A Comprehensive Collection Of Balancing Chemical Equations ... - Free Printable

Let's go through each of the chemical equations and balance them step by step. Balancing chemical equations means making sure that the number of atoms of each element is the same on both sides (reactants and products) by adjusting coefficients (the numbers in front of the compounds).

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1. CH₄ + O₂ → CO₂ + H₂O

Unbalanced:
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
- Now oxygen: Right side has 2 (from CO₂) + 1×2 = 2 from 2H₂O → total 4 O
- Left: O₂ → need 2 O₂ to give 4 O

So:
CH₄ + 2O₂ → CO₂ + 2H₂O

✔ Balanced.

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2. Na⁺ + Cl⁻ → NaCl

This is an ionic compound formation. But note: it's not a reaction between elements — it's just combining ions.

But since Na⁺ and Cl⁻ combine in a 1:1 ratio to form NaCl:

Na⁺ + Cl⁻ → NaCl

We can write it as:
1 Na⁺ + 1 Cl⁻ → 1 NaCl

✔ Already balanced.

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3. Al + O₂ → Al₂O₃

Unbalanced:
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

So:
- 3 O₂ gives 6 O
- 2 Al₂O₃ gives 6 O and 4 Al

So:
- Need 4 Al on left
- 3 O₂ on left

4Al + 3O₂ → 2Al₂O₃

✔ Balanced.

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4. N₂ + H₂ → NH₃

Unbalanced:
N₂ + H₂ → NH₃

- N: 2 on left, 1 on right → need 2 NH₃
- H: 2 on left, 3×2 = 6 on right → need 3 H₂

So:
N₂ + 3H₂ → 2NH₃

✔ Balanced.

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5. CO(g) + H₂(g) → C₂H₅OH(l) + H₂O

Wait! The product is C₂H₅OH, which is ethanol — contains 2 carbon atoms, but CO only has 1 C.

So we need to balance accordingly.

Unbalanced:
CO + H₂ → C₂H₅OH + H₂O

Left: C=1, O=1, H=2
Right: C=2, H=6+2=8, O=1+1=2

We need:
- 2 CO for 2 C
- Then check oxygen: 2 O from CO → need 2 O on right → one in C₂H₅OH and one in H₂O → OK
- H: C₂H₅OH has 6 H, H₂O has 2 → total 8 H → need 4 H₂ molecules (since each gives 2 H)

Try:
2CO + 4H₂ → C₂H₅OH + H₂O

Check:
- C: 2 = 2 → OK
- O: 2 = 1 (in C₂H₅OH) + 1 (in H₂O) = 2 → OK
- H: 4×2 = 8 → 6 (in C₂H₅OH) + 2 (in H₂O) = 8 → OK

✔ Balanced.

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6. Fe₂O₃(s) + CO(g) → Fe(l) + CO₂(g)

Unbalanced:
Fe₂O₃ + CO → Fe + CO₂

- Fe: 2 on left → need 2 Fe on right
- O: 3 from Fe₂O₃ + 1 from CO → total 4 O? Wait — let’s think.

Actually, Fe₂O₃ has 3 O atoms. Each CO takes one O to become CO₂.

So to remove 3 O atoms from Fe₂O₃, we need 3 CO → produce 3 CO₂

Then Fe: 2 → so 2 Fe

So:
Fe₂O₃ + 3CO → 2Fe + 3CO₂

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

✔ Balanced.

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7. H₂SO₄ + Pb(OH)₄ → Pb(SO₄)₂ + H₂O

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

Note: Pb(SO₄)₂ has 2 SO₄ groups, so need 2 H₂SO₄

Also, Pb(OH)₄ has 4 OH, so when it reacts with acid, forms water.

So:
- 2 H₂SO₄ provides 2 SO₄²⁻ → needed for Pb(SO₄)₂
- Pb(OH)₄ has 4 OH⁻ → will react with 4 H⁺ → form 4 H₂O

Each H₂SO₄ provides 2 H⁺ → so 2 H₂SO₄ gives 4 H⁺ → perfect

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

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

✔ Balanced.

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8. Al + HCl → AlCl₃ + H₂

Unbalanced:
Al + HCl → AlCl₃ + H₂

- Al: 1 = 1 → OK
- Cl: 1 on left, 3 on right → need 3 HCl
- H: 3 on left → need 3/2 H₂ → better use whole numbers

So:
2Al + 6HCl → 2AlCl₃ + 3H₂

Check:
- Al: 2 = 2
- Cl: 6 = 6
- H: 6 = 6 → 3H₂ has 6 H

✔ Balanced.

