Balancing Chemical Equations Worksheet #2 featuring 13 problems to practice balancing chemical reactions.

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 → 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 O on right → put 2 in front of N₂O? But then N becomes 4.

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

Better:
We need even oxygen on right. Try 2 N₂O → needs 4 N and 2 O → so left: 2 N₂ and 1 O₂? Wