Zeal Academy Ch-1 Worksheet: Balance the chemical equations.

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.

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1) N₂ + H₂ → NH₃

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

To balance:
- Use 2 NH₃ → 2 N, 6 H
- So need 3 H₂ to give 6 H
- And 1 N₂ gives 2 N

✔ Balanced:
1 N₂ + 3 H₂ → 2 NH₃

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2) KClO₃ → KCl + O₂

- Left: K=1, Cl=1, O=3
- Right: K=1, Cl=1, O=2 (in O₂)

Need to balance oxygen:
- 3 O on left → so need 3/2 O₂ on right → but we avoid fractions.
- Multiply whole equation by 2:

→ 2 KClO₃ → 2 KCl + 3 O₂

✔ Balanced:
2 KClO₃ → 2 KCl + 3 O₂

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3) NaCl + F₂ → NaF + Cl₂

- This is a single displacement: fluorine displaces chlorine.

Left: Na=1, Cl=1, F=2
Right: Na=1, F=1, Cl=2

Balance:
- Need 2 NaCl → 2 Na, 2 Cl
- Need 2 NaF → 2 Na, 2 F
- F₂ provides 2 F → good
- Cl₂ → 2 Cl → good

✔ Balanced:
2 NaCl + F₂ → 2 NaF + Cl₂

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4) H₂ + O₂ → H₂O

- Left: H=2, O=2
- Right: H=2, O=1

So need 2 H₂O → 2 O, 4 H → need 2 H₂

✔ Balanced:
2 H₂ + O₂ → 2 H₂O

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5) Pb(OH)₂ + HCl → H₂O + PbCl₂

- Left: Pb=1, O=2, H=2+1=3, Cl=1
- Right: H=2, O=1, Pb=1, Cl=2

Balance:
- PbCl₂ needs 2 Cl → so 2 HCl
- Then H from HCl = 2, plus OH has 2 H → total H = 4
- H₂O has 2 H per molecule → need 2 H₂O
- O: Pb(OH)₂ has 2 O, H₂O has 2 O → balanced

✔ Balanced:
Pb(OH)₂ + 2 HCl → 2 H₂O + PbCl₂

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6) AlBr₃ + K₂SO₄ → KBr + Al₂(SO₄)₃

- Left: Al=1, Br=3, K=2, S=1, O=4
- Right: K=1, Br=1, Al=2, S=3, O=12

Need to balance:
- Al₂(SO₄)₃ → 2 Al, 3 SO₄ → so need 3 K₂SO₄ → 6 K, 3 S, 12 O
- AlBr₃ → need 2 AlBr₃ → 2 Al, 6 Br
- KBr → need 6 KBr → 6 K, 6 Br

✔ Balanced:
2 AlBr₃ + 3 K₂SO₄ → 6 KBr + Al₂(SO₄)₃

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7) CH₄ + O₂ → CO₂ + H₂O

- Combustion of methane
- C: 1, H: 4, O: 2 on left
- Right: CO₂ has 1 C, 2 O; H₂O has 2 H, 1 O

Balance H: 4 H → need 2 H₂O → 4 H, 2 O
Now total O on right: CO₂ (2) + H₂O (2) = 4 O → need 2 O₂

✔ Balanced:
CH₄ + 2 O₂ → CO₂ + 2 H₂O

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8) C₃H₈ + O₂ → CO₂ + H₂O

Propane combustion:
- C: 3, H: 8
- CO₂: 1 C → need 3 CO₂
- H₂O: 8 H → need 4 H₂O

O on right: 3×2 = 6 (CO₂) + 4×1 = 4 (H₂O) = 10 O → need 5 O₂

✔ Balanced:
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

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9) C₈H₁₈ + O₂ → CO₂ + H₂O

Octane combustion:
- C: 8 → 8 CO₂
- H: 18 → 9 H₂O
- O: 8×2 = 16 (CO₂) + 9×1 = 9 (H₂O) = 25 O → need 25/2 O₂ → multiply all by 2

→ 2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O

✔ Balanced:
2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O

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10) FeCl₃ + NaOH → Fe(OH)₃ + NaCl

- Fe: 1, Cl: 3, Na: 1, O: 1, H: 1
- Fe(OH)₃ → 3 OH → need 3 NaOH
- Then Na: 3 → need 3 NaCl
- Cl: 3 → matches

✔ Balanced:
FeCl₃ + 3 NaOH → Fe(OH)₃ + 3 NaCl

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11) P + O₂ → P₂O₅

- P: 1 → need 2 P
- O: 2 → P₂O₅ has 5 O → need 5/2 O₂ → use 2× for no fraction

→ 4 P + 5 O₂ → 2 P₂O₅

✔ Balanced:
4 P + 5 O₂ → 2 P₂O₅

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12) Na + H₂O → NaOH + H₂

- Na: 1, H: 2, O: 1
- Right: NaOH has 1 Na, 1 O, 1 H; H₂ has 2 H → total H = 3 → not balanced

Try 2 Na:
- 2 Na + 2 H₂O → 2 NaOH + H₂
- Left: Na=2, H=4, O=2
- Right: 2 NaOH → 2 Na, 2 O, 2 H; H₂ → 2 H → total H=4

✔ Balanced:
2 Na + 2 H₂O → 2 NaOH + H₂

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13) Ag₂O → Ag + O₂

- Left: Ag=2, O=1
- Right: Ag=1, O=2

Need 2 Ag on right → 2 Ag
O: 1 on left → need 1/2 O₂ → multiply all by 2:

→ 2 Ag₂O → 4 Ag + O₂

✔ Balanced:
2 Ag₂O → 4 Ag + O₂

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14) S₈ + O₂ → SO₃

- S₈ → 8 S → need 8 SO₃
- Each SO₃ has 3 O → 8×3 = 24 O → need 12 O₂

✔ Balanced:
S₈ + 12 O₂ → 8 SO₃

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15) CO₂ + H₂O → C₆H₁₂O₆ + O₂

Photosynthesis:
- Left: C=1, O=3, H=2
- Right: C=6, H=12, O=6+2=8

So scale up:
- 6 CO₂ → 6 C, 12 O
- 6 H₂O → 12 H, 6 O → total O = 18
- C₆H₁₂O₆ → 6 C, 12 H, 6 O
- O₂ → need 6 O → 3 O₂

✔ Balanced:
6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂

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16) K + MgBr₂ → KBr + Mg

Single displacement:
- MgBr₂ → 2 Br → need 2 KBr
- So 2 K needed
- Mg: 1 → 1 Mg

✔ Balanced:
2 K + MgBr₂ → 2 KBr + Mg

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17) HCl + CaCO₃ → CaCl₂ + H₂O + CO₂

- CaCO₃ reacts with acid
- Ca: 1 → CaCl₂ → need 2 Cl → 2 HCl
- H: 2 → H₂O → 2 H
- CO₃ → CO₂ + H₂O

So:
- 2 HCl + CaCO₃ → CaCl₂ + H₂O + CO₂

Check atoms:
- Left: H=2, Cl=2, Ca=1, C=1, O=3
- Right: Ca=1, Cl=2, H=2, O=1 (H₂O) + 2 (CO₂) = 3 → yes

✔ Balanced:
2 HCl + CaCO₃ → CaCl₂ + H₂O + CO₂

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18) HNO₃ + NaHCO₃ → NaNO₃ + H₂O + CO₂

Acid-base reaction:
- HNO₃ → H⁺, NO₃⁻
- NaHCO₃ → Na⁺, HCO₃⁻ → forms H₂CO₃ → H₂O + CO₂

So:
- HNO₃ + NaHCO₃ → NaNO₃ + H₂O + CO₂

All atoms balanced as written:
- H: 1+1=2 → H₂O has 2 H
- N:1, O:3+3=6 → NaNO₃ has 3, H₂O has 1, CO₂ has 2 → total 6
- Na:1, C:1

✔ Balanced:
HNO₃ + NaHCO₃ → NaNO₃ + H₂O + CO₂

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19) H₂O + O₂ → H₂O₂

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

Not balanced.

But H₂O₂ has 2 H, 2 O → need 1 O from O₂? But O₂ has 2 O.

Wait:
H₂O + O₂ → H₂O₂ → O atoms: left=3, right=2 → imbalance

Actually, this reaction is not typically balanced like this. But let's see:

Suppose:
2 H₂O + O₂ → 2 H₂O₂

Left: H=4, O=2+2=4
Right: H=4, O=4 → yes!

✔ Balanced:
2 H₂O + O₂ → 2 H₂O₂

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20) NaBr + CaF₂ → NaF + CaBr₂

Double displacement:
- CaBr₂ → needs 2 Br → so 2 NaBr
- NaF → need 2 NaF
- CaF₂ → 2 F → matches

✔ Balanced:
2 NaBr + CaF₂ → 2 NaF + CaBr₂

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21) H₂SO₄ + NaNO₂ → HNO₂ + Na₂SO₄

- Left: H=2, S=1, O=4+2=6, Na=1, N=1
- Right: H=1, N=1, O=2+4=6, Na=2, S=1 → Na unbalanced

Need 2 Na on left → 2 NaNO₂
Then N=2 → need 2 HNO₂
H: left=2, right=2 → ok
O: H₂SO₄ → 4 O, 2 NaNO₂ → 4 O → total 8 O
Right: 2 HNO₂ → 4 O, Na₂SO₄ → 4 O → total 8 O → ok

✔ Balanced:
H₂SO₄ + 2 NaNO₂ → 2 HNO₂ + Na₂SO₄

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✔ Final Answers:

1) N₂ + 3 H₂ → 2 NH₃
2) 2 KClO₃ → 2 KCl + 3 O₂
3) 2 NaCl + F₂ → 2 NaF + Cl₂
4) 2 H₂ + O₂ → 2 H₂O
5) Pb(OH)₂ + 2 HCl → 2 H₂O + PbCl₂
6) 2 AlBr₃ + 3 K₂SO₄ → 6 KBr + Al₂(SO₄)₃
7) CH₄ + 2 O₂ → CO₂ + 2 H₂O
8) C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
9) 2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O
10) FeCl₃ + 3 NaOH → Fe(OH)₃ + 3 NaCl
11) 4 P + 5 O₂ → 2 P₂O₅
12) 2 Na + 2 H₂O → 2 NaOH + H₂
13) 2 Ag₂O → 4 Ag + O₂
14) S₈ + 12 O₂ → 8 SO₃
15) 6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂
16) 2 K + MgBr₂ → 2 KBr + Mg
17) 2 HCl + CaCO₃ → CaCl₂ + H₂O + CO₂
18) HNO₃ + NaHCO₃ → NaNO₃ + H₂O + CO₂
19) 2 H₂O + O₂ → 2 H₂O₂
20) 2 NaBr + CaF₂ → 2 NaF + CaBr₂
21) H₂SO₄ + 2 NaNO₂ → 2 HNO₂ + Na₂SO₄

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Let me know if you'd like an explanation for any specific one!