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

Let’s go through each equation one by one and balance them step by step. Balancing means making sure the number of atoms of each element is the same on both sides of the arrow.

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1. CH₄ + O₂ → CO₂ + H₂O

Left: C=1, H=4, O=2
Right: C=1, H=2, O=3 (2 from CO₂ + 1 from H₂O)

→ Start with H: 4 on left, so need 2 H₂O on right → now H=4, O=2+2=4 on right
→ Now O: 4 on right, so need 2 O₂ on left → O=4

Balanced: 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O

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2. Na⁺ + Cl⁻ → NaCl

This is already balanced — 1 Na and 1 Cl on each side.

Balanced: 1 Na⁺ + 1 Cl⁻ → 1 NaCl

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3. Al + O₂ → Al₂O₃

Left: Al=1, O=2
Right: Al=2, O=3

→ Find LCM for O: 6 → use 3 O₂ (gives 6 O) and 2 Al₂O₃ (also gives 6 O)
→ Now Al: 4 on right → need 4 Al on left

Balanced: 4 Al + 3 O₂ → 2 Al₂O₃

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4. N₂⁺ + H₂ → NH₃

Wait — N₂⁺? That’s unusual. Probably a typo — should be N₂ (neutral). Let’s assume it’s N₂.

N₂ + H₂ → NH₃

Left: N=2, H=2
Right: N=1, H=3

→ Make N equal: 2 NH₃ → now N=2, H=6
→ Need 3 H₂ to get 6 H

Balanced: 1 N₂ + 3 H₂ → 2 NH₃

*(If it really is N₂⁺, we’d have to account for charge — but in basic balancing, we usually ignore charges unless specified. Since product is neutral NH₃, likely typo. We’ll proceed with N₂.)*

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5. CO(g) + H₂(g) → C₈H₁₈(l) + H₂O

Left: C=1, O=1, H=2
Right: C=8, H=18+2=20, O=1

→ Need 8 CO for 8 C → now O=8 → need 8 H₂O for 8 O → now H from water = 16
→ Total H needed on right: 18 (from C₈H₁₈) + 16 (from 8 H₂O) = 34 H → so need 17 H₂ on left

Check:
Left: 8 CO → 8C, 8O; 17 H₂ → 34H
Right: 1 C₈H₁₈ → 8C, 18H; 8 H₂O → 16H, 8O → total H=34, O=8, C=8 ✔

Balanced: 8 CO + 17 H₂ → 1 C₈H₁₈ + 8 H₂O

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6. FeO₃(s) + CO(g) → Fe(l) + CO₂(g)

Wait — FeO₃? Iron doesn’t form FeO₃. Common iron oxides are FeO or Fe₂O₃. Likely typo — probably Fe₂O₃.

Assume: Fe₂O₃ + CO → Fe + CO₂

Left: Fe=2, O=3+1=4? Wait — Fe₂O₃ has 3 O, CO has 1 O per molecule.

Better to write:

Fe₂O₃ + CO → Fe + CO₂

Balance Fe: 2 Fe on right → 2 Fe
Balance O: Fe₂O₃ has 3 O, each CO takes one O to become CO₂ → so need 3 CO to take 3 O → makes 3 CO₂
Now check C: 3 on left, 3 on right ✔

Balanced: 1 Fe₂O₃ + 3 CO → 2 Fe + 3 CO₂

*(Note: Original said FeO₃ — if taken literally, it would be FeO₃ + CO → Fe + CO₂ → then Fe=1, O=4 on left, O=2 on right → not possible without changing formulas. So definitely typo — standard reaction uses Fe₂O₃.)*

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7. H₂SO₄ + Pb(OH)₄ → Pb(SO₄)₂ + H₂O

Left: H=2+4=6, S=1, O=4+4=8, Pb=1
Right: Pb=1, S=2, O=8+1=9? Wait — Pb(SO₄)₂ has 2 SO₄ → 2 S, 8 O; plus H₂O → 1 O and 2 H → total O=9, H=2

Not matching.

Actually, Pb(OH)₄ has Pb⁴⁺, so sulfate should be Pb(SO₄)₂ — correct.

But H₂SO₄ provides 2 H⁺ and SO₄²⁻. To make Pb(SO₄)₂, need 2 SO₄ → so 2 H₂SO₄.

