Balancing Equations Worksheet for chemistry practice.

Let's solve the Balancing Equations Worksheet step by step. I'll go through each chemical equation, balance it using the law of conservation of mass, and provide the balanced coefficients.

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🔧 How to Balance Chemical Equations:

1. Count atoms of each element on both sides.
2. Use coefficients (whole numbers) to balance atoms.
3. Start with elements that appear in only one compound on each side.
4. Save H and O for last if they appear in multiple compounds.
5. Check all atoms after balancing.

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Now let’s solve each one:

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1)

H₃PO₄ + KOH → K₃PO₄ + H₂O

- P: 1 = 1 ✔
- K: 1 on left → need 3 on right → put 3 KOH
- H: 3 from H₃PO₄ + 3 from KOH = 6 H; right: 2 H₂O → 2×2 = 4 H → not balanced
- Try:
H₃PO₄ + 3KOH → K₃PO₄ + 3H₂O
H: 3+3=6 on left; 3×2=6 on right ✔
O: 4+3=7 vs 4+3=7 ✔
K: 3 = 3 ✔

✔ Answer:
1 H₃PO₄ + 3 KOH → 1 K₃PO₄ + 3 H₂O

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2)

K + B₂O₃ → K₂O + B

- B: 2 on left → need 2 B on right → 2 B
- O: 3 on left → K₂O has 1 O → need 3 K₂O → 6 K
- So: 6 K → 3 K₂O → 6 K atoms
- Now: 6 K + B₂O₃ → 3 K₂O + 2 B

✔ Answer:
6 K + 1 B₂O₃ → 3 K₂O + 2 B

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3)

HCl + NaOH → NaCl + H₂O

This is an acid-base neutralization:
- HCl + NaOH → NaCl + H₂O
All atoms are already balanced.

✔ Answer:
1 HCl + 1 NaOH → 1 NaCl + 1 H₂O

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4)

Na + NaNO₃ → Na₂O + N₂

- Na: 1 + 1 = 2 on left → right: Na₂O has 2 Na → good
- But N: 1 on left → N₂ has 2 → need 2 NO₃ → so use 2 NaNO₃
- Then: 2 NaNO₃ → 2 Na + 2 NO₃ → but we have Na on both sides

Try:
Let’s suppose:
a Na + b NaNO₃ → c Na₂O + d N₂

Balance:
- Na: a + b = 2c
- N: b = 2d
- O: 3b = c

From O: c = 3b
From N: d = b/2 → so b must be even

Try b = 2 → then d = 1, c = 6
Then Na: a + 2 = 2×6 = 12 → a = 10

So:
10 Na + 2 NaNO₃ → 6 Na₂O + 1 N₂

Check:
- Na: 10 + 2 = 12; right: 6×2 = 12 ✔
- N: 2 = 2 ✔
- O: 2×3 = 6; right: 6×1 = 6 ✔

✔ Answer:
10 Na + 2 NaNO₃ → 6 Na₂O + 1 N₂

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5)

C + S₈ → CS₂

S₈ has 8 S atoms → CS₂ has 2 S → need 4 CS₂ → 8 S
→ C: 4 needed

So: 4 C + 1 S₈ → 4 CS₂

✔ Answer:
4 C + 1 S₈ → 4 CS₂

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6)

Na + O₂ → Na₂O

- O₂ → 2 O → Na₂O has 1 O → need 2 Na₂O → 4 Na
- So: 4 Na + O₂ → 2 Na₂O

✔ Answer:
4 Na + 1 O₂ → 2 Na₂O

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7)

N₂ + O₂ → N₂O₅

- N₂: 2 N → N₂O₅ has 2 N → good
- O: 2 on left → 5 on right → need 5/2 O₂ → multiply whole by 2

→ 2 N₂ + 5 O₂ → 2 N₂O₅

✔ Answer:
2 N₂ + 5 O₂ → 2 N₂O₅

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8)

H₃PO₄ + Mg(OH)₂ → Mg₃(PO₄)₂ + H₂O

Mg₃(PO₄)₂ → needs 3 Mg and 2 PO₄

So:
- Need 2 H₃PO₄ → gives 2 PO₄
- Need 3 Mg(OH)₂ → gives 3 Mg

Left:
- H: 2×3 = 6 from H₃PO₄ + 3×2 = 6 from Mg(OH)₂ → total 12 H
- O: many, but H₂O will take care

