Chemistry worksheet for balancing chemical equations.

Let's solve each of these chemical equations by balancing them. We'll go step-by-step and explain the logic for each.

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1.

Na(s) + Cl₂(g) → NaCl(s)

- Left: 1 Na, 2 Cl
- Right: 1 Na, 1 Cl
→ Need 2 NaCl to balance Cl, then 2 Na on left.

✔ 2 Na + 1 Cl₂ → 2 NaCl

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2.

Na(s) + O₂(g) → Na₂O(s)

- Left: 1 Na, 2 O
- Right: 2 Na, 1 O
→ Need 4 Na and 2 O₂ to make 2 Na₂O (since Na₂O has 2 Na and 1 O)

Try:
- 4 Na + 1 O₂ → 2 Na₂O
→ Check:
Left: 4 Na, 2 O
Right: 4 Na, 2 O ✔

✔ 4 Na + 1 O₂ → 2 Na₂O

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3.

Mg(s) + O₂(g) → MgO(s)

- Left: 1 Mg, 2 O
- Right: 1 Mg, 1 O
→ Need 2 MgO → need 2 Mg and 1 O₂

✔ 2 Mg + 1 O₂ → 2 MgO

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4.

Mg(s) + HCl(aq) → MgCl₂(aq) + H₂(g)

- MgCl₂ has 2 Cl → need 2 HCl
- Then H: 2 H on left → H₂ on right is fine

So:
- Mg + 2 HCl → MgCl₂ + H₂

✔ 1 Mg + 2 HCl → 1 MgCl₂ + 1 H₂

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5.

H₂O₂(aq) → H₂(g) + O₂(g)

Decomposition of hydrogen peroxide.

- Left: 2 H, 2 O
- Right: 2 H in H₂, 2 O in O₂ → but O₂ is diatomic

But if we write:
- 2 H₂O₂ → 2 H₂ + 1 O₂
Check:
Left: 4 H, 4 O
Right: 4 H, 2 O ✘ (only 2 O from O₂)

Wait — O₂ gives 2 O atoms. So 2 H₂O₂ → 2 H₂ + 1 O₂
→ 4 H, 4 O on both sides?
No:
- 2 H₂O₂ → 4 H, 4 O
- 2 H₂ → 4 H
- 1 O₂ → 2 O → only 2 O ✘

We need 2 O₂? But that would be 4 O.

Actually:
- 2 H₂O₂ → 2 H₂ + 1 O₂ → 4 H, 4 O → 4 H, 2 O ✘

Wait — O₂ has 2 oxygen atoms. So:

2 H₂O₂ → 2 H₂ + 1 O₂ → 4 H, 4 O → 4 H, 2 O ✘

Wait! Mistake: 2 H₂O₂ has 4 H and 4 O
2 H₂ has 4 H
1 O₂ has 2 O → only 2 O → not balanced.

So need 2 O₂ → 4 O

But then:
2 H₂O₂ → 2 H₂ + 2 O₂ → 4 H, 4 O → 4 H, 4 O ✔

But wait — this is not correct because H₂O₂ decomposition is:

2 H₂O₂ → 2 H₂O + O₂, not H₂.

But here it says → H₂ and O₂ — so it's not standard.

But let’s assume it’s possible.

So:
H₂O₂ → H₂ + O₂
Atoms:
Left: 2 H, 2 O
Right: 2 H, 2 O → seems balanced?

But O₂ is molecule with 2 O atoms → yes.

So:
1 H₂O₂ → 1 H₂ + 1 O₂ → H: 2=2, O: 2=2 ✔

But this is not chemically accurate — but for balancing purposes, if the reaction is written as such, it can be balanced.

Wait: 1 H₂O₂ → 1 H₂ + 1 O₂
→ H: 2 = 2, O: 2 = 2 ✔

But this implies breaking into H₂ and O₂ — which is not typical, but mathematically possible.

So:
✔ 1 H₂O₂ → 1 H₂ + 1 O₂

But actually, in reality, it's 2 H₂O₂ → 2 H₂O + O₂, but since the problem says H₂ and O₂, we’ll go with it.

So: 1 H₂O₂ → 1 H₂ + 1 O₂

But check:
- Left: H=2, O=2
- Right: H=2, O=2 ✔

Yes, it works.

✔ 1 H₂O₂ → 1 H₂ + 1 O₂

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6.

Al(s) + HF(aq) → AlF₃(aq) + H₂(g)

- AlF₃ has 3 F → need 3 HF
- Then H: 3 H → H₂ needs even number → so need 6 H → 6 HF

Try:
- 2 Al + 6 HF → 2 AlF₃ + 3 H₂
Check:
- Al: 2 = 2
- F: 6 = 6
- H: 6 = 6 (in 3 H₂)

✔ 2 Al + 6 HF → 2 AlF₃ + 3 H₂

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7.

