Balancing Equations worksheet with ten chemical equations to balance, including CH4 + O2 → CO2 + H2O, Na+ + Cl- → NaCl, Al + O2 → Al2O3, N2+ + H2 → NH3, CO(g) + H2(g) → C8H18(l) + H2O, FeO3(s) + CO(g) → Fe(l) + CO2(g), H2SO4 + Pb(OH)4 → Pb(SO4)2 + H2O, Al + HCl → AlCl3 + H2, Ca3(PO4)2 + H2SO4 → CaSO4 + Ca(H2PO4)2, H3PO4 + HCl → PCl5 + H2O, with spaces for coefficients and a header for name and date.

Let's solve each of these chemical equations by balancing them. Balancing a chemical equation means ensuring the number of atoms of each element is equal on both sides (reactants and products), following the Law of Conservation of Mass.

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1. CH₄ + O₂ → CO₂ + H₂O

Step-by-step:
- Carbon: 1 on both sides → OK.
- Hydrogen: 4 on left, 2 on right → need 2 H₂O to get 4 H.
- Oxygen: 2 on left (from O₂), but now we have 2 (CO₂) + 1 (from 2 H₂O) = 3 on right → not balanced.
- Try 2 O₂ → 4 oxygen atoms.
- Now: CO₂ has 2 O, 2 H₂O has 2 O → total 4 O → good.
- So:
- CH₄ + 2O₂ → CO₂ + 2H₂O

✔ Balanced:
CH₄ + 2O₂ → CO₂ + 2H₂O

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2. Na⁺ + Cl⁻ → NaCl

This is an ionic combination. Each Na⁺ combines with one Cl⁻ to form one NaCl.

So:
Na⁺ + Cl⁻ → NaCl

But since it's already 1:1, just add coefficients:

✔ Balanced:
Na⁺ + Cl⁻ → NaCl

(Note: Often written as Na(s) + Cl₂(g) → 2NaCl, but here it’s ions, so this is acceptable.)

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3. Al + O₂ → Al₂O₃

- Al: 1 on left, 2 on right → need 2 Al
- O: 2 on left, 3 on right → LCM of 2 and 3 is 6 → need 3 O₂ and 2 Al₂O₃

Try:
- 4Al + 3O₂ → 2Al₂O₃

Check:
- Al: 4 = 4 → OK
- O: 6 = 6 → OK

✔ Balanced:
4Al + 3O₂ → 2Al₂O₃

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4. N₂ + H₂ → NH₃

- N₂ has 2 N → NH₃ has 1 N → need 2 NH₃
- So: N₂ + H₂ → 2NH₃
- Now H: 2 on left, 6 on right → need 3 H₂

✔ Balanced:
N₂ + 3H₂ → 2NH₃

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5. CO + H₂ → C₈H₁₈ + H₂O

This is synthesis of octane from CO and H₂ — unusual, but let's balance.

We want:
- 8 C → need 8 CO
- 18 H → but H₂ gives 2 H per molecule → need 9 H₂
- But also produces H₂O → which has H and O

Let’s try:
- 8CO + 9H₂ → C₈H₁₈ + H₂O

Now check:
- C: 8 = 8 → OK
- O: 8 on left → only 1 in H₂O → need 8 H₂O
- H: left: 9×2 = 18 → right: C₈H₁₈ has 18 H, plus 8×2 = 16 H from H₂O → total 34 H → too many

Wait — this reaction isn't standard. Let’s think again.

Actually, CO and H₂ can produce hydrocarbons via Fischer-Tropsch, but for C₈H₁₈, we need:

We want to make C₈H₁₈ and H₂O.

Let’s suppose:
- 8CO + x H₂ → C₈H₁₈ + y H₂O

Balance:
- C: 8 = 8 → OK
- O: 8 = y → so y = 8
- H: 2x = 18 + 2×8 = 18 + 16 = 34 → x = 17

So:
8CO + 17H₂ → C₈H₁₈ + 8H₂O

✔ Balanced:
8CO + 17H₂ → C₈H₁₈ + 8H₂O

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6. Fe₂O₃ + CO → Fe + CO₂

This is a reduction of iron ore.

Fe₂O₃ → 2Fe → so need 2Fe on right

CO → CO₂ → each CO gains one O atom

Fe₂O₃ has 3 O atoms → needs 3 CO → gives 3 CO₂

So:
Fe₂O₃ + 3CO → 2Fe + 3CO₂

Check:
- Fe: 2 = 2 → OK
- O: 3 + 3 = 6; right: 3×2 = 6 → OK
- C: 3 = 3 → OK

✔ Balanced:
Fe₂O₃ + 3CO → 2Fe + 3CO₂

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7. H₂SO₄ + Pb(OH)₄ → Pb(SO₄)₂ + H₂O

This is an acid-base neutralization.

Pb(OH)₄ is lead(IV) hydroxide, and Pb(SO₄)₂ has two SO₄²⁻ → so need 2 H₂SO₄

So:
2H₂SO₄ + Pb(OH)₄ → Pb(SO₄)₂ + H₂O

Now H: left: 2×2 = 4 H from H₂SO₄ + 4 H from Pb(OH)₄ = 8 H
Right: H₂O → each has 2 H → need 4 H₂O

O: Check later.