Right: H₂O → need 6 H₂O to get 12 H

Check:
- 2 H₃PO₄ + 3 Mg(OH)₂ → 1 Mg₃(PO₄)₂ + 6 H₂O

Atoms:
- P: 2 = 2 ✔
- Mg: 3 = 3 ✔
- O: 2×4 = 8 + 3×2 = 6 → 14 O left; right: 8 (in PO₄) + 6 = 14 ✔
- H: 6 + 6 = 12 → 6×2 = 12 ✔

✔ Answer:
2 H₃PO₄ + 3 Mg(OH)₂ → 1 Mg₃(PO₄)₂ + 6 H₂O

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9)

NaOH + H₂CO₃ → Na₂CO₃ + H₂O

- CO₃²⁻: 1 → 1 → good
- Na: 1 on left → Na₂CO₃ has 2 → need 2 NaOH
- H: 2 NaOH → 2 H; H₂CO₃ → 2 H → total 4 H
- Right: H₂O → 2 H → need 2 H₂O

So: 2 NaOH + H₂CO₃ → Na₂CO₃ + 2 H₂O

✔ Answer:
2 NaOH + 1 H₂CO₃ → 1 Na₂CO₃ + 2 H₂O

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10)

KOH + HBr → KBr + H₂O

Simple acid-base:
KOH + HBr → KBr + H₂O → already balanced

✔ Answer:
1 KOH + 1 HBr → 1 KBr + 1 H₂O

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11)

Na + O₂ → Na₂O

Same as #6:
4 Na + O₂ → 2 Na₂O

✔ Answer:
4 Na + 1 O₂ → 2 Na₂O

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12)

Al(OH)₃ + H₂CO₃ → Al₂(CO₃)₃ + H₂O

Al₂(CO₃)₃ → needs 2 Al and 3 CO₃

So:
- 2 Al(OH)₃ → 2 Al
- 3 H₂CO₃ → 3 CO₃

H: 2×3 = 6 from Al(OH)₃ + 3×2 = 6 from H₂CO₃ → 12 H
→ Need 6 H₂O

So: 2 Al(OH)₃ + 3 H₂CO₃ → Al₂(CO₃)₃ + 6 H₂O

✔ Answer:
2 Al(OH)₃ + 3 H₂CO₃ → 1 Al₂(CO₃)₃ + 6 H₂O

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13)

Al + S₈ → Al₂S₃

S₈ → 8 S → Al₂S₃ has 3 S → LCM of 8 and 3 is 24

So: 8 Al₂S₃ → 16 Al, 24 S
S₈ → 3 S₈ → 24 S
Al: 16 → so 16 Al

So: 16 Al + 3 S₈ → 8 Al₂S₃

✔ Answer:
16 Al + 3 S₈ → 8 Al₂S₃

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14)

Cs + N₂ → Cs₃N

N₂ → 2 N → Cs₃N has 1 N → need 2 Cs₃N → 6 Cs

So: 6 Cs + N₂ → 2 Cs₃N

✔ Answer:
6 Cs + 1 N₂ → 2 Cs₃N

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15)

Mg + Cl₂ → MgCl₂

Already balanced: 1 Mg + 1 Cl₂ → 1 MgCl₂

✔ Answer:
1 Mg + 1 Cl₂ → 1 MgCl₂

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16)

Rb + RbNO₃ → Rb₂O + N₂

This is a redox reaction.

Rb metal reduces NO₃⁻ to N₂.

Let’s balance:
- N: RbNO₃ → N₂ → 2 N → need 2 RbNO₃
- Rb: 2 RbNO₃ → 2 Rb
- But also Rb metal → so total Rb on left: Rb + 2 Rb = 3 Rb? Wait

Let’s write:
a Rb + b RbNO₃ → c Rb₂O + d N₂

- Rb: a + b = 2c
- N: b = 2d
- O: 3b = c

From O: c = 3b
From N: d = b/2 → b even

Try b = 2 → d = 1, c = 6
Then Rb: a + 2 = 12 → a = 10

So: 10 Rb + 2 RbNO₃ → 6 Rb₂O + 1 N₂

Check:
- Rb: 10 + 2 = 12; right: 6×2 = 12 ✔
- N: 2 = 2 ✔
- O: 2×3 = 6; right: 6×1 = 6 ✔

✔ Answer:
10 Rb + 2 RbNO₃ → 6 Rb₂O + 1 N₂

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17)

C₆H₆ + O₂ → CO₂ + H₂O

Combustion of benzene.

C₆H₆ → 6 C → 6 CO₂
H: 6 → 3 H₂O

So: C₆H₆ + O₂ → 6 CO₂ + 3 H₂O

O: right: 6×2 + 3×1 = 12 + 3 = 15 O → need 15/2 O₂ → ×2

→ 2 C₆H₆ + 15 O₂ → 12 CO₂ + 6 H₂O

✔ Answer:
2 C₆H₆ + 15 O₂ → 12 CO₂ + 6 H₂O

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18)

N₂ + H₂ → NH₃

Classic: N₂ + 3 H₂ → 2 NH₃

✔ Answer:
1 N₂ + 3 H₂ → 2 NH₃

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19)

C₁₀H₂₂ + O₂ → CO₂ + H₂O

C₁₀H₂₂ → 10 C → 10 CO₂
H: 22 → 11 H₂O

O: right: 10×2 + 11×1 = 20 + 11 = 31 O → need 31/2 O₂ → ×2

→ 2 C₁₀H₂₂ + 31 O₂ → 20 CO₂ + 22 H₂O

✔ Answer:
2 C₁₀H₂₂ + 31 O₂ → 20 CO₂ + 22 H₂O

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20)

Al(OH)₃ + HBr → AlBr₃ + H₂O

Al(OH)₃ → Al³⁺, OH⁻
HBr → H⁺, Br⁻

Need 3 Br → 3 HBr
Also: 3 H₂O from 3 OH and 3 H

So: Al(OH)₃ + 3 HBr → AlBr₃ + 3 H₂O

✔ Answer:
1 Al(OH)₃ + 3 HBr → 1 AlBr₃ + 3 H₂O

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21)

CH₃CH₂CH₂CH₃ (butane) + O₂ → CO₂ + H₂O

Butane: C₄H₁₀

→ 4 CO₂ + 5 H₂O

O: 4×2 + 5×1 = 8 + 5 = 13 → 13/2 O₂ → ×2

→ 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

✔ Answer:
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

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22)

C₃H₈ + O₂ → CO₂ + H₂O

Propane: C₃H₈

→ 3 CO₂ + 4 H₂O

O: 3×2 + 4×1 = 6 + 4 = 10 → 5 O₂

So: C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

✔ Answer:
1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

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23)

Li + AlCl₃ → LiCl + Al

Redox: Li displaces Al

AlCl₃ → 3 Cl⁻ → need 3 LiCl → 3 Li

So: 3 Li + AlCl₃ → 3 LiCl + Al

✔ Answer:
3 Li + 1 AlCl₃ → 3 LiCl + 1 Al

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24)

C₂H₆ + O₂ → CO₂ + H₂O

Ethane: C₂H₆

→ 2 CO₂ + 3 H₂O

O: 4 + 3 = 7 → 7/2 O₂ → ×2

→ 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

✔ Answer:
2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

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25)

NH₄OH + H₃PO₄ → (NH₄)₃PO₄ + H₂O

(NH₄)₃PO₄ → needs 3 NH₄⁺ → 3 NH₄OH

H₃PO₄ → 1

Now: 3 NH₄OH + H₃PO₄ → (NH₄)₃PO₄ + 3 H₂O

H: left: 3×5 = 15 H (NH₄OH has 5 H), H₃PO₄ has 3 → total 18 H
Right: (NH₄)₃PO₄ has 12 H, 3 H₂O has 6 → 18 ✔

O: 3×1 + 4 = 7; right: 4 + 3 = 7 ✔

✔ Answer:
3 NH₄OH + 1 H₃PO₄ → 1 (NH₄)₃PO₄ + 3 H₂O

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26)

Rb + P → Rb₃P

Need 3 Rb per P → 3 Rb + P → Rb₃P

✔ Answer:
3 Rb + 1 P → 1 Rb₃P

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27)

CH₄ + O₂ → CO₂ + H₂O

Methane combustion:

CH₄ → CO₂ + 2 H₂O

O: 2 + 1 = 3 → 3/2 O₂ → ×2

→ 2 CH₄ + 3 O₂ → 2 CO₂ + 4 H₂O

✔ Answer:
2 CH₄ + 3 O₂ → 2 CO₂ + 4 H₂O

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28)

Al(OH)₃ + H₂SO₄ → Al₂(SO₄)₃ + H₂O

Al₂(SO₄)₃ → needs 2 Al and 3 SO₄

So:
- 2 Al(OH)₃
- 3 H₂SO₄

H: 2×3 = 6 from Al(OH)₃ + 3×2 = 6 from H₂SO₄ → 12 H → 6 H₂O

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

✔ Answer:
2 Al(OH)₃ + 3 H₂SO₄ → 1 Al₂(SO₄)₃ + 6 H₂O

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29)

Na + Cl₂ → NaCl

2 Na + Cl₂ → 2 NaCl

✔ Answer:
2 Na + 1 Cl₂ → 2 NaCl

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30)

