Chemistry worksheet for balancing chemical equations.

Let's solve each of these chemical equations by balancing them. We'll go step-by-step and provide the correct coefficients for each reaction.

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1.

Na(s) + Cl₂(g) → NaCl(s)

- Na: 1 on left, 1 on right (in NaCl)
- Cl: 2 on left (Cl₂), 1 on right (in NaCl)

To balance Cl, need 2 NaCl → so 2 Na on left.

✔ 2 Na + 1 Cl₂ → 2 NaCl

Answer: `2 1 2`

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2.

Na(s) + O₂(g) → Na₂O(s)

- O: 2 on left (O₂), 1 on right (in Na₂O)
- Na: 1 on left, 2 on right

So, need 2 Na on left → but then we have 2 Na and 1 O₂ → makes 2 Na₂O? No — wait:

We need 2 Na to make one Na₂O, but O₂ has 2 oxygen atoms → can make 2 Na₂O, which needs 4 Na

So:
- 4 Na + 1 O₂ → 2 Na₂O

✔ 4 Na + 1 O₂ → 2 Na₂O

Answer: `4 1 2`

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3.

Mg(s) + O₂(g) → MgO(s)

- O: 2 on left, 1 on right → need 2 MgO
- Then Mg: 2 on right → need 2 Mg on left

✔ 2 Mg + 1 O₂ → 2 MgO

Answer: `2 1 2`

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4.

Mg(s) + HCl(aq) → MgCl₂(aq) + H₂(g)

- Mg: 1 on both sides → OK
- Cl: 1 in HCl, 2 in MgCl₂ → need 2 HCl
- H: 2 in 2 HCl → gives 1 H₂

✔ 1 Mg + 2 HCl → 1 MgCl₂ + 1 H₂

Answer: `1 2 1 1`

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5.

H₂O₂(aq) → H₂(g) + O₂(g)

Decomposition of hydrogen peroxide.

- H: 2 on left, 2 on right (in H₂) → OK
- O: 2 on left, 2 on right (in O₂) → OK

But: H₂O₂ → H₂ + O₂

Wait: left: 2 H, 2 O
Right: 2 H, 2 O → seems balanced?

But this is not correct because H₂O₂ decomposes into H₂O and O₂, not H₂ and O₂ directly.

But the equation given is: H₂O₂ → H₂ + O₂

Let’s check atoms:
- Left: H=2, O=2
- Right: H=2, O=2 → appears balanced

But it's not chemically accurate, but we are just balancing as written.

So:
H₂O₂ → H₂ + O₂

Atoms:
H: 2 = 2 ✔
O: 2 = 2 ✔

So it's already balanced?

Wait — but oxygen is diatomic, so O₂ is fine.

But let's see if we need coefficients.

Try:
2 H₂O₂ → 2 H₂ + 1 O₂

Now:
- H: 4 on left, 4 on right → OK
- O: 4 on left, 2 on right → NO

Wait! So if we do:

2 H₂O₂ → 2 H₂ + 1 O₂ → O: 4 vs 2 → imbalance

Actually, correct decomposition is:

2 H₂O₂ → 2 H₂O + O₂

But here it says H₂O₂ → H₂ + O₂

That would require:
2 H₂O₂ → 2 H₂ + O₂ → now check:
- H: 4 = 4 ✔
- O: 4 = 2 ✘

No. So only way to balance is:

2 H₂O₂ → 2 H₂ + 2 O₂ → O: 4 = 4, H: 4 = 4 → but that's not real.

Wait — no, the correct stoichiometry for H₂O₂ → H₂ + O₂ is not possible without changing oxidation states.

But again, we are just balancing as written.

Let’s suppose:
a H₂O₂ → b H₂ + c O₂

H: 2a = 2b → a = b
O: 2a = 2c → a = c

So a = b = c

So: H₂O₂ → H₂ + O₂ → balanced with coefficient 1 for all?

But:
Left: H=2, O=2
Right: H=2, O=2 → YES!

So it is balanced as written.

✔ 1 H₂O₂ → 1 H₂ + 1 O₂

Answer: `1 1 1`

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6.