NaClO₃(aq) → NaCl(aq) + O₂(g)

This is a decomposition.

- Left: Na, Cl, 3 O
- Right: Na, Cl, 2 O in O₂ → need to balance O

So need multiple O₂.

Let’s suppose:
- 2 NaClO₃ → 2 NaCl + 3 O₂
Check:
- Na: 2 = 2
- Cl: 2 = 2
- O: 6 = 6 (3×2) ✔

✔ 2 NaClO₃ → 2 NaCl + 3 O₂

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8.

ZnS(s) + AlP → Zn₃P₂(s) + Al₂S₃(s)

This is a double displacement.

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

So we need:
- ZnS: provides Zn and S
- AlP: provides Al and P

To get:
- 3 Zn → 3 ZnS
- 2 P → 2 AlP
- 2 Al → 2 AlP
- 3 S → 3 ZnS

So:
- 3 ZnS + 2 AlP → Zn₃P₂ + Al₂S₃

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

✔ 3 ZnS + 2 AlP → 1 Zn₃P₂ + 1 Al₂S₃

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9.

Ag₂S(s) → Ag(s) + S₈(s)

- Ag₂S → 2 Ag, 1 S
- But S product is S₈ → 8 S atoms

So need 8 S atoms → need 8 Ag₂S → 16 Ag

Then:
- 8 Ag₂S → 16 Ag + 1 S₈

Check:
- Ag: 16 = 16
- S: 8 = 8 ✔

✔ 8 Ag₂S → 16 Ag + 1 S₈

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10.

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

Double displacement.

- Ba(NO₃)₂ → Ba²⁺ and 2 NO₃⁻
- H₂CO₃ → 2 H⁺ and CO₃²⁻
→ Form BaCO₃ (insoluble) and 2 HNO₃

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

Check:
- Ba: 1 = 1
- N: 2 = 2
- O: many, but count atoms:
- Left: Ba(NO₃)₂: 1 Ba, 2 N, 6 O; H₂CO₃: 2 H, 1 C, 3 O → total: 1 Ba, 2 N, 2 H, 1 C, 9 O
- Right: BaCO₃: 1 Ba, 1 C, 3 O; 2 HNO₃: 2 H, 2 N, 6 O → total: 1 Ba, 2 N, 2 H, 1 C, 9 O ✔

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

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11.

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

Balance atoms.

First, look at Pb: 1 on both sides.

Now Cu: Cu₂O → 2 Cu → so need 2 CuOH on right.

So:
- Pb(OH)₄ + Cu₂O → PbO₂ + 2 CuOH

Now check O and H.

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

Right:
- PbO₂: Pb, 2 O
- 2 CuOH: 2 Cu, 2 O, 2 H → total: Pb, 2 Cu, 4 O, 2 H

Not balanced: O: 5 vs 4, H: 4 vs 2

Need more H and O on right.

Maybe 2 Pb(OH)₄?

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

Left:
- 2 Pb, 8 O, 8 H from OH; Cu₂O: 2 Cu, 1 O → total: 2 Pb, 2 Cu, 9 O, 8 H

Right:
- 2 PbO₂: 2 Pb, 4 O
- 2 CuOH: 2 Cu, 2 O, 2 H → total: 2 Pb, 2 Cu, 6 O, 2 H → still off

H: 8 vs 2 → need more CuOH

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

Right: 4 CuOH → 4 Cu, 4 O, 4 H → but we have only 2 Cu from Cu₂O → too many Cu

So must keep Cu₂O → 2 Cu → so max 2 CuOH

But H: left has 8 H, right has 2 H → need 8 H → so 8 CuOH → 8 Cu → need 4 Cu₂O

Try:
- 2 Pb(OH)₄ + 4 Cu₂O → 2 PbO₂ + 8 CuOH

Now check:
Left:
- Pb: 2
- O: from 2 Pb(OH)₄: 8 O (from OH), 4 O (from OH groups? Wait: Pb(OH)₄ has 4 O and 4 H)
Actually: Pb(OH)₄ = Pb, 4 O, 4 H
So 2 Pb(OH)₄ → 2 Pb, 8 O, 8 H
- 4 Cu₂O → 8 Cu, 4 O → total: 2 Pb, 8 Cu, 12 O, 8 H

Right:
- 2 PbO₂ → 2 Pb, 4 O
- 8 CuOH → 8 Cu, 8 O, 8 H → total: 2 Pb, 8 Cu, 12 O, 8 H ✔

Perfect!

✔ 2 Pb(OH)₄ + 4 Cu₂O → 2 PbO₂ + 8 CuOH

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12.

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

This is not a standard reaction — but let’s balance.

