H, 2S, 8O
Right: Cu, S (in CuSO₄), S (in SO₂), 2H₂O → 4H, 2O → total O: 4 (CuSO₄) + 2 (SO₂) + 2 (H₂O) = 8 → OK
H: 4 → 4 → OK
S: 2 → 2 → OK

Yes.

Answer:
> Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O

---

7. (NH₄)₂SO₄ + KOH → K₂SO₄ + NH₃ + H₂O

Double displacement.

(NH₄)₂SO₄ + 2KOH → K₂SO₄ + 2NH₃ + 2H₂O

Because NH₄⁺ + OH⁻ → NH₃ + H₂O

So:

Left: 2N, 8H, S, 4O, 2K, 2O, 2H → wait

(NH₄)₂SO₄: 2N, 8H, S, 4O
2KOH: 2K, 2O, 2H → total: 2N, 10H, S, 6O, 2K

Right: K₂SO₄: 2K, S, 4O
2NH₃: 2N, 6H
2H₂O: 4H, 2O → total: 2N, 10H, S, 6O, 2K → OK

✔

Answer:
> (NH₄)₂SO₄ + 2KOH → K₂SO₄ + 2NH₃ + 2H₂O

---

8. Ferric sulfide + oxygen gas → ferric oxide + sulfur dioxide

Ferric sulfide = Fe₂S₃
Ferric oxide = Fe₂O₃
Sulfur dioxide = SO₂

So:

Fe₂S₃ + O₂ → Fe₂O₃ + SO₂

Balance:

Fe: 2 → 2 → OK
S: 3 → 3 SO₂ → so 3 SO₂

So:

Fe₂S₃ + O₂ → Fe₂O₃ + 3SO₂

Now O: right: 3 (in Fe₂O₃) + 6 (in 3SO₂) = 9 O → need 9/2 O₂

So:

Fe₂S₃ + 9/2 O₂ → Fe₂O₃ + 3SO₂

Multiply by 2:

> 2Fe₂S₃ + 9O₂ → 2Fe₂O₃ + 6SO₂

Check:

Fe: 4 → 4
S: 6 → 6
O: 18 → 6 (Fe₂O₃) + 12 (SO₂) = 18 → OK

✔

Answer:
> 2Fe₂S₃ + 9O₂ → 2Fe₂O₃ + 6SO₂

---

9. Ammonia + oxygen gas → nitrogen monoxide + water

NH₃ + O₂ → NO + H₂O

Redox: N from -3 to +2 → loses 5e⁻
O from 0 to -2 → gains 2e⁻ per O

So:

4NH₃ → 4NO → 4N from -3 to +2 → 4×5 = 20e⁻ lost
5O₂ → 10 O → each gains 2e⁻ → 20e⁻ gained

So:

4NH₃ + 5O₂ → 4NO + 6H₂O

Check H: 12 → 12 → OK
O: 10 → 4 + 6 = 10 → OK

✔

Answer:
> 4NH₃ + 5O₂ → 4NO + 6H₂O

---

10. Calcium hydroxide + nitrous acid → calcium nitrite + water

Ca(OH)₂ + HNO₂ → Ca(NO₂)₂ + H₂O

Ca(OH)₂ has 2 OH⁻, HNO₂ has H⁺ → neutralization

So:

Ca(OH)₂ + 2HNO₂ → Ca(NO₂)₂ + 2H₂O

Check:

Ca: 1 → 1
O: 2 + 2 = 4 → 4 in Ca(NO₂)₂ + 2 in H₂O → 4 + 2 = 6? Wait.

Ca(OH)₂: 2 O, 2 H
2HNO₂: 2 H, 2 N, 4 O → total: 2H, 2H = 4H, 2N, 6O, Ca

Right: Ca(NO₂)₂: Ca, 2N, 4O
2H₂O: 4H, 2O → total: 4H, 2N, 6O, Ca → OK

✔

Answer:
> Ca(OH)₂ + 2HNO₂ → Ca(NO₂)₂ + 2H₂O

---

11. Aluminum + plumbous nitrate → aluminum nitrate + lead

Plumbous = Pb²⁺
Aluminum = Al
Aluminum nitrate = Al(NO₃)₃
Lead = Pb

So:

Al + Pb(NO₃)₂ → Al(NO₃)₃ + Pb

Balance:

Al: 1 → 1
Pb: 1 → 1
NO₃: 2 → 3 → need 3 Pb(NO₃)₂ and 2 Al(NO₃)₃

So:

2Al + 3Pb(NO₃)₂ → 2Al(NO₃)₃ + 3Pb

Check:

Al: 2 → 2
Pb: 3 → 3
NO₃: 6 → 6 → OK

✔

Answer:
> 2Al + 3Pb(NO₃)₂ → 2Al(NO₃)₃ + 3Pb

---

12. Methane + oxygen gas → carbon dioxide + water vapor

CH₄ + O₂ → CO₂ + H₂O

C: 1 → 1
H: 4 → 2 → need 2 H₂O
O: 2 → 2 (CO₂) + 1 (H₂O) = 3 → need 3/2 O₂

So:

CH₄ + 3/2 O₂ → CO₂ + 2H₂O

×2:

