Balancing Chemical Equations Worksheet #2 featuring 13 equations to balance.

Let’s go through each equation one by one and balance them. Balancing means making sure the number of atoms of each element is the same on both sides of the arrow.

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1) CH₄ + O₂ → CO₂ + H₂O

Left: C=1, H=4, O=2
Right: C=1, H=2, O=3 (2 from CO₂ + 1 from H₂O)

→ Put 2 in front of H₂O to get 4 H on right:
CH₄ + O₂ → CO₂ + 2H₂O
Now Right: H=4, O=4 (2 from CO₂ + 2 from 2H₂O)
Left: O=2 → need 4 O → put 2 in front of O₂

✔ Balanced: 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O

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2) Na + I₂ → NaI

Left: Na=1, I=2
Right: Na=1, I=1 → need 2 NaI

→ Put 2 in front of NaI:
Na + I₂ → 2NaI
Now Right: Na=2, I=2 → need 2 Na on left

✔ Balanced: 2 Na + 1 I₂ → 2 NaI

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3) N₂ + O₂ → N₂O

Left: N=2, O=2
Right: N=2, O=1 → need 2 N₂O to get O=2? But then N=4 → not good.

Try:
N₂ + O₂ → 2N₂O → Right: N=4, O=2 → too many N

Better:
Make O match first. Right has 1 O per N₂O → so for 2 O on left, need 2 N₂O → but that gives 4 N → so left needs 2 N₂

Wait — let’s try coefficients:

Set N₂ coefficient = 2 → Left: N=4
Then right: need 2 N₂O → N=4, O=2 → matches O₂ on left!

✔ Balanced: 2 N₂ + 1 O₂ → 2 N₂O

Wait — check:
Left: N=4, O=2
Right: 2×N₂O = N=4, O=2 → YES!

But actually, standard way:
We can also do:
N₂ + ½O₂ → N₂O → but we want whole numbers.

So multiply all by 2:
2 N₂ + 1 O₂ → 2 N₂O ← correct.

Actually, wait — let me double-check with smallest integers.

If I write:
N₂ + O₂ → 2 N₂O → no, that’s 4N vs 2N.

Correct balancing:

To get even oxygen:
Let’s set product as 2 N₂O → requires 4 N and 2 O → so reactants: 2 N₂ and 1 O₂.

Yes.

✔ Final: 2 N₂ + 1 O₂ → 2 N₂O

But actually — common mistake. Let me think again.

Standard balanced form for N₂ + O₂ → N₂O is:

It should be:
2 N₂ + O₂ → 2 N₂O

Yes, that’s correct.

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4) N₂ + H₂ → NH₃

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

Need to make N and H match.

Put 2 in front of NH₃ → Right: N=2, H=6
Then left: H₂ must be 3 → because 3×2=6

✔ Balanced: 1 N₂ + 3 H₂ → 2 NH₃

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5) KI + Cl₂ → KCl + I₂

Left: K=1, I=1, Cl=2
Right: K=1, Cl=1, I=2

Iodine: right has 2, left has 1 → put 2 in front of KI
→ 2KI + Cl₂ → KCl + I₂
Now Left: K=2, I=2, Cl=2
Right: K=1, Cl=1, I=2 → need 2 KCl

→ 2KI + Cl₂ → 2KCl + I₂
Now Right: K=2, Cl=2, I=2 → matches left!

✔ Balanced: 2 KI + 1 Cl₂ → 2 KCl + 1 I₂

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6) HCl + Ca(OH)₂ → CaCl₂ + H₂O

Left: H=1+2=3? Wait — Ca(OH)₂ has 2 OH → so 2 H and 2 O from it, plus HCl.

Better to count atoms:

Ca(OH)₂: Ca=1, O=2, H=2
HCl: H=1, Cl=1

Total left: Ca=1, O=2, H=3, Cl=1

Right: CaCl₂ → Ca=1, Cl=2; H₂O → H=2, O=1

Not matching.

We need 2 Cl on right → so 2 HCl on left.

Try:
2HCl + Ca(OH)₂ → CaCl₂ + H₂O

Left: H=2 (from HCl) + 2 (from Ca(OH)₂) = 4 H; Cl=2; Ca=1; O=2
Right: CaCl₂ → Ca=1, Cl=2; H₂O → H=2, O=1 → only 2 H and 1 O → need more water.

Put 2 H₂O on right:
→ 2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O

Right: H=4, O=2 → matches left!

✔ Balanced: 2 HCl + 1 Ca(OH)₂ → 1 CaCl₂ + 2 H₂O

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7) KClO₃ → KCl + O₂

Left: K=1, Cl=1, O=3
Right: K=1, Cl=1, O=2

Oxygen doesn’t match. Need multiple of 3 and 2 → LCM is 6.

