Balancing Chemical Equations - Worksheet 1 | Science 9th Grade - Free Printable

Let's go through each of these chemical equations and balance them step by step. Balancing chemical equations involves ensuring that the number of atoms of each element is the same on both sides of the equation (conservation of mass).

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1.

H₃PO₄ + KOH → K₃PO₄ + H₂O

- Left: H, P, O, K
- Right: K, P, O, H

We see:
- 3 K on right → need 3 KOH on left
- 3 H from H₃PO₄ + 3 from KOH = 6 H on left; 3 H₂O has 6 H → good
- So:

H₃PO₄ + 3KOH → K₃PO₄ + 3H₂O

✔ Balanced.

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2.

HCl + NaOH → NaCl + H₂O

This is a neutralization reaction.

- One H⁺ from HCl reacts with one OH⁻ from NaOH to form H₂O.
- So:

HCl + NaOH → NaCl + H₂O

✔ Already balanced.

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3.

Na + NaNO₃ → Na₂O + N₂

This is a redox reaction. Sodium metal reduces nitrate.

Let’s analyze:
- On the right: Na₂O has 2 Na, N₂ has 2 N
- Left: Na and NaNO₃ — we need to determine stoichiometry.

Assume:
- Let’s suppose x Na + y NaNO₃ → z Na₂O + w N₂

But note: NaNO₃ contains Na, so total Na atoms = x + y

Right: 2z Na in Na₂O

Also, nitrogen: y → 2w (since N₂ has 2 N)

Oxygen: 3y → z (since Na₂O has 1 O)

So:
- O: 3y = z → z = 3y
- N: y = 2w → w = y/2
- Na: x + y = 2z = 2(3y) = 6y → x = 5y

Try y = 2 → then w = 1, z = 6, x = 10

So:

10Na + 2NaNO₃ → 6Na₂O + N₂

Check:
- Left: Na: 10 + 2 = 12; N: 2; O: 6
- Right: Na: 6×2 = 12; O: 6; N: 2

✔ Balanced.

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4.

N₂ + O₂ → N₂O₅

Balance O: 5 on right → need 5/2 O₂ → multiply whole equation by 2

So:
2N₂ + 5O₂ → 2N₂O₅

✔ Balanced.

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5.

H₃PO₄ + Mg(OH)₂ → Mg₃(PO₄)₂ + H₂O

Mg₃(PO₄)₂ has:
- 3 Mg
- 2 PO₄³⁻

So need:
- 2 H₃PO₄
- 3 Mg(OH)₂

Left:
- H: 2×3 = 6 from H₃PO₄ + 3×2 = 6 from Mg(OH)₂ → 12 H
- O: many, but let's count water

Right:
- Mg₃(PO₄)₂ → no H or O directly
- Water: must account for 12 H → 6 H₂O

Now check O:
- Left: 2×4 = 8 from H₃PO₄ + 3×2 = 6 from Mg(OH)₂ → 14 O
- Right: 8 from PO₄ (each PO₄ has 4 O → 2×4=8) + 6 from H₂O → 14 O ✔

So:

2H₃PO₄ + 3Mg(OH)₂ → Mg₃(PO₄)₂ + 6H₂O

✔ Balanced.

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6.

NaOH + H₂CO₃ → Na₂CO₃ + H₂O

- CO₃²⁻ comes from H₂CO₃, Na⁺ from NaOH
- Need 2 NaOH for Na₂CO₃

So:
2NaOH + H₂CO₃ → Na₂CO₃ + 2H₂O

Check:
- Na: 2 = 2
- O: 2+3 = 5 left; 3+2 = 5 right
- H: 2+2 = 4 left; 4 right

✔ Balanced.

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7.

H₂ + O₂ → H₂O₂

Hydrogen peroxide formation.

H₂ + O₂ → H₂O₂

Already balanced? Yes — 2H, 2O on both sides.

✔ Balanced.

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8.

Na + O₂ → Na₂O

Na₂O has 2 Na and 1 O

O₂ has 2 O → so need 2 Na₂O → 4 Na

So:
4Na + O₂ → 2Na₂O

✔ Balanced.

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9.

Al + S₈ → Al₂S₃

S₈ is elemental sulfur (ring of 8 S atoms)

Need to make Al₂S₃ → smallest common multiple of Al and S

Al₂S₃ has 2 Al, 3 S

S₈ has 8 S → LCM of 3 and 8 is 24

So:
- 24 S → 8 S₈ (8×3 = 24)
- 24 S requires 16 Al (since 2 Al per 3 S → 24 S → 16 Al)
- So 16 Al

Then Al₂S₃: 16 Al → 8 Al₂S₃

So:

16Al + 8S₈ → 8Al₂S₃

But simplify by dividing by 8:

2Al + S₈ → Al₂S₃

Wait: 2 Al and 1 S₈ → S₈ has 8 S, Al₂S₃ has 3 S → not balanced.

Better:

We want:
- Al₂S₃: 2 Al, 3 S
- S₈: 8 S → so use 8 Al₂S₃ → needs 16 Al and 24 S
- 24 S → 3 S₈ (since 3×8 = 24)

So:

16Al + 3S₈ → 8Al₂S₃

✔ Balanced.

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10.

Cs + N₂ → Cs₃N

N₂ → 2 N atoms → each Cs₃N needs 1 N → so 2 Cs₃N

Then: 2 Cs₃N → 6 Cs

So:

6Cs + N₂ → 2Cs₃N

✔ Balanced.

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11.

Mg + Cl₂ → MgCl₂

Simple combination.

Mg + Cl₂ → MgCl₂

✔ Already balanced.

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12.

Rb + RbNO₃ → Rb₂O + N₂

Redox: Rb metal reduces nitrate.

Let’s assume:
- Rb is oxidized to Rb⁺
- NO₃⁻ reduced to N₂

Nitrogen goes from +5 in NO₃⁻ to 0 in N₂ → gain of 5 electrons per N atom → 10 e⁻ per N₂

Each Rb loses 1 electron → need 10 Rb to lose 10 e⁻

So 10 Rb atoms from metal

But also Rb in RbNO₃: each RbNO₃ has 1 Rb and 1 NO₃⁻

To get N₂, need 2 NO₃⁻ → so 2 RbNO₃ → gives 2 Rb⁺ and 2 NO₃⁻ → produces N₂

Total Rb atoms: 10 (metal) + 2 (from salt) = 12 Rb

On right: Rb₂O → 2 Rb per formula unit

So 12 Rb → 6 Rb₂O

Check oxygen: 2 RbNO₃ → 6 O → 6 Rb₂O → 6 O ✔

So:

10Rb + 2RbNO₃ → 6Rb₂O + N₂

Check:
- Rb: 10 + 2 = 12; right: 6×2 = 12 ✔
- N: 2 → 2 ✔
- O: 6 → 6 ✔

✔ Balanced.

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13.

C₆H₆ + O₂ → CO₂ + H₂O

Combustion of benzene.

C₆H₆ → 6 C → 6 CO₂
→ 6 H → 3 H₂O

So:
C₆H₆ → 6CO₂ + 3H₂O

Now O atoms:
- Right: 6×2 = 12 + 3×1 = 3 → 15 O
- So need 15/2 O₂ → multiply entire equation by 2:

2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O

✔ Balanced.

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14.

N₂ + H₂ → NH₃

Classic Haber process.

N₂ + 3H₂ → 2NH₃

✔ Balanced.

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15.

C + O₂ → CO₂

C + O₂ → CO₂

✔ Already balanced.

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16.

C₃H₈ + O₂ → CO₂ + H₂O

Propane combustion.

C₃H₈ → 3 CO₂ + 4 H₂O

Now O:
- Right: 3×2 = 6 + 4×1 = 4 → 10 O → need 5 O₂

So:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

✔ Balanced.

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17.

Li + AlCl₃ → LiCl + Al

Single displacement.

AlCl₃ → needs 3 Li to displace Al

So:

3Li + AlCl₃ → 3LiCl + Al

✔ Balanced.

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18.

Rb + P → Rb₃P

Phosphide formation.

3Rb + P → Rb₃P

✔ Balanced.

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19.

Al(OH)₃ + H₂SO₄ → Al₂(SO₄)₃ + H₂O

Al₂(SO₄)₃ has 2 Al, 3 SO₄

So need:
- 2 Al(OH)₃
- 3 H₂SO₄

Now H and O:

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

Total H: 6 + 6 = 12 → need 6 H₂O

Right: 6 H₂O → 12 H, 6 O

But also Al₂(SO₄)₃ → 3×4 = 12 O from SO₄

Total O on right: 12 (SO₄) + 6 (H₂O) = 18

Left: 6 (from OH) + 12 (from SO₄) = 18 ✔

So:

2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O

✔ Balanced.