Al(s) + HF(aq) → AlF₃(aq) + H₂(g)

- Al: 1 on both sides → OK
- F: 1 in HF, 3 in AlF₃ → need 3 HF
- H: 3 in 3 HF → produces 3/2 H₂ → so use 3/2 H₂ → multiply whole equation by 2

Start:
Al + 3 HF → AlF₃ + 3/2 H₂

Multiply by 2:
2 Al + 6 HF → 2 AlF₃ + 3 H₂

✔ 2 Al + 6 HF → 2 AlF₃ + 3 H₂

Answer: `2 6 2 3`

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7.

NaClO₃(aq) → NaCl(aq) + O₂(g)

This is decomposition of sodium chlorate.

- Na: 1 on both sides → OK
- Cl: 1 on both → OK
- O: 3 on left, 2 on right → need to balance O

Let’s try:
2 NaClO₃ → 2 NaCl + 3 O₂

Check:
- Na: 2 = 2 ✔
- Cl: 2 = 2 ✔
- O: 6 = 6 ✔

✔ 2 NaClO₃ → 2 NaCl + 3 O₂

Answer: `2 2 3`

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8.

ZnS(s) + AlP → Zn₃P₂(s) + Al₂S₃(s)

This is a double displacement.

Look at products:
- Zn₃P₂ → needs 3 Zn and 2 P
- Al₂S₃ → needs 2 Al and 3 S

Reactants:
- ZnS → provides Zn and S
- AlP → provides Al and P

So we need:
- For Zn₃P₂: 3 Zn and 2 P
- For Al₂S₃: 2 Al and 3 S

So total:
- Zn: 3 → from 3 ZnS
- S: 3 → from 3 ZnS
- P: 2 → from 2 AlP
- Al: 2 → from 2 AlP

So:
3 ZnS + 2 AlP → 1 Zn₃P₂ + 1 Al₂S₃

Check atoms:
- Zn: 3 = 3 ✔
- S: 3 = 3 ✔
- Al: 2 = 2 ✔
- P: 2 = 2 ✔

✔ 3 ZnS + 2 AlP → 1 Zn₃P₂ + 1 Al₂S₃

Answer: `3 2 1 1`

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9.

Ag₂S(s) → Ag(s) + S₈(s)

- Ag: 2 on left, 1 on right → need 2 Ag
- S: 1 on left, 8 on right → need 8 Ag₂S to get 8 S

So:
8 Ag₂S → 16 Ag + 1 S₈

Check:
- Ag: 16 = 16 ✔
- S: 8 = 8 ✔

✔ 8 Ag₂S → 16 Ag + 1 S₈

Answer: `8 16 1`

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10.

Ba(NO₃)₂(aq) + H₂CO₃(aq) → BaCO₃(s) + HNO₃(aq)

Double displacement.

- Ba: 1 on both → OK
- CO₃: 1 on both → OK
- NO₃: 2 on left, 1 on right → need 2 HNO₃
- H: 2 on left (H₂CO₃), 2 on right (2 HNO₃) → OK

So:
1 Ba(NO₃)₂ + 1 H₂CO₃ → 1 BaCO₃ + 2 HNO₃

✔ 1 Ba(NO₃)₂ + 1 H₂CO₃ → 1 BaCO₃ + 2 HNO₃

Answer: `1 1 1 2`

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11.

Pb(OH)₄(aq) + Cu₂O(aq) → PbO₂(s) + CuOH(aq)

Check atoms:

Left:
- Pb: 1
- O: 4 from OH + 1 from Cu₂O → total O: 5?
Wait: Pb(OH)₄ → Pb, 4 O, 4 H
Cu₂O → 2 Cu, 1 O

Total left:
- Pb: 1
- Cu: 2
- O: 4 + 1 = 5
- H: 4

Right:
- PbO₂: Pb, 2 O
- CuOH: Cu, O, H

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

Check:
- Pb: 1 = 1 ✔
- Cu: 2 = 2 ✔
- O: left = 4 (from OH) + 1 (from Cu₂O) = 5
Right: 2 (PbO₂) + 2×1 (from 2 CuOH) = 4 ✘

Not balanced.

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

Still O: 5 vs 4

Need more O on right?

Wait: maybe coefficient change.

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

H: 4 = 2 × 1 = 2 → no, 4 H on left, only 2 on right

So need 4 CuOH → 4 H

But Cu: left has 2 Cu → so need 2 CuOH

Contradiction.

Wait: Cu₂O has 2 Cu → so must produce 2 CuOH

But CuOH has one Cu → so 2 CuOH → 2 Cu

H: 2 H → but left has 4 H from Pb(OH)₄

So H mismatch.