Rb + S₈ → Rb₂S

S₈ → 8 S → Rb₂S has 1 S → need 8 Rb₂S → 16 Rb

So: 16 Rb + S₈ → 8 Rb₂S

✔ Answer:
16 Rb + 1 S₈ → 8 Rb₂S

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31)

H₃PO₄ + Ca(OH)₂ → Ca₃(PO₄)₂ + H₂O

Ca₃(PO₄)₂ → needs 3 Ca and 2 PO₄

So:
- 2 H₃PO₄ → 2 PO₄
- 3 Ca(OH)₂ → 3 Ca

H: 2×3 = 6 from H₃PO₄ + 3×2 = 6 → 12 H → 6 H₂O

So: 2 H₃PO₄ + 3 Ca(OH)₂ → Ca₃(PO₄)₂ + 6 H₂O

✔ Answer:
2 H₃PO₄ + 3 Ca(OH)₂ → 1 Ca₃(PO₄)₂ + 6 H₂O

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32)

NH₃ + HCl → NH₄Cl

Already balanced: 1:1

✔ Answer:
1 NH₃ + 1 HCl → 1 NH₄Cl

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33)

Li + H₂O → LiOH + H₂

Li + H₂O → LiOH + H₂

Balance:
- Li: 1 = 1
- H: 2 on left → right: 1 in LiOH + 2 in H₂ → 3 → too much

Try: 2 Li + 2 H₂O → 2 LiOH + H₂

H: 4 on left → 2 in LiOH + 2 in H₂ → 4 ✔

O: 2 = 2 ✔

✔ Answer:
2 Li + 2 H₂O → 2 LiOH + 1 H₂

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34)

Ca₃(PO₄)₂ + SiO₂ + C → CaSiO₃ + CO + P

This is a complex reduction.

We want to extract P from phosphate.

Assume:
- Ca₃(PO₄)₂ → P
- SiO₂ → CaSiO₃
- C → CO

Let’s balance:

Start with P: 1 P → need 1 P

Ca: 3 → 3 CaSiO₃ → 3 SiO₂

So: Ca₃(PO₄)₂ + 3 SiO₂ + C → 3 CaSiO₃ + CO + P

Now O: left: 8 (from PO₄) + 3×2 = 6 → 14 O
Right: 3×3 = 9 in CaSiO₃ + 1 in CO → 10 O → need more CO?

Wait: CO has 1 O → need 14 - 9 = 5 O → 5 CO

But C: 5 → so 5 C

So: Ca₃(PO₄)₂ + 3 SiO₂ + 5 C → 3 CaSiO₃ + 5 CO + P

Check:
- Ca: 3 = 3 ✔
- P: 1 = 1 ✔
- Si: 3 = 3 ✔
- O: 8 (PO₄) + 6 (SiO₂) = 14; right: 3×3 = 9 + 5 = 14 ✔
- C: 5 = 5 ✔
- H: none

✔ Answer:
1 Ca₃(PO₄)₂ + 3 SiO₂ + 5 C → 3 CaSiO₃ + 5 CO + 1 P

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35)

NH₃ + O₂ → N₂ + H₂O

Ammonia oxidation:

2 NH₃ → N₂ + 3 H₂O

H: 6 → 3 H₂O → 6 H ✔

O: 3 → need 3/2 O₂ → ×2

→ 4 NH₃ + 3 O₂ → 2 N₂ + 6 H₂O

✔ Answer:
4 NH₃ + 3 O₂ → 2 N₂ + 6 H₂O

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36)

FeS₂ + O₂ → Fe₂O₃ + SO₂

Pyrite roasting.

FeS₂ → Fe₂O₃ → need 2 Fe → 2 FeS₂
S: 2×2 = 4 → 4 SO₂

So: 2 FeS₂ + O₂ → Fe₂O₃ + 4 SO₂

O: right: 3 + 8 = 11 → 11/2 O₂ → ×2

→ 4 FeS₂ + 11 O₂ → 2 Fe₂O₃ + 8 SO₂

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

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37)

C + SO₂ → CS₂ + CO

Carbon reduces SO₂ to CS₂.

