Worksheet for practicing balancing chemical equations at three difficulty levels.

Let’s solve this step by step.

We are given three chemical equations to balance. Balancing means making sure the number of atoms of each element is the same on both sides (left and right) of the arrow.

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Level: Beginner

Equation 1:
H₂ + O₂ → H₂O

Left side:
- H: 2 atoms (from H₂)
- O: 2 atoms (from O₂)

Right side:
- H: 2 atoms (from H₂O)
- O: 1 atom (from H₂O)

Oxygen is not balanced. We have 2 on left, 1 on right. Let’s put a 2 in front of H₂O:

H₂ + O₂ → 2H₂O

Now right side has:
- H: 4 atoms (2 × 2)
- O: 2 atoms (2 × 1)

Left side still has:
- H: 2
- O: 2

Hydrogen is now unbalanced. Put a 2 in front of H₂:

2H₂ + O₂ → 2H₂O

Check again:
Left: H = 4, O = 2
Right: H = 4, O = 2 ✔ Balanced!

Number of Atoms:
H: _4_ : _4_
O: _2_ : _2_

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Equation 2:
N₂ + H₂ → NH₃

Left:
- N: 2
- H: 2

Right:
- N: 1
- H: 3

Not balanced. Let’s start with nitrogen. Put a 2 in front of NH₃:

N₂ + H₂ → 2NH₃

Now right:
- N: 2
- H: 6

Left:
- N: 2
- H: 2 → need 6 hydrogens → so put 3 in front of H₂:

N₂ + 3H₂ → 2NH₃

Check:
Left: N=2, H=6
Right: N=2, H=6 ✔ Balanced!

Number of Atoms:
N: _2_ : _2_
H: _6_ : _6_

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Level: Intermediate

Equation 3:
Fe + O₂ → Fe₂O₃

Left:
- Fe: 1
- O: 2

Right:
- Fe: 2
- O: 3

Need to balance both. Start with oxygen. Least common multiple of 2 and 3 is 6.

So make O₂ into 3O₂ (gives 6 O), and Fe₂O₃ into 2Fe₂O₃ (also gives 6 O):

Fe + 3O₂ → 2Fe₂O₃

Now right: Fe = 4 (2×2), O = 6
Left: Fe = 1, O = 6

Put 4 in front of Fe:

4Fe + 3O₂ → 2Fe₂O₃

Check:
Left: Fe=4, O=6
Right: Fe=4, O=6 ✔ Balanced!

Number of Atoms:
Fe: _4_ : _4_
O: _6_ : _6_

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Equation 4:
Al + HCl → AlCl₃ + H₂

Left:
- Al: 1
- H: 1
- Cl: 1

Right:
- Al: 1
- Cl: 3
- H: 2

Chlorine and hydrogen are off. Start with Cl. Right has 3 Cl, left has 1 → put 3 in front of HCl:

Al + 3HCl → AlCl₃ + H₂

Now left: H=3, Cl=3
Right: H=2, Cl=3 → hydrogen mismatch.

We have 3 H on left, 2 on right. LCM of 3 and 2 is 6.

Multiply HCl by 6? Wait — let’s think differently.

Try putting 2 in front of AlCl₃? No, that would mess up Al.

Better approach:

We want H to be even on both sides. Currently left has 3H (odd), right has 2H (even). So multiply HCl by 2? But then Cl becomes 2, but right has 3 Cl in AlCl₃.

Wait — let’s try:

Set coefficient for HCl to 6 → gives 6H and 6Cl.

Then right: AlCl₃ needs 6 Cl → so 2AlCl₃ (since 2×3=6 Cl)

And H₂: 6H → so 3H₂ (3×2=6H)

Now Al: right has 2Al (from 2AlCl₃), so left needs 2Al.

So:

2Al + 6HCl → 2AlCl₃ + 3H₂

Check:
Left: Al=2, H=6, Cl=6
Right: Al=2, Cl=6, H=6 ✔ Balanced!

Number of Atoms:
Al: _2_ : _2_
H: _6_ : _6_
Cl: _6_ : _6_

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Level: Advanced

Equation 5:
C₃H₈ + O₂ → CO₂ + H₂O

This is combustion of propane.

Left:
- C: 3
- H: 8
- O: 2

Right:
- C: 1 (in CO₂)
- H: 2 (in H₂O)
- O: 3 (2 from CO₂ + 1 from H₂O? Wait no — depends on coefficients)

Start with carbon. Put 3 in front of CO₂:

C₃H₈ + O₂ → 3CO₂ + H₂O

Now C: 3=3 ✔

Hydrogen: left has 8H → right has 2H per H₂O → so need 4H₂O (4×2=8H)

C₃H₈ + O₂ → 3CO₂ + 4H₂O

Now count oxygen on right:
3CO₂ → 3×2 = 6 O
4H₂O → 4×1 = 4 O
Total O = 10

Left: O₂ → each molecule has 2 O → so we need 5 O₂ (5×2=10)

Final:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Check:
Left: C=3, H=8, O=10
Right: C=3, H=8, O=6+4=10 ✔ Balanced!

Number of Atoms:
C: _3_ : _3_
H: _8_ : _8_
O: _10_ : _10_

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

This is acid-base reaction.

Left:
- Ca: 1
- O: 2 (from OH) + 4 (from PO₄)? Wait — better to break down.

Actually, Ca(OH)₂ has: Ca, 2O, 2H
H₃PO₄ has: 3H, P, 4O

But when combined, it's messy. Better to treat polyatomic ions as groups if possible.

Note: PO₄ appears on both sides. In product: Ca₃(PO₄)₂ → so 2 PO₄ groups.

So we need 2 H₃PO₄ on left to get 2 PO₄.

Also, Ca₃(PO)₂ has 3 Ca → so need 3 Ca(OH)₂ on left.

Try:

3Ca(OH)₂ + 2H₃PO₄ → Ca(PO₄)₂ + H₂O

Now check atoms:

Left:
Ca: 3
O from OH: 3×2 = 6 O
H from OH: 3×2 = 6 H
Plus from 2H₃PO₄: H=6, P=2, O=8
Total left:
Ca=3, P=2, H=6+6=12, O=6+8=14

Right:
Ca₃(PO₄)₂: Ca=3, P=2, O=8 (since PO₄ has 4 O, two of them = 8 O)
Plus H₂O: say x H₂O → H=2x, O=x

Total right:
Ca=3, P=2, H=2x, O=8+x

Set equal:

H: 12 = 2x → x=6
O: 14 = 8 + x → 14 = 8+6 → 14=14 ✔

So add 6 H₂O:

3Ca(OH)₂ + 2H₃PO₄ → Ca₃(PO₄)₂ + 6H₂O

Check all:

Left:
Ca: 3
O: from 3Ca(OH)₂ → 3×2=6 O; from 2H₃PO₄ → 2×4=8 O → total O=14
H: from 3Ca(OH)₂ → 3×2=6 H; from 2H₃PO₄ → 2×3=6 H → total H=12
P: 2

Right:
Ca₃(PO₄)₂: Ca=3, P=2, O=8
6H₂O: H=12, O=6 → total O=8+6=14, H=12 ✔ Perfect!

Number of Atoms:
Ca: _3_ : _3_
P: _2_ : _2_
O: _14_ : _14_
H: _12_ : _12_

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Final Answer:

Beginner Level:

1. 2H₂ + O₂ → 2H₂O
H: 4 : 4
O: 2 : 2

2. N₂ + 3H₂ → 2NH₃
N: 2 : 2
H: 6 : 6

Intermediate Level:

3. 4Fe + 3O₂ → 2Fe₂O₃
Fe: 4 : 4
O: 6 : 6

4. 2Al + 6HCl → 2AlCl₃ + 3H₂
Al: 2 : 2
H: 6 : 6
Cl: 6 : 6

Advanced Level:

5. C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
C: 3 : 3
H: 8 : 8
O: 10 : 10

6. 3Ca(OH)₂ + 2H₃PO₄ → Ca₃(PO₄)₂ + 6H₂O
Ca: 3 : 3
P: 2 : 2
O: 14 : 14
H: 12 : 12