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

Left: H=4+4=8, S=2, O=8+4=12, Pb=1
Right: Pb=1, S=2, O=8 (from sulfates) + ? from water → need to balance H and O.

H on left: 8 → so need 4 H₂O on right → that’s 8 H and 4 O → total O on right: 8 (sulfate) + 4 (water) = 12 ✔

Balanced: 2 H₂SO₄ + 1 Pb(OH)₄ → 1 Pb(SO₄)₂ + 4 H₂O

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8. Al + HCl → AlCl₃ + H₂

Left: Al=1, H=1, Cl=1
Right: Al=1, Cl=3, H=2

→ Need 3 Cl on left → 3 HCl → now H=3, Cl=3
→ Right: H₂ has 2 H → need even H → try 2 AlCl₃ → needs 6 Cl → so 6 HCl → H=6 → need 3 H₂

Then Al: 2 on right → 2 Al on left

Balanced: 2 Al + 6 HCl → 2 AlCl₃ + 3 H₂

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

Left: Ca=3, P=2, O=8+4=12? Wait — Ca₃(PO₄)₂ has 3 Ca, 2 P, 8 O; H₂SO₄ has 2H, S, 4O

Right: CaSO₄ has Ca, S, 4O; Ca(H₂PO₄)₂ has Ca, 4H, 2P, 8O

Total right: Ca=1+1=2, S=1, P=2, H=4, O=4+8=12

Left: Ca=3, P=2, H=2, S=1, O=8+4=12

Ca mismatch: 3 vs 2

Try coefficients:

Let’s say: 1 Ca₃(PO₄)₂ + x H₂SO₄ → y CaSO₄ + z Ca(H₂PO₄)₂

Ca: 3 = y + z
P: 2 = 2z → so z=1 → then y=2
S: x = y = 2
H: 2x = 4z → 2*2=4, 4*1=4 ✔
O: check later — should work.

So: 1 Ca₃(PO₄)₂ + 2 H₂SO₄ → 2 CaSO₄ + 1 Ca(H₂PO₄)₂

Check atoms:

Left: Ca=3, P=2, H=4, S=2, O=8 (phosphate) + 8 (sulfuric) = 16? Wait — Ca₃(PO₄)₂: PO₄ is 4 O each → 2×4=8 O; H₂SO₄: 4 O each → 2×4=8 O → total O=16

Right: 2 CaSO₄ → 2×4=8 O; Ca(H₂PO₄)₂ → 2×4=8 O (since two H₂PO₄ groups, each has 4 O) → total O=16 ✔

H: left 4, right: Ca(H₂PO₄)₂ has 4 H ✔

Perfect.

Balanced: 1 Ca₃(PO₄)₂ + 2 H₂SO₄ → 2 CaSO₄ + 1 Ca(H₂PO₄)₂

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10. H₃PO₄ + HCl → PCl₅ + H₂O

Left: H=3+1=4, P=1, O=4, Cl=1
Right: P=1, Cl=5, H=2, O=1

→ Need 5 Cl on left → 5 HCl → now H=3+5=8, Cl=5
→ Right: PCl₅ ok, H₂O → need 4 H₂O to use 8 H → then O=4 on right → matches left O=4

Balanced: 1 H₃PO₄ + 5 HCl → 1 PCl₅ + 4 H₂O

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Final Answer:

1. 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O
2. 1 Na⁺ + 1 Cl⁻ → 1 NaCl
3. 4 Al + 3 O₂ → 2 Al₂O₃
4. 1 N₂ + 3 H₂ → 2 NH₃ *(assuming N₂ instead of N₂⁺)*
5. 8 CO + 17 H₂ → 1 C₈H₁₈ + 8 H₂O
6. 1 Fe₂O₃ + 3 CO → 2 Fe + 3 CO₂ *(assuming Fe₂O₃ instead of FeO₃)*
7. 2 H₂SO₄ + 1 Pb(OH)₄ → 1 Pb(SO₄)₂ + 4 H₂O
8. 2 Al + 6 HCl → 2 AlCl₃ + 3 H₂
9. 1 Ca₃(PO₄)₂ + 2 H₂SO₄ → 2 CaSO₄ + 1 Ca(H₂PO₄)₂
10. 1 H₃PO₄ + 5 HCl → 1 PCl₅ + 4 H₂O