Balancing Chemical Equations Worksheet 24 | Chemistry worksheets ... - Free Printable

Let's solve each of these chemical equations by balancing them. Balancing means adjusting coefficients (numbers in front of the compounds) so that the number of atoms of each element is the same on both sides of the equation.

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1. N₂ + H₂ → NH₃

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

To balance:
- Put 2 in front of NH₃ → 2N, 6H
- Then need 3 H₂ to get 6 H on left
- Need 1 N₂ to give 2 N

✔ Balanced:
N₂ + 3H₂ → 2NH₃

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2. S₈ + O₂ → SO₃

- S₈ has 8 S atoms
- Each SO₃ has 1 S and 3 O
- So we need 8 SO₃ → 8 S, 24 O
- O₂ provides 2 O per molecule → need 12 O₂

✔ Balanced:
S₈ + 12O₂ → 8SO₃

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3. HgO → Hg + O₂

- Left: 1 Hg, 1 O
- Right: 1 Hg, 2 O → oxygen not balanced

Need even number of O on left → use 2 HgO
→ 2 Hg, 2 O
→ Right: 2 Hg, O₂ (which is 2 O)

✔ Balanced:
2HgO → 2Hg + O₂

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4. Zn + HCl → ZnCl₂ + H₂

- Zn: 1 on both sides
- Cl: 1 on left, 2 on right → need 2 HCl
- H: 2 on left, 2 on right → good

✔ Balanced:
Zn + 2HCl → ZnCl₂ + H₂

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5. SiCl₄ + H₂O → H₄SiO₄ + HCl

- Left: Si=1, Cl=4, H=2, O=1
- Right: Si=1, O=4, H=4, Cl=1 → not balanced

We need 4 Cl on right → 4 HCl
→ 4 H on right from HCl, plus 4 H in H₄SiO₄ → total 8 H
Left: H₂O → must have 4 H₂O → 8 H, 4 O

Now check:
- Si: 1 = 1
- Cl: 4 = 4
- O: 4 = 4
- H: 8 = 8

✔ Balanced:
SiCl₄ + 4H₂O → H₄SiO₄ + 4HCl

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6. Na + H₂O → NaOH + H₂

- Left: Na=1, H=2, O=1
- Right: Na=1, O=1, H=1+2=3 → H not balanced

Try 2 Na → 2 NaOH → needs 2 H from H₂O and 2 H for H₂?

Better:
- Try 2 Na → 2 NaOH → need 2 H and 1 O → but water gives 2 H and 1 O per molecule
- So 2 H₂O → 4 H, 2 O
- But only need 2 H for 2 NaOH and 2 H for H₂

Wait:

Let’s try:
- 2Na + 2H₂O → 2NaOH + H₂

Check:
- Na: 2 = 2
- O: 2 = 2
- H: 4 on left → 2 in 2NaOH + 2 in H₂ → 4 → good

✔ Balanced:
2Na + 2H₂O → 2NaOH + H₂

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7. H₃PO₄ → H₄P₂O₇ + H₂O

- Left: H=3, P=1, O=4
- Right: H=4+2=6, P=2, O=7+1=8 → not balanced

We want to form H₄P₂O₇ from H₃PO₄
- H₄P₂O₇ has 2 P → need 2 H₃PO₄ → 6 H, 2 P, 8 O
- Right: H₄P₂O₇ has 4 H, 2 P, 7 O → so 1 H₂O has 2 H, 1 O → total H=6, O=8

So:
- 2 H₃PO₄ → H₄P₂O₇ + H₂O

✔ Balanced:
2H₃PO₄ → H₄P₂O₇ + H₂O

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8. Si₂H₆ + O₂ → SiO₂ + H₂O

- Left: Si=2, H=6, O=?
- Right: SiO₂ has 1 Si, 2 O; H₂O has 2 H, 1 O

Need 2 SiO₂ → 2 Si, 4 O
Need 3 H₂O → 6 H, 3 O → total O = 4 + 3 = 7 O → need 3.5 O₂ → not integer

Try:
- 2 Si₂H₆ → 4 Si, 12 H
- Need 4 SiO₂ → 4 Si, 8 O
- Need 6 H₂O → 12 H, 6 O → total O = 14 → need 7 O₂

So:
- 2Si₂H₆ + 7O₂ → 4SiO₂ + 6H₂O

Check:
- Si: 4 = 4
- H: 12 = 12
- O: 14 = 8 + 6 = 14 ✔

✔ Balanced:
2Si₂H₆ + 7O₂ → 4SiO₂ + 6H₂O

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9. Al(OH)₃ + H₂SO₄ → Al₂(SO₄)₃ + H₂O

- Al: 1 on left, 2 on right → need 2 Al(OH)₃
- SO₄: 1 on left, 3 on right → need 3 H₂SO₄
- Now H: 2×3 = 6 H from Al(OH)₃, 3×2 = 6 H from H₂SO₄ → total 12 H
- O: from OH and SO₄ — better count products

Right: Al₂(SO₄)₃ → 3 SO₄ → 12 O, 2 Al
H₂O → let's say x H₂O → 2x H, x O

Left:
- 2Al(OH)₃ → 2 Al, 6 O, 6 H
- 3H₂SO₄ → 6 H, 3 S, 12 O → total H = 12, O = 6 + 12 = 18

Right:
- Al₂(SO₄)₃ → 2 Al, 3 S, 12 O
- H₂O → x molecules → 2x H, x O → total O = 12 + x, H = 2x

Set equal:
- H: 12 = 2x → x = 6
- O: 18 = 12 + 6 → yes

✔ Balanced:
2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O

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10. Fe + O₂ → Fe₂O₃

- Fe: 1 on left, 2 on right → need 2 Fe
- O: 2 on left, 3 on right → LCM of 2 and 3 is 6 → need 3 O₂ (6 O), 2 Fe₂O₃ (6 O, 4 Fe)

So:
- 4 Fe + 3 O₂ → 2 Fe₂O₃

✔ Balanced:
4Fe + 3O₂ → 2Fe₂O₃

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11. Fe₂(SO₄)₃ + KOH → K₂SO₄ + Fe(OH)₃

- Fe: 2 on left → need 2 Fe(OH)₃
- SO₄: 3 on left → need 3 K₂SO₄ → 6 K
- K: 6 K → need 6 KOH
- OH: 6 KOH → 6 OH → 2 Fe(OH)₃ uses 6 OH → good

So:
- Fe₂(SO₄)₃ + 6KOH → 3K₂SO₄ + 2Fe(OH)₃

Check:
- Fe: 2 = 2
- S: 3 = 3
- O: many, but check K: 6 = 6
- H: 6 = 6

✔ Balanced:
Fe₂(SO₄)₃ + 6KOH → 3K₂SO₄ + 2Fe(OH)₃

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12. FeS₂ + O₂ → Fe₂O₃ + SO₂

- Fe: 1 on left, 2 on right → need 2 FeS₂
- S: 2×2 = 4 S → need 4 SO₂
- O: right: Fe₂O₃ has 3 O, 4 SO₂ has 8 O → total 11 O
- Left: O₂ → need 11/2 = 5.5 → multiply all by 2

Start over:
- 2 FeS₂ → 2 Fe, 4 S
- → Fe₂O₃ (2 Fe), 4 SO₂ (4 S)
- O needed: 3 (from Fe₂O₃) + 8 (from SO₂) = 11 O → need 11/2 O₂

Multiply whole equation by 2:
- 4 FeS₂ + 11 O₂ → 2 Fe₂O₃ + 8 SO₂

Check:
- Fe: 4 = 4
- S: 8 = 8
- O: 22 = 6 + 16 = 22 ✔

✔ Balanced:
4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂

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13. Al + FeO → Al₂O₃ + Fe

- Al: 1 on left, 2 on right → need 2 Al
- Fe: 1 on left, 1 on right → OK
- O: 1 on left, 3 on right → need 3 FeO

So:
- 2Al + 3FeO → Al₂O₃ + 3Fe

Check:
- Al: 2 = 2
- Fe: 3 = 3
- O: 3 = 3 ✔

✔ Balanced:
2Al + 3FeO → Al₂O₃ + 3Fe

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14. Na₂CO₃ + HCl → NaCl + H₂O + CO₂

- Na: 2 on left → need 2 NaCl
- C: 1 = 1
- O: 3 on left → CO₂ has 2, H₂O has 1 → total 3 → OK
- H: 1 on left, 2 on right → need 2 HCl

So:
- Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

Check:
- Na: 2 = 2
- C: 1 = 1
- O: 3 = 1 (H₂O) + 2 (CO₂) = 3
- H: 2 = 2
- Cl: 2 = 2 ✔

✔ Balanced:
Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

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15. K + Br₂ → KBr

- K: 1 on left, 1 on right
- Br: 2 on left, 1 on right → need 2 KBr → 2 K

So:
- 2K + Br₂ → 2KBr

✔ Balanced:
2K + Br₂ → 2KBr

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16. P₄ + O₂ → P₂O₅

- P: 4 on left, 2 on right → need 2 P₂O₅ → 4 P
- O: 2×5 = 10 O → need 5 O₂

So:
- P₄ + 5O₂ → 2P₂O₅

✔ Balanced:
P₄ + 5O₂ → 2P₂O₅

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17. C₂H₂ + O₂ → CO₂ + H₂O

- C: 2 on left → need 2 CO₂
- H: 2 on left → need 1 H₂O? But H₂O has 2 H → so 1 H₂O → H=2
- But O: 2 CO₂ → 4 O, 1 H₂O → 1 O → total 5 O → need 5/2 O₂

Multiply by 2:
- 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O

Check:
- C: 4 = 4
- H: 4 = 4
- O: 10 = 8 + 2 = 10 ✔

✔ Balanced:
2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O

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✔ Final Answers:

1. N₂ + 3H₂ → 2NH₃
2. S₈ + 12O₂ → 8SO₃
3. 2HgO → 2Hg + O₂
4. Zn + 2HCl → ZnCl₂ + H₂
5. SiCl₄ + 4H₂O → H₄SiO₄ + 4HCl
6. 2Na + 2H₂O → 2NaOH + H₂
7. 2H₃PO₄ → H₄P₂O₇ + H₂O
8. 2Si₂H₆ + 7O₂ → 4SiO₂ + 6H₂O
9. 2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O
10. 4Fe + 3O₂ → 2Fe₂O₃
11. Fe₂(SO₄)₃ + 6KOH → 3K₂SO₄ + 2Fe(OH)₃
12. 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
13. 2Al + 3FeO → Al₂O₃ + 3Fe
14. Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂
15. 2K + Br₂ → 2KBr
16. P₄ + 5O₂ → 2P₂O₅
17. 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O

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