- PCl₅ has 5 Cl → need 5 HCl
- H₃PO₄ has 1 P, 4 O, 3 H
- HCl: H and Cl

Products:
- PCl₅: P, 5 Cl
- H₂O: H and O

From H₃PO₄: 3 H, 4 O
From 5 HCl: 5 H, 5 Cl

Total reactants: 8 H, 4 O, 1 P, 5 Cl

Products:
- PCl₅: 1 P, 5 Cl
- H₂O: say x H₂O → 2x H, x O

Set:
- H: 8 = 2x → x = 4
- O: 4 = x → x = 4 ✔

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

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13.

C₂H₆(s) + O₂(g) → CO₂(g) + H₂O(g)

Combustion of ethane.

C₂H₆ → 2 C, 6 H

Products:
- CO₂: 1 C, 2 O
- H₂O: 2 H, 1 O

So:
- 2 CO₂ for 2 C
- 3 H₂O for 6 H

So:
- C₂H₆ + O₂ → 2 CO₂ + 3 H₂O

Now O:
- Right: 2×2 + 3×1 = 4 + 3 = 7 O → so need 7/2 O₂

Multiply whole equation by 2:

- 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

Check:
- C: 4 = 4
- H: 12 = 12
- O: 14 = 8 + 6 = 14 ✔

So coefficients:
- 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

But the original is one C₂H₆ → so we can write fractional or use integers.

Since question asks for coefficients, use integers.

✔ 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

But the blank is for one molecule? No — just fill coefficient.

So:
- First blank: 2
- Second: 7
- Third: 4
- Fourth: 6

✔ 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

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14.

C₃H₈(s) + O₂(g) → CO₂(g) + H₂O(g)

Propane combustion.

C₃H₈ → 3 C, 8 H

→ 3 CO₂, 4 H₂O (since 8 H → 4 H₂O)

So:
- C₃H₈ + O₂ → 3 CO₂ + 4 H₂O

O:
- Right: 3×2 + 4×1 = 6 + 4 = 10 O → so 5 O₂

✔ 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

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15.

C₅H₁₀(s) + O₂(g) → CO₂(g) + H₂O(g)

C₅H₁₀ → 5 C, 10 H

→ 5 CO₂, 5 H₂O (since 10 H → 5 H₂O)

O:
- Right: 5×2 + 5×1 = 10 + 5 = 15 O → so 15/2 O₂

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

But if we want smallest integers:
- C₅H₁₀ + 15/2 O₂ → 5 CO₂ + 5 H₂O

But usually write whole numbers.

So:
- 2 C₅H₁₀ + 15 O₂ → 10 CO₂ + 10 H₂O

But since the equation has single molecules, we can use:

✔ 1 C₅H₁₀ + 15/2 O₂ → 5 CO₂ + 5 H₂O — but fractions not allowed.

So better to write:
- 2 C₅H₁₀ + 15 O₂ → 10 CO₂ + 10 H₂O

But the blanks are for each species — so likely expect smallest integers.

But if only one molecule of C₅H₁₀, then use fraction — but instructions say "type a number" — no fractions.

So best to use:

✔ 2 C₅H₁₀ + 15 O₂ → 10 CO₂ + 10 H₂O

But since the first blank is for C₅H₁₀, and it's singular, perhaps they accept:

Wait — the problem says "type the correct coefficient" — so we can use 2.

So:

- C₅H₁₀: 2
- O₂: 15
- CO₂: 10
- H₂O: 10

✔ 2 C₅H₁₀ + 15 O₂ → 10 CO₂ + 10 H₂O

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Final Answers:

1. 2 Na + 1 Cl₂ → 2 NaCl
2. 4 Na + 1 O₂ → 2 Na₂O
3. 2 Mg + 1 O₂ → 2 MgO
4. 1 Mg + 2 HCl → 1 MgCl₂ + 1 H₂
5. 1 H₂O₂ → 1 H₂ + 1 O₂
6. 2 Al + 6 HF → 2 AlF₃ + 3 H₂
7. 2 NaClO₃ → 2 NaCl + 3 O₂
8. 3 ZnS + 2 AlP → 1 Zn₃P₂ + 1 Al₂S₃
9. 8 Ag₂S → 16 Ag + 1 S₈
10. 1 Ba(NO₃)₂ + 1 H₂CO₃ → 1 BaCO₃ + 2 HNO₃
11. 2 Pb(OH)₄ + 4 Cu₂O → 2 PbO₂ + 8 CuOH
12. 1 H₃PO₄ + 5 HCl → 1 PCl₅ + 4 H₂O
13. 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O
14. 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
15. 2 C₅H₁₀ + 15 O₂ → 10 CO₂ + 10 H₂O

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Now, list the coefficients as numbers for each blank:

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. 2, 4, 2, 8
12. 1, 5, 1, 4
13. 2, 7, 4, 6
14. 1, 5, 3, 4
15. 2, 15, 10, 10

✔ These are the correct coefficients to type in.