Grade 10 Science homework sheet focused on balancing chemical equations, with 17 problems listed for practice.

Let's solve each of these chemical equations by balancing them. Balancing a chemical equation means ensuring the number of atoms of each element is equal on both sides (reactants and products), following the Law of Conservation of Mass.

We’ll go through each one step-by-step and provide the balanced equation.

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1. N₂ + H₂ → NH₃

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

To balance:
- Put 2 in front of NH₃ → now N = 2, H = 6
- So we need 3 H₂ → gives 6 H
- N₂ already has 2 N

✔ Balanced:
N₂ + 3H₂ → 2NH₃

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2. S₈ + O₂ → SO₃

- Left: S = 8, O = 2
- Right: S = 1, O = 3

We need 8 S on right → so 8 SO₃ → S = 8, O = 24
Now O₂ must supply 24 O → so 12 O₂

✔ Balanced:
S₈ + 12O₂ → 8SO₃

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3. HgO → Hg + O₂

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

Need 2 O on left → use 2 HgO
Then get 2 Hg and 1 O₂

✔ Balanced:
2HgO → 2Hg + O₂

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4. Zn + HCl → ZnCl₂ + H₂

- Left: Zn = 1, H = 1, Cl = 1
- Right: Zn = 1, Cl = 2, H = 2

So need 2 HCl to give 2 Cl and 2 H

✔ 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
But that gives 4 H from HCl → total H = 4 (from H₄SiO₄) + 4 (from HCl) = 8 H

Left: H₂O → only 2 H per molecule → need 4 H₂O → gives 8 H and 4 O

Now check O: left = 4 O, right = 4 O in H₄SiO₄ → good

✔ 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 (in NaOH) + 2 (in H₂) = 3 → unbalanced

Try 2Na → 2NaOH → needs 2H₂O → gives 4H and 2O
Right: 2NaOH → 2Na, 2O, 2H; H₂ → 2H → total H = 4

Left: 2H₂O → 4H, 2O → matches

✔ Balanced:
2Na + 2H₂O → 2NaOH + H₂

Wait — H₂ is diatomic, but only 2H needed? Let's double-check:

2Na + 2H₂O → 2NaOH + H₂
Atoms:
Left: Na=2, H=4, O=2
Right: Na=2, O=2, H=2 (in NaOH) + 2 (in H₂) = 4 → ✔

✔ Final: 2Na + 2H₂O → 2NaOH + H₂

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7. H₃PO₄ → H₄P₂O₇ + H₂O

- Left: H=3, P=1, O=4
- Right: H=4, P=2, O=7 (in H₄P₂O₇) + 1 (in H₂O) = 8 → not balanced

We want to form H₄P₂O₇ → needs 2 P → so 2 H₃PO₄
Left: 2H₃PO₄ → H=6, P=2, O=8

Right: H₄P₂O₇ → H=4, P=2, O=7; H₂O → H=2, O=1 → total H=6, O=8 → perfect!

✔ Balanced:
2H₃PO₄ → H₄P₂O₇ + H₂O

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8. Si₂H₆ + O₂ → SiO₂ + H₂O

- Left: Si=2, H=6, O=2
- Right: Si=1, O=2 (in SiO₂) + 1 (in H₂O) = 3, H=2 → unbalanced

Use 2 SiO₂ → Si=2, O=4
Need H₂O for H: 6 H → 3 H₂O → H=6, O=3 → total O = 4+3 = 7

So O₂ must supply 7 O → 3.5 O₂ → multiply all by 2 to eliminate fraction

Start over with 2 Si₂H₆ → Si=4, H=12
→ 4 SiO₂ → Si=4, O=8
→ 6 H₂O → H=12, O=6 → total O = 14

So O₂ needed: 14/2 = 7 O₂

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

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9. Al(OH)₃ + H₂SO₄ → Al₂(SO₄)₃ + H₂O

- Left: Al=1, O=3+4=7, H=3+2=5, S=1
- Right: Al=2, S=3, O=12+1=13, H=2 → unbalanced

Need 2 Al(OH)₃ → Al=2, O=6, H=6
Need 3 H₂SO₄ → S=3, H=6, O=12 → total H=6+6=12

Right: Al₂(SO₄)₃ → Al=2, S=3, O=12
H₂O → need 6 H₂O → H=12, O=6 → total O = 12+6=18