So put 2 KClO₃ → Left: O=6
Then right: need 3 O₂ → O=6

Also, 2 KClO₃ → 2 KCl

✔ Balanced: 2 KClO₃ → 2 KCl + 3 O₂

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8) K₃PO₄ + HCl → KCl + H₃PO₄

Left: K=3, P=1, O=4, H=1, Cl=1
Right: K=1, Cl=1, H=3, P=1, O=4

Potassium: left 3, right 1 → put 3 in front of KCl
Phosphoric acid: H₃PO₄ has 3 H → so need 3 HCl to provide 3 H

Try:
K₃PO₄ + 3HCl → 3KCl + H₃PO₄

Check:
Left: K=3, P=1, O=4, H=3, Cl=3
Right: K=3, Cl=3, H=3, P=1, O=4 → perfect!

✔ Balanced: 1 K₃PO₄ + 3 HCl → 3 KCl + 1 H₃PO₄

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9) S + O₂ → SO₃

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

Need to balance oxygen. LCM of 2 and 3 is 6.

So:
Put 2 SO₃ → Right: S=2, O=6
Then left: need 2 S and 3 O₂ (since 3×2=6 O)

✔ Balanced: 2 S + 3 O₂ → 2 SO₃

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10) KI + Pb(NO₃)₂ → KNO₃ + PbI₂

Left: K=1, I=1, Pb=1, N=2, O=6
Right: K=1, N=1, O=3, Pb=1, I=2

Iodine: right has 2, left has 1 → put 2 KI
Lead iodide: PbI₂ already has 2 I → good.

Now: 2KI + Pb(NO₃)₂ → KNO₃ + PbI₂

Left: K=2, I=2, Pb=1, N=2, O=6
Right: K=1, N=1, O=3 → need 2 KNO₃

→ 2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂

Check:
Right: K=2, N=2, O=6, Pb=1, I=2 → matches left!

✔ Balanced: 2 KI + 1 Pb(NO₃)₂ → 2 KNO₃ + 1 PbI₂

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11) CaSO₄ + AlBr₃ → CaBr₂ + Al₂(SO₄)₃

This one is trickier.

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

Sulfate group: SO₄ appears as unit. On right, Al₂(SO₄)₃ has 3 SO₄ → so need 3 CaSO₄ on left.

Try:
3CaSO₄ + AlBr₃ → CaBr₂ + Al₂(SO₄)₃

Left: Ca=3, S=3, O=12, Al=1, Br=3
Right: Ca=1, Br=2, Al=2, S=3, O=12

Calcium: left 3, right 1 → put 3 CaBr₂
Aluminum: left 1, right 2 → put 2 AlBr₃
Bromine: if 2 AlBr₃ → Br=6; 3 CaBr₂ → Br=6 → good.

So:
3 CaSO₄ + 2 AlBr₃ → 3 CaBr₂ + 1 Al₂(SO₄)₃

Check atoms:

Left:
Ca=3, S=3, O=12, Al=2, Br=6
Right:
Ca=3 (from 3 CaBr₂), Br=6, Al=2, S=3, O=12 → perfect!

✔ Balanced: 3 CaSO₄ + 2 AlBr₃ → 3 CaBr₂ + 1 Al₂(SO₄)₃

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12) H₂O₂ → H₂O + O₂

Left: H=2, O=2
Right: H=2, O=1 (from H₂O) + 2 (from O₂) = 3 → not balanced.

Try 2 H₂O₂ → Left: H=4, O=4
Right: need 2 H₂O → H=4, O=2; and 1 O₂ → O=2 → total O=4

✔ Balanced: 2 H₂O₂ → 2 H₂O + 1 O₂

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13) Na + H₂O → NaOH + H₂

Left: Na=1, H=2, O=1
Right: Na=1, O=1, H=1 (from NaOH) + 2 (from H₂) = 3 H → not balanced.

Hydrogen: left 2, right 3 → need even numbers.

Try 2 Na + 2 H₂O → ?
Left: Na=2, H=4, O=2
Right: if 2 NaOH → Na=2, O=2, H=2; and H₂ → H=2 → total H=4 → perfect!

✔ Balanced: 2 Na + 2 H₂O → 2 NaOH + 1 H₂

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Now, compiling all answers with coefficients:

Final Answer:
1) 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O
2) 2 Na + 1 I₂ → 2 NaI
3) 2 N₂ + 1 O₂ → 2 N₂O
4) 1 N₂ + 3 H₂ → 2 NH₃
5) 2 KI + 1 Cl₂ → 2 KCl + 1 I₂
6) 2 HCl + 1 Ca(OH)₂ → 1 CaCl₂ + 2 H₂O
7) 2 KClO₃ → 2 KCl + 3 O₂
8) 1 K₃PO₄ + 3 HCl → 3 KCl + 1 H₃PO₄
9) 2 S + 3 O₂ → 2 SO₃
10) 2 KI + 1 Pb(NO₃)₂ → 2 KNO₃ + 1 PbI₂
11) 3 CaSO₄ + 2 AlBr₃ → 3 CaBr₂ + 1 Al₂(SO₄)₃
12) 2 H₂O₂ → 2 H₂O + 1 O₂
13) 2 Na + 2 H₂O → 2 NaOH + 1 H₂