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

This looks like a double displacement or acid-base reaction.

Note:
- Ca₃(PO₄)₂ has 3 Ca 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:
- We want to convert PO₄³⁻ to H₂PO₄⁻ → needs 2 H⁺ per PO₄³⁻ → so 2 H⁺ per PO₄ → 2 H₂SO₄ per PO₄ → 4 H₂SO₄ for two PO₄

But also need CaSO₄ for sulfate.

Let’s see:

Ca₃(PO₄)₂ → 3 Ca²⁺ and 2 PO₄³⁻

To make Ca(H₂PO₄)₂, we need 2 H₂PO₄⁻ → so we can use 2 PO₄³⁻ to make 2 H₂PO₄⁻ → requires 4 H⁺ → 2 H₂SO₄

But now we have 3 Ca²⁺ → one goes to Ca(H₂PO₄)₂ (uses 1 Ca), so remaining 2 Ca²⁺ must go to CaSO₄

So need 2 CaSO₄ → requires 2 SO₄²⁻ → 2 H₂SO₄

Total H₂SO₄: 2 (for CaSO₄) + 2 (for H⁺) = 4 H₂SO₄

So:
Ca₃(PO₄)₂ + 4H₂SO₄ → 2CaSO₄ + Ca(H₂PO₄)₂

Wait — check Ca:
Left: 3
Right: 2 (from CaSO₄) + 1 (from Ca(H₂PO₄)₂) = 3 → OK

P: 2 = 2 → OK

S: 4 = 2 (from CaSO₄) → no! Only 2 S on right, but 4 on left → problem

Wait — we used 4 H₂SO₄ → 4 SO₄²⁻

But we only made 2 CaSO₄ → uses 2 SO₄²⁻

Where are the other 2 SO₄²⁻?

Ah — mistake.

We need to account for all SO₄.

But Ca(H₂PO₄)₂ doesn’t have sulfate.

So if we use 4 H₂SO₄ → 4 SO₄²⁻ → must form 4 CaSO₄ → but we only have 3 Ca²⁺

Too many sulfates.

Alternative approach:

The correct reaction is:
Ca₃(PO₄)₂ + 2H₂SO₄ → 2CaSO₄ + Ca(H₂PO₄)₂

Now check:
- Ca: 3 = 2 + 1 → OK
- P: 2 = 2 → OK
- S: 2 = 2 → OK
- O: complex, but let’s check H

H: 2×2 = 4 H → Ca(H₂PO₄)₂ has 4 H → OK

But what about the phosphorus?

PO₄³⁻ + 2H⁺ → H₂PO₄⁻ → so each PO₄³⁻ needs 2H⁺ → 2 PO₄ → need 4 H⁺ → 2 H₂SO₄ → yes

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

Yes!

Check:
- Ca: 3 = 2 + 1 → OK
- P: 2 = 2 → OK
- S: 2 = 2 → OK
- H: 4 = 4 → OK
- O: too many to count, but it balances

✔ Balanced.

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10. H₃PO₄ + HCl → PCl₅ + H₂O

This seems odd — H₃PO₄ is phosphoric acid, HCl is hydrochloric acid. PCl₅ is phosphorus pentachloride.

But this is not a typical reaction — usually PCl₅ is made from P₄ and Cl₂.

But assuming this is a hypothetical or a substitution reaction.

Let’s try to balance:

H₃PO₄ + HCl → PCl₅ + H₂O

Look at P: 1 = 1 → OK

Cl: 1 on left → 5 on right → need 5 HCl

H: H₃PO₄ has 3 H, 5 HCl has 5 H → total 8 H → H₂O has 2 H → need 4 H₂O

O: H₃PO₄ has 4 O → 4 H₂O has 4 O → OK

So:
H₃PO₄ + 5HCl → PCl₅ + 4H₂O

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

✔ Balanced.

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✔ Final Answers:

1. CH₄ + 2O₂ → CO₂ + 2H₂O
2. Na⁺ + Cl⁻ → NaCl
3. 4Al + 3O₂ → 2Al₂O₃
4. N₂ + 3H₂ → 2NH₃
5. 2CO + 4H₂ → C₂H₅OH + H₂O
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

All equations are now balanced. Let me know if you'd like explanations for any specific one!