> 2CH₄ + 3O₂ → 2CO₂ + 4H₂O

✔

Answer:
> 2CH₄ + 3O₂ → 2CO₂ + 4H₂O

---

13. Aluminum nitrate + sulfuric acid → aluminum sulfate + nitric acid

Al(NO₃)₃ + H₂SO₄ → Al₂(SO₄)₃ + HNO₃

Al: 2 → so 2 Al(NO₃)₃
SO₄: 3 → so 3 H₂SO₄
HNO₃: 6 → so 6 HNO₃

So:

2Al(NO₃)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6HNO₃

Check:

Al: 2 → 2
N: 6 → 6
O: many → skip
H: 6 → 6
S: 3 → 3
SO₄: 3 → 3
NO₃: 6 → 6

✔

Answer:
> 2Al(NO₃)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6HNO₃

---

14. Iron + hydrochloric acid → ferric chloride + hydrogen

Iron = Fe
Hydrochloric acid = HCl
Ferric chloride = FeCl₃
Hydrogen = H₂

Fe → Fe³⁺ → loses 3e⁻
2H⁺ → H₂ + 2e⁻ → so need 3 H⁺ per Fe → 3/2 H₂

So:

2Fe + 6HCl → 2FeCl₃ + 3H₂

Check:

Fe: 2 → 2
Cl: 6 → 6
H: 6 → 6 → OK

✔

Answer:
> 2Fe + 6HCl → 2FeCl₃ + 3H₂

---

15. Phosphoric acid + magnesium hydroxide → magnesium phosphate + water

H₃PO₄ + Mg(OH)₂ → Mg₃(PO₄)₂ + H₂O

Mg₃(PO₄)₂ → 3 Mg, 2 PO₄

So need 3 Mg(OH)₂ and 2 H₃PO₄

2H₃PO₄ + 3Mg(OH)₂ → Mg₃(PO₄)₂ + 6H₂O

Check:

H: 6 + 6 = 12 → 12 in 6H₂O → OK
O: 8 + 6 = 14 → 8 in Mg₃(PO₄)₂ + 6 in H₂O = 14 → OK
P: 2 → 2
Mg: 3 → 3

✔

Answer:
> 2H₃PO₄ + 3Mg(OH)₂ → Mg₃(PO₄)₂ + 6H₂O

---

16. Ferric bromide + ammonium sulfide → ferric sulfide + ammonium bromide

FeBr₃ + (NH₄)₂S → Fe₂S₃ + NH₄Br

Fe: 2 → so 2 FeBr₃
S: 3 → so 3 (NH₄)₂S
Br: 6 → so 6 NH₄Br

So:

2FeBr₃ + 3(NH₄)₂S → Fe₂S₃ + 6NH₄Br

Check:

Fe: 2 → 2
Br: 6 → 6
S: 3 → 3
N: 6 → 6
H: 24 → 24 → OK

✔

Answer:
> 2FeBr₃ + 3(NH₄)₂S → Fe₂S₃ + 6NH₄Br

---

17. Benzene (C₆H₆) + oxygen gas → carbon dioxide + water vapor

C₆H₆ + O₂ → CO₂ + H₂O

C: 6 → 6 CO₂
H: 6 → 3 H₂O

So:

C₆H₆ + O₂ → 6CO₂ + 3H₂O

O: right: 12 + 3 = 15 → need 15/2 O₂

So:

C₆H₆ + 15/2 O₂ → 6CO₂ + 3H₂O

×2:

> 2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O

✔

Answer:
> 2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O

---

## ✔ Final Answers:

1. Ca(AlO₂)₂ + 8HCl → 2AlCl₃ + CaCl₂ + 4H₂O
2. 2CeO₂ + 2KI + 8HCl → 2KCl + 2CeCl₃ + I₂ + 4H₂O
3. 2Ho + 6H₂O → 2Ho(OH)₃ + 3H₂
4. 2IrCl₃ + 6NaOH → Ir₂O₃ + 6NaCl + 3H₂O *(assuming HCl is typo)*
5. 2MoO₃ + 3Zn + 3H₂SO₄ → Mo₂O₃ + 3ZnSO₄ + 3H₂O
6. Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O
7. (NH₄)₂SO₄ + 2KOH → K₂SO₄ + 2NH₃ + 2H₂O
8. 2Fe₂S₃ + 9O₂ → 2Fe₂O₃ + 6SO₂
9. 4NH₃ + 5O₂ → 4NO + 6H₂O
10. Ca(OH)₂ + 2HNO₂ → Ca(NO₂)₂ + 2H₂O
11. 2Al + 3Pb(NO₃)₂ → 2Al(NO₃)₃ + 3Pb
12. 2CH₄ + 3O₂ → 2CO₂ + 4H₂O
13. 2Al(NO₃)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6HNO₃
14. 2Fe + 6HCl → 2FeCl₃ + 3H₂
15. 2H₃PO₄ + 3Mg(OH)₂ → Mg₃(PO₄)₂ + 6H₂O
16. 2FeBr₃ + 3(NH₄)₂S → Fe₂S₃ + 6NH₄Br
17. 2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O

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