Left: O from Al(OH)₃: 2×3 = 6; from H₂SO₄: 3×4 = 12 → total O = 18 → good

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

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10. Fe + O₂ → Fe₂O₃

- Left: Fe=1, O=2
- Right: Fe=2, O=3

Need 2 Fe → 2Fe
Need 3/2 O₂ → multiply whole equation by 2

→ 4Fe + 3O₂ → 2Fe₂O₃

✔ Balanced:
4Fe + 3O₂ → 2Fe₂O₃

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11. Fe₂(SO₄)₃ + KOH → K₂SO₄ + Fe(OH)₃

- Left: Fe=2, S=3, O=12+1=13 (from SO₄), K=1, O=1, H=1
- Right: K=2, S=1, O=4+3=7, Fe=1, O=3, H=3 → messy

Fe₂(SO₄)₃ → 2 Fe, 3 SO₄
Each SO₄ needs 2 K⁺ → so 6 K⁺ → 3 K₂SO₄
So 6 KOH → 6 K, 6 O, 6 H

Fe(OH)₃ → each Fe needs 3 OH → 2 Fe → 6 OH → so 6 H₂O? No — OH comes from KOH

So 6 KOH → provides 6 K and 6 OH

Products: 3 K₂SO₄ (K=6, S=3, O=12), 2 Fe(OH)₃ (Fe=2, O=6, H=6)

Check left:
Fe₂(SO₄)₃ → Fe=2, S=3, O=12
6 KOH → K=6, O=6, H=6 → total O = 12+6=18

Right: 3 K₂SO₄ → O=12, 2 Fe(OH)₃ → O=6 → total O=18 → good

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

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12. FeS₂ + O₂ → Fe₂O₃ + SO₂

- Left: Fe=1, S=2, O=2
- Right: Fe=2, O=3, S=1, O=2 → total O=5

Need 2 Fe → 2 FeS₂ → Fe=2, S=4
→ 2 Fe₂O₃ → Fe=4? No — wait: 2 FeS₂ → 2 Fe

So need 2 FeS₂ → 2 Fe, 4 S
→ 2 Fe₂O₃ → 4 Fe → too many

So 2 FeS₂ → 2 Fe → need 1 Fe₂O₃ → Fe=2 → so 2 FeS₂ → 1 Fe₂O₃ → Fe=2

S=4 → 4 SO₂ → S=4, O=8
Fe₂O₃ → O=3 → total O = 3+8=11

So O₂ must supply 11 O → 5.5 O₂ → ×2

Multiply all by 2:

4 FeS₂ → 2 Fe₂O₃ + 8 SO₂ → O from products: 2×3 + 8×2 = 6+16=22

So O₂ needed: 22/2 = 11 O₂

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

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13. Al + FeO → Al₂O₃ + Fe

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

Need 2 Al → 2 Al
Need 3 FeO → Fe=3, O=3 → then 3 Fe produced

So: 2Al + 3FeO → Al₂O₃ + 3Fe

✔ Balanced:
2Al + 3FeO → Al₂O₃ + 3Fe

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14. Na₂CO₃ + HCl → NaCl + H₂O + CO₂

- Left: Na=2, C=1, O=3, H=1, Cl=1
- Right: Na=1, Cl=1, H=2, O=1+2=3, C=1 → unbalanced

Need 2 NaCl → so 2 HCl → H=2, Cl=2
Left: Na₂CO₃ + 2HCl → Na=2, C=1, O=3, H=2, Cl=2

Right: 2NaCl → Na=2, Cl=2; H₂O → H=2, O=1; CO₂ → C=1, O=2 → total O=3 → good

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

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15. K + Br₂ → KBr

- Left: K=1, Br=2
- Right: K=1, Br=1 → unbalanced

Need 2 KBr → so 2 K and 1 Br₂

✔ Balanced:
2K + Br₂ → 2KBr

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16. P₄ + O₂ → P₂O₅

- Left: P=4, O=2
- Right: P=2, O=5

Need 2 P₂O₅ → P=4, O=10
So O₂ needed: 10/2 = 5 O₂

✔ Balanced:
P₄ + 5O₂ → 2P₂O₅

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17. C₂H₂ + O₂ → CO₂ + H₂O

- Left: C=2, H=2, O=2
- Right: C=1, O=2+1=3, H=2 → unbalanced

Need 2 CO₂ → C=2, O=4
H₂O → H=2 → 1 H₂O → H=2, O=1 → total O=5

So O₂ needed: 5/2 → ×2

Multiply all by 2:

2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O

Check:
Left: C=4, H=4, O=10
Right: C=4, O=8+2=10, H=4 → ✔

✔ Balanced:
2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O

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✔ Final Answers (Balanced Equations):

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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