C + SO₂ → CS₂ + CO

But S: 1 → 2 → need 2 SO₂ → 2 S → CS₂

C: left: 1 + ? → right: 1 (CS₂) + 1 (CO) = 2 → need 2 C

So: 2 C + 2 SO₂ → CS₂ + 2 CO

But S: 2 → 2 ✔
C: 2 = 1 + 1 ✔
O: 4 → 2 in CO → 2 O → missing

Wait: 2 SO₂ → 4 O → right: 2 CO → 2 O → need 2 more → not balanced

Try:
2 C + 2 SO₂ → CS₂ + 2 CO → O: 4 → 2 → no

Alternative:
Use 2 C + 2 SO₂ → CS₂ + 2 CO → O: 4 → 2 → imbalance

Wait: SO₂ has 2 O → 2 SO₂ → 4 O → need 4 O on right

CO has 1 O → need 4 CO → 4 C

So: 4 C + 2 SO₂ → CS₂ + 4 CO

C: 4 → 1 + 4 = 5 → too many

Try:
CS₂ needs 1 S → SO₂ provides 1 S → so 1 SO₂ → CS₂

But CS₂ has 2 C → need 2 C

Then: 2 C + SO₂ → CS₂ + ?

O: 2 → need product with 2 O → CO₂? But CO is given.

Wait: maybe:
2 C + SO₂ → CS₂ + 2 CO → O: 2 → 2 → good

C: 2 → 2 → good

S: 1 → 1 → good

Yes!

✔ Answer:
2 C + 1 SO₂ → 1 CS₂ + 2 CO

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## ✔ Final Balanced Answers:

| # | Balanced Equation |
|---|-------------------|
| 1 | 1 H₃PO₄ + 3 KOH → 1 K₃PO₄ + 3 H₂O |
| 2 | 6 K + 1 B₂O₃ → 3 K₂O + 2 B |
| 3 | 1 HCl + 1 NaOH → 1 NaCl + 1 H₂O |
| 4 | 10 Na + 2 NaNO₃ → 6 Na₂O + 1 N₂ |
| 5 | 4 C + 1 S₈ → 4 CS₂ |
| 6 | 4 Na + 1 O₂ → 2 Na₂O |
| 7 | 2 N₂ + 5 O₂ → 2 N₂O₅ |
| 8 | 2 H₃PO₄ + 3 Mg(OH)₂ → 1 Mg₃(PO₄)₂ + 6 H₂O |
| 9 | 2 NaOH + 1 H₂CO₃ → 1 Na₂CO₃ + 2 H₂O |
|10 | 1 KOH + 1 HBr → 1 KBr + 1 H₂O |
|11 | 4 Na + 1 O₂ → 2 Na₂O |
|12 | 2 Al(OH)₃ + 3 H₂CO₃ → 1 Al₂(CO₃)₃ + 6 H₂O |
|13 | 16 Al + 3 S₈ → 8 Al₂S₃ |
|14 | 6 Cs + 1 N₂ → 2 Cs₃N |
|15 | 1 Mg + 1 Cl₂ → 1 MgCl₂ |
|16 | 10 Rb + 2 RbNO₃ → 6 Rb₂O + 1 N₂ |
|17 | 2 C₆H₆ + 15 O₂ → 12 CO₂ + 6 H₂O |
|18 | 1 N₂ + 3 H₂ → 2 NH₃ |
|19 | 2 C₁₀H₂₂ + 31 O₂ → 20 CO₂ + 22 H₂O |
|20 | 1 Al(OH)₃ + 3 HBr → 1 AlBr₃ + 3 H₂O |
|21 | 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O |
|22 | 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O |
|23 | 3 Li + 1 AlCl₃ → 3 LiCl + 1 Al |
|24 | 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O |
|25 | 3 NH₄OH + 1 H₃PO₄ → 1 (NH₄)₃PO₄ + 3 H₂O |
|26 | 3 Rb + 1 P → 1 Rb₃P |
|27 | 2 CH₄ + 3 O₂ → 2 CO₂ + 4 H₂O |
|28 | 2 Al(OH)₃ + 3 H₂SO₄ → 1 Al₂(SO₄)₃ + 6 H₂O |
|29 | 2 Na + 1 Cl₂ → 2 NaCl |
|30 | 16 Rb + 1 S₈ → 8 Rb₂S |
|31 | 2 H₃PO₄ + 3 Ca(OH)₂ → 1 Ca₃(PO₄)₂ + 6 H₂O |
|32 | 1 NH₃ + 1 HCl → 1 NH₄Cl |
|33 | 2 Li + 2 H₂O → 2 LiOH + 1 H₂ |
|34 | 1 Ca₃(PO₄)₂ + 3 SiO₂ + 5 C → 3 CaSiO₃ + 5 CO + 1 P |
|35 | 4 NH₃ + 3 O₂ → 2 N₂ + 6 H₂O |
|36 | 4 FeS₂ + 11 O₂ → 2 Fe₂O₃ + 8 SO₂ |
|37 | 2 C + 1 SO₂ → 1 CS₂ + 2 CO |

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