So need to adjust.

Let’s write:

Pb(OH)₄ → PbO₂ + 2 H₂O

But here it's reacting with Cu₂O.

Maybe:

Pb(OH)₄ + Cu₂O → PbO₂ + 2 CuOH

Now check:
- Pb: 1 = 1 ✔
- Cu: 2 = 2 ✔
- O: left: 4 (from OH) + 1 (Cu₂O) = 5
Right: 2 (PbO₂) + 2 (from 2 CuOH) = 4 ✘
- H: 4 = 2 ✘

So not balanced.

Alternative idea: maybe water is formed?

But equation doesn't show water.

Wait — perhaps it's:

Pb(OH)₄ + Cu₂O → PbO₂ + 2 CuOH

But atoms don’t match.

Try balancing:

Let’s suppose:
a Pb(OH)₄ + b Cu₂O → c PbO₂ + d CuOH

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

From H: d = 4a
From Cu: d = 2b → so 4a = 2b → b = 2a
From Pb: c = a

Now O:
Left: 4a + b = 4a + 2a = 6a
Right: 2c + d = 2a + 4a = 6a → ✔

Perfect!

So a = 1 → b = 2, c = 1, d = 4

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

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

Answer: `1 2 1 4`

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12.

H₃PO₄(aq) + HCl(aq) → PCl₅(s) + H₂O(l)

Wait — this looks wrong.

H₃PO₄ has P, H, O
HCl has H, Cl
PCl₅ has P, Cl
H₂O has H, O

So:
P: 1 = 1 → OK
But Cl: 1 in HCl, 5 in PCl₅ → need 5 HCl
H: 3 from H₃PO₄ + 5 from HCl = 8 H → H₂O has 2 H → need 4 H₂O
O: 4 from H₃PO₄ → 4 H₂O has 4 O → OK

So:
1 H₃PO₄ + 5 HCl → 1 PCl₅ + 4 H₂O

Check:
- P: 1 = 1 ✔
- Cl: 5 = 5 ✔
- H: 3 + 5 = 8; 4 H₂O → 8 H ✔
- O: 4 = 4 ✔

✔ 1 H₃PO₄ + 5 HCl → 1 PCl₅ + 4 H₂O

Answer: `1 5 1 4`

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13.

C₂H₆(s) + O₂(g) → CO₂(g) + H₂O(g)

Combustion of ethane.

C₂H₆ + O₂ → CO₂ + H₂O

Balance C: 2 → 2 CO₂
H: 6 → 3 H₂O
Now O: right: 2×2 + 3×1 = 4 + 3 = 7 O → need 7/2 O₂

So:
C₂H₆ + 7/2 O₂ → 2 CO₂ + 3 H₂O

Multiply by 2:
2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

✔ 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

Answer: `2 7 4 6`

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14.

C₃H₈(s) + O₂(g) → CO₂(g) + H₂O(g)

Propane combustion.

C₃H₈ + O₂ → CO₂ + H₂O

C: 3 → 3 CO₂
H: 8 → 4 H₂O
O: right: 3×2 + 4×1 = 6 + 4 = 10 → need 5 O₂

So:
1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

✔ 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

Answer: `1 5 3 4`

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15.

C₅H₁₀(s) + O₂(g) → CO₂(g) + H₂O(g)

C₅H₁₀ (pentene or cyclopentane) combustion.

C: 5 → 5 CO₂
H: 10 → 5 H₂O
O: right: 5×2 + 5×1 = 10 + 5 = 15 → need 15/2 O₂

So:
C₅H₁₀ + 15/2 O₂ → 5 CO₂ + 5 H₂O

Multiply by 2:
2 C₅H₁₀ + 15 O₂ → 10 CO₂ + 10 H₂O

✔ 2 C₅H₁₀ + 15 O₂ → 10 CO₂ + 10 H₂O

Answer: `2 15 10 10`

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✔ Final Answers:

1. `2 1 2`
2. `4 1 2`
3. `2 1 2`
4. `1 2 1 1`
5. `1 1 1`
6. `2 6 2 3`
7. `2 2 3`
8. `3 2 1 1`
9. `8 16 1`
10. `1 1 1 2`
11. `1 2 1 4`
12. `1 5 1 4`
13. `2 7 4 6`
14. `1 5 3 4`
15. `2 15 10 10`

Let me know if you'd like these explained further!