So:
2H₂SO₄ + Pb(OH)₄ → Pb(SO₄)₂ + 4H₂O

Check:
- S: 2 = 2 → OK
- Pb: 1 = 1 → OK
- H: 4 (from H₂SO₄) + 4 (from OH) = 8 → 4 H₂O → 8 H → OK
- O: left: 2×4 = 8 (from H₂SO₄) + 4 (from OH) = 12
Right: Pb(SO₄)₂ has 8 O, 4 H₂O has 4 O → total 12 → OK

✔ Balanced:
2H₂SO₄ + Pb(OH)₄ → Pb(SO₄)₂ + 4H₂O

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8. Al + HCl → AlCl₃ + H₂

Aluminum reacts with acid.

Al → AlCl₃ → needs 3 Cl → so need 3 HCl

But HCl provides H → H₂ gas → 2 H → so need even number of HCl

So use 2Al → 2AlCl₃ → needs 6 Cl → so 6 HCl → gives 6 H → makes 3 H₂

So:
2Al + 6HCl → 2AlCl₃ + 3H₂

Check:
- Al: 2 = 2 → OK
- Cl: 6 = 6 → OK
- H: 6 = 6 → OK

✔ Balanced:
2Al + 6HCl → 2AlCl₃ + 3H₂

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9. Ca₃(PO₄)₂ + H₂SO₄ → CaSO₄ + Ca(H₂PO₄)₂

This is a double displacement reaction.

Note: Ca₃(PO₄)₂ has 3 Ca²⁺ and 2 PO₄³⁻

H₂SO₄ provides SO₄²⁻ and H⁺

Products: CaSO₄ and Ca(H₂PO₄)₂

Ca(H₂PO₄)₂ is calcium dihydrogen phosphate → contains Ca²⁺ and 2 H₂PO₄⁻

So:
- One Ca²⁺ goes to CaSO₄
- Two Ca²⁺ go to Ca(H₂PO₄)₂

So need:
- 1 CaSO₄
- 1 Ca(H₂PO₄)₂ → uses 2 Ca²⁺
- Total 3 Ca²⁺ → matches Ca₃(PO₄)₂

Now, PO₄³⁻ → becomes H₂PO₄⁻ → so each PO₄³⁻ takes 2 H⁺ → 2 H₂PO₄⁻

So 2 PO₄³⁻ → need 4 H⁺ → from 2 H₂SO₄

So:
Ca₃(PO₄)₂ + 2H₂SO₄ → CaSO₄ + Ca(H₂PO₄)₂

But wait: we have 2 SO₄²⁻ → so should get 2 CaSO₄?

No! Only one CaSO₄? But we have 2 SO₄²⁻ → need 2 CaSO₄

But only 3 Ca²⁺ available.

Let’s rework.

Let’s say:
- Ca₃(PO₄)₂ → 3 Ca²⁺ and 2 PO₄³⁻
- We want to make CaSO₄ and Ca(H₂PO₄)₂

Suppose:
- x CaSO₄
- y Ca(H₂PO₄)₂

Then:
- Ca: x + y = 3
- PO₄: 2y = 2 → y = 1 → then x = 2

So:
- 2 CaSO₄
- 1 Ca(H₂PO₄)₂

Now SO₄²⁻ needed: 2 → so 2 H₂SO₄

H⁺ needed: 2 H₂PO₄⁻ → each needs 2 H⁺ → 4 H⁺ → from 2 H₂SO₄ → yes

So:
Ca₃(PO₄)₂ + 2H₂SO₄ → 2CaSO₄ + Ca(H₂PO₄)₂

Check:
- Ca: 3 = 2 + 1 → OK
- P: 2 = 2 → OK
- O: complicated, but let’s check H and S
- S: 2 = 2 → OK
- H: 2×2 = 4 → Ca(H₂PO₄)₂ has 4 H → OK

✔ Balanced:
Ca₃(PO₄)₂ + 2H₂SO₄ → 2CaSO₄ + Ca(H₂PO₄)₂

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10. H₃PO₄ + HCl → PCl₅ + H₂O

This seems unusual — phosphoric acid reacting with HCl to give PCl₅?

But PCl₅ is typically made from P₄ and Cl₂.

However, maybe it's a hypothetical or lab synthesis.

Let’s try balancing.

PCl₅ has 1 P and 5 Cl → so need 5 Cl from HCl → 5 HCl

H₃PO₄ has 3 H and 1 P → so one H₃PO₄

Now, H: left: 3 (from H₃PO₄) + 5 (from HCl) = 8 H → right: H₂O → need 4 H₂O

O: H₃PO₄ has 4 O → all go to H₂O → 4 H₂O → 4 O → OK

So:
H₃PO₄ + 5HCl → PCl₅ + 4H₂O

Check:
- P: 1 = 1 → OK
- Cl: 5 = 5 → OK
- H: 3 + 5 = 8 → 4×2 = 8 → OK
- O: 4 = 4 → OK

✔ Balanced:
H₃PO₄ + 5HCl → PCl₅ + 4H₂O

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✔ Final Balanced Equations:

1. CH₄ + 2O₂ → CO₂ + 2H₂O
2. Na⁺ + Cl⁻ → NaCl
3. 4Al + 3O₂ → 2Al₂O₃
4. N₂ + 3H₂ → 2NH₃
5. 8CO + 17H₂ → C₈H₁₈ + 8H₂O
6. Fe₂O₃ + 3CO → 2Fe + 3CO₂
7. 2H₂SO₄ + Pb(OH)₄ → Pb(SO₄)₂ + 4H₂O
8. 2Al + 6HCl → 2AlCl₃ + 3H₂
9. Ca₃(PO₄)₂ + 2H₂SO₄ → 2CaSO₄ + Ca(H₂PO₄)₂
10. H₃PO₄ + 5HCl → PCl₅ + 4H₂O

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