Balancing Chemical Equations interactive worksheet - Free Printable

Absolutely! Let’s balance each chemical equation step by step. The goal is to make sure the number of atoms of each element is the same on both sides of the reaction arrow (law of conservation of mass).

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1. ___ H₂O → ___ H₂ + ___ O₂

This is the decomposition of water.

- Left: H=2, O=1
- Right: H=2, O=2 → Oxygen is unbalanced.

We need 2 oxygen atoms on the left → put 2 in front of H₂O.

→ 2H₂O → ? H₂ + ? O₂

Now: Left: H=4, O=2
Right: To balance O, we need 1 O₂ (since O₂ has 2 O atoms).
To balance H, we need 2 H₂ (since each H₂ has 2 H atoms → 2×2=4 H).

✔ Balanced:
2 H₂O → 2 H₂ + 1 O₂

*(Note: We usually omit “1” as a coefficient, so it’s written as just O₂)*

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2. ___ Fe₂O₃ → ___ Fe + ___ O₂

Decomposition of iron(III) oxide.

- Left: Fe=2, O=3
- Right: Fe=1, O=2 → Unbalanced.

We need even number of O atoms on right (since O₂), so multiply Fe₂O₃ by 2 → gives 6 O atoms.

→ 2Fe₂O₃ → ? Fe + ? O₂

Left: Fe=4, O=6
Right: Need 3 O₂ (3×2=6 O) and 4 Fe.

✔ Balanced:
2 Fe₂O₃ → 4 Fe + 3 O₂

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3. ___ Na + ___ Cl₂ → ___ NaCl

Sodium reacts with chlorine gas to form sodium chloride.

- Left: Na=1, Cl=2
- Right: Na=1, Cl=1 → Cl unbalanced.

Need 2 NaCl to get 2 Cl atoms → then need 2 Na on left.

✔ Balanced:
2 Na + 1 Cl₂ → 2 NaCl

*(Again, “1” is often omitted for Cl₂)*

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4. ___ HgO → ___ Hg + ___ O₂

Decomposition of mercury(II) oxide.

- Left: Hg=1, O=1
- Right: Hg=1, O=2 → O unbalanced.

Multiply HgO by 2 → 2 HgO → ? Hg + ? O₂

Left: Hg=2, O=2
Right: Need 2 Hg and 1 O₂.

✔ Balanced:
2 HgO → 2 Hg + 1 O₂

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5. ___ TiCl₄ + ___ O₂ → ___ TiO₂ + ___ Cl₂

Titanium tetrachloride reacts with oxygen to form titanium dioxide and chlorine gas.

Let’s count:

Left: Ti=1, Cl=4, O=2
Right: Ti=1, O=2, Cl=2 → Cl unbalanced.

We need 4 Cl on right → so 2 Cl₂ (2×2=4 Cl).

Now: Right: Ti=1, O=2, Cl=4
Left: TiCl₄ gives Ti=1, Cl=4 — good. But O₂ only gives 2 O — matches TiO₂.

✔ Already balanced? Wait — let’s check again.

TiCl₄ + O₂ → TiO₂ + 2Cl₂

Left: Ti=1, Cl=4, O=2
Right: Ti=1, O=2, Cl=4 → YES!

✔ Balanced:
1 TiCl₄ + 1 O₂ → 1 TiO₂ + 2 Cl₂

*(Coefficients of 1 are usually not written)*

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6. ___ Al + ___ O₂ → ___ Al₂O₃

Aluminum + oxygen → aluminum oxide.

Left: Al=1, O=2
Right: Al=2, O=3 → both unbalanced.

Find LCM for O: 2 and 3 → LCM = 6.

So, need 3 O₂ (3×2=6 O) on left → gives 6 O.

On right, Al₂O₃ has 3 O → so need 2 Al₂O₃ to get 6 O → 2 Al₂O₃ has 4 Al.

So left needs 4 Al.

✔ Balanced:
4 Al + 3 O₂ → 2 Al₂O₃

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7. ___ HCl + ___ Zn → ___ H₂ + ___ ZnCl₂

Hydrochloric acid + zinc → hydrogen gas + zinc chloride.

Left: H=1, Cl=1, Zn=1
Right: H=2, Cl=2, Zn=1 → H and Cl unbalanced.

ZnCl₂ has 2 Cl → need 2 HCl to provide 2 Cl and 2 H.

Then H₂ will have 2 H — perfect.

✔ Balanced:
2 HCl + 1 Zn → 1 H₂ + 1 ZnCl₂

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8. ___ Al + ___ CuCl₂ → ___ AlCl₃ + ___ Cu

Single displacement: Aluminum displaces copper.

Left: Al=1, Cu=1, Cl=2
Right: Al=1, Cl=3, Cu=1 → Cl unbalanced.

LCM of Cl: 2 and 3 → 6.

So, need 3 CuCl₂ (3×2=6 Cl) → gives 3 Cu.

On right, need 2 AlCl₃ (2×3=6 Cl) → gives 2 Al.

So left needs 2 Al.

✔ Balanced:
2 Al + 3 CuCl₂ → 2 AlCl₃ + 3 Cu

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## ✔ Final Balanced Equations:

1. 2 H₂O → 2 H₂ + O₂
2. 2 Fe₂O₃ → 4 Fe + 3 O₂
3. 2 Na + Cl₂ → 2 NaCl
4. 2 HgO → 2 Hg + O₂
5. TiCl₄ + O₂ → TiO₂ + 2 Cl₂
6. 4 Al + 3 O₂ → 2 Al₂O₃
7. 2 HCl + Zn → H₂ + ZnCl₂
8. 2 Al + 3 CuCl₂ → 2 AlCl₃ + 3 Cu

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Let me know if you’d like to see the atom counts for each one verified!