Balancing Chemical Equations worksheet with ten problems for students to practice balancing chemical reactions.

Let’s solve each chemical equation by balancing the atoms on both sides. We’ll go one by one, making sure the number of each type of atom is equal on the left (reactants) and right (products).

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1. Fe + H₂SO₄ → Fe₂(SO₄)₃ + H₂

- Right side has 2 Fe → put 2 in front of Fe on left.
- Right side has 3 SO₄ groups → put 3 in front of H₂SO₄ on left.
- Now left has 6 H (from 3 H₂SO₄), so right needs 3 H₂ → put 3 in front of H₂.

✔ Balanced:
2 Fe + 3 H₂SO₄ → 1 Fe₂(SO₄)₃ + 3 H₂

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2. CH₄ + O₂ → CO₂ + H₂O

- Left: 1 C, 4 H; Right: 1 C, 2 H (in H₂O) → need 2 H₂O to get 4 H → put 2 in front of H₂O.
- Now right has 2 O (from CO₂) + 2 O (from 2 H₂O) = 4 O → so need 2 O₂ on left (since each O₂ has 2 O atoms).

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

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3. SiCl₄(l) + H₂O(l) → SiO₂(s) + HCl(aq)

- Left: 1 Si, 4 Cl, 2 H, 1 O; Right: 1 Si, 2 O, 1 H, 1 Cl → not balanced.
- Need 4 HCl on right to match 4 Cl from SiCl₄ → put 4 in front of HCl.
- That gives 4 H on right → need 2 H₂O on left (each has 2 H) → put 2 in front of H₂O.
- Check O: left has 2 O (from 2 H₂O), right has 2 O (in SiO₂) → good.

✔ Balanced:
1 SiCl₄ + 2 H₂O → 1 SiO₂ + 4 HCl

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4. AgI + Na₂S → Ag₂S + NaI

- Right has 2 Ag → put 2 in front of AgI on left.
- Right has 2 I → but now left has 2 I (from 2 AgI), so right needs 2 NaI → put 2 in front of NaI.
- Left has 2 Na (from Na₂S), right has 2 Na (from 2 NaI) → good.
- Sulfur: 1 on each side → good.

✔ Balanced:
2 AgI + 1 Na₂S → 1 Ag₂S + 2 NaI

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5. NH₃ + O₂ → NO + H₂O

- Let’s try coefficients: a NH₃ + b O₂ → c NO + d H₂O
- N: a = c
- H: 3a = 2d → let’s pick a=4 → then d=6
- Then c=4 (for N)
- O: 2b = c + d = 4 + 6 = 10 → b=5

✔ Balanced:
4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

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6. FeO₃(s) + CO(g) → Fe(l) + CO₂(g)

Wait — FeO₃? That’s unusual. Typically iron oxide is Fe₂O₃ or Fe₃O₄. But we’ll work with what’s given.

Assume it’s FeO₃ as written.

Left: 1 Fe, 3 O (from FeO₃) + 1 O (from CO) = 4 O total? Wait — better to balance per compound.

Actually, let’s treat it as:

FeO₃ + CO → Fe + CO₂

- Fe: 1 on each side → ok for now.
- O: left has 3 (from FeO₃) + 1 (from CO) = 4 O; right has 2 O (from CO₂) → not balanced.
- Try putting 3 CO on left → makes 3 CO₂ on right → that uses 3 O from CO and 3 O from FeO₃? No.

Better approach:

FeO₃ has 3 O atoms. Each CO can take 1 O to become CO₂. So to remove 3 O from FeO₃, we need 3 CO → produces 3 CO₂.

So:

FeO₃ + 3 CO → Fe + 3 CO₂

Check:
Fe: 1 = 1
O: 3 + 3 = 6 on left; 3×2=6 on right → good.
C: 3 = 3

✔ Balanced:
1 FeO₃ + 3 CO → 1 Fe + 3 CO₂

*(Note: In reality, FeO₃ isn’t common — likely meant Fe₂O₃, but we solved as written.)*

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7. SiO₂ + HF → SiF₄ + H₂O

- Si: 1 each side → ok.
- F: left has 1 (HF), right has 4 (SiF₄) → need 4 HF on left.
- H: 4 H from 4 HF → need 2 H₂O on right (each has 2 H).
- O: left has 2 O (SiO₂), right has 2 O (from 2 H₂O) → good.

✔ Balanced:
1 SiO₂ + 4 HF → 1 SiF₄ + 2 H₂O

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8. NaBr + Cl₂ → NaCl + Br₂

- Br: left 1, right 2 → need 2 NaBr on left.
- Na: 2 on left → need 2 NaCl on right.
- Cl: 2 on left (Cl₂), 2 on right (2 NaCl) → good.

✔ Balanced:
2 NaBr + 1 Cl₂ → 2 NaCl + 1 Br₂

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9. (NH₄)₃PO₄ + Pb(NO₃)₄ → Pb₃(PO₄)₄ + NH₄NO₃

This one is tricky.

Left:
- (NH₄)₃PO₄ has 3 NH₄, 1 PO₄
- Pb(NO₃)₄ has 1 Pb, 4 NO₃

Right:
- Pb₃(PO₄)₄ has 3 Pb, 4 PO₄
- NH₄NO₃ has 1 NH₄, 1 NO₃

We need 4 PO₄ on right → so need 4 (NH₄)₃PO₄ on left → gives 12 NH₄ and 4 PO₄.

We need 3 Pb on right → so need 3 Pb(NO₃)₄ on left → gives 12 NO₃.

Now, NH₄: 12 on left → need 12 NH₄NO₃ on right → which also takes 12 NO₃ → matches!

✔ Balanced:
4 (NH₄)₃PO₄ + 3 Pb(NO₃)₄ → 1 Pb₃(PO₄)₄ + 12 NH₄NO₃

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10. Mg(OH)₂ + HCl → MgCl₂ + H₂O

- Mg: 1 each side → ok.
- Cl: left 1 (HCl), right 2 (MgCl₂) → need 2 HCl on left.
- H: left: 2 from Mg(OH)₂ + 2 from 2 HCl = 4 H; right: 2 H per H₂O → need 2 H₂O.
- O: left: 2 O from Mg(OH)₂; right: 2 O from 2 H₂O → good.

✔ Balanced:
1 Mg(OH)₂ + 2 HCl → 1 MgCl₂ + 2 H₂O

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Final Answer:
1. 2, 3, 1, 3
2. 1, 2, 1, 2
3. 1, 2, 1, 4
4. 2, 1, 1, 2
5. 4, 5, 4, 6
6. 1, 3, 1, 3
7. 1, 4, 1, 2
8. 2, 1, 2, 1
9. 4, 3, 1, 12
10. 1, 2, 1, 2