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20.

Rb + S₈ → Rb₂S

S₈ → 8 S atoms → each Rb₂S has 1 S → so 8 Rb₂S → 16 Rb

So:

16Rb + S₈ → 8Rb₂S

✔ Balanced.

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21.

CaSO₄ + KOH → Ca(OH)₂ + K₂SO₄

Double displacement.

Swap ions: Ca²⁺ with K⁺

So:

CaSO₄ + 2KOH → Ca(OH)₂ + K₂SO₄

✔ Balanced.

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22.

FeBr₃ + Na → Fe + NaBr

Displacement.

FeBr₃ → Fe³⁺ → needs 3 Na to reduce to Fe

So:

FeBr₃ + 3Na → Fe + 3NaBr

✔ Balanced.

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23.

Al₄C₃ + H₂O → CH₄ + Al(OH)₃

Aluminum carbide hydrolysis.

Al₄C₃ → 3 C → 3 CH₄

Each C becomes CH₄ → need 4 H per CH₄ → total 12 H

From water: each H₂O provides 2 H → need 6 H₂O

Also, Al₄C₃ → 4 Al → 4 Al(OH)₃

Now check O:
- 6 H₂O → 6 O
- 4 Al(OH)₃ → 4×3 = 12 O → need more?

Wait: Al(OH)₃ has 3 O → 4×3 = 12 O

But only 6 O from water → not enough

But each Al(OH)₃ has 3 O and 3 H → total 6 H and 3 O

But H₂O provides H and O.

Let’s write:

Al₄C₃ → 4 Al and 3 C

3 C → 3 CH₄ → need 12 H

4 Al → 4 Al(OH)₃ → each needs 3 OH → 12 OH → need 12 H and 12 O

Total H needed: 12 (for CH₄) + 12 (for OH) = 24 H → 12 H₂O

Total O needed: 12 O → 12 H₂O → yes

So:

Al₄C₃ + 12H₂O → 3CH₄ + 4Al(OH)₃

Check:
- Al: 4 = 4
- C: 3 = 3
- H: 24 = 3×4 + 4×3 = 12 + 12 = 24
- O: 12 = 4×3 = 12

✔ Balanced.

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24.

Ba + HNO₃ → H₂ + Ba(NO₃)₂

Barium displaces hydrogen.

Ba → Ba²⁺ → needs 2 H⁺ → produces H₂

So:

Ba + 2HNO₃ → H₂ + Ba(NO₃)₂

✔ Balanced.

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25.

NH₄OH + HBr → H₂O + NH₄Br

Neutralization.

NH₄OH + HBr → NH₄Br + H₂O

✔ Already balanced.

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## ✔ Final Balanced Equations:

1. H₃PO₄ + 3KOH → K₃PO₄ + 3H₂O
2. HCl + NaOH → NaCl + H₂O
3. 10Na + 2NaNO₃ → 6Na₂O + N₂
4. 2N₂ + 5O₂ → 2N₂O₅
5. 2H₃PO₄ + 3Mg(OH)₂ → Mg₃(PO₄)₂ + 6H₂O
6. 2NaOH + H₂CO₃ → Na₂CO₃ + 2H₂O
7. H₂ + O₂ → H₂O₂
8. 4Na + O₂ → 2Na₂O
9. 16Al + 3S₈ → 8Al₂S₃
10. 6Cs + N₂ → 2Cs₃N
11. Mg + Cl₂ → MgCl₂
12. 10Rb + 2RbNO₃ → 6Rb₂O + N₂
13. 2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O
14. N₂ + 3H₂ → 2NH₃
15. C + O₂ → CO₂
16. C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
17. 3Li + AlCl₃ → 3LiCl + Al
18. 3Rb + P → Rb₃P
19. 2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O
20. 16Rb + S₈ → 8Rb₂S
21. CaSO₄ + 2KOH → Ca(OH)₂ + K₂SO₄
22. FeBr₃ + 3Na → Fe + 3NaBr
23. Al₄C₃ + 12H₂O → 3CH₄ + 4Al(OH)₃
24. Ba + 2HNO₃ → H₂ + Ba(NO₃)₂
25. NH₄OH + HBr → H₂O + NH₄Br

All equations are now balanced! Let me know if you'd like a printable version or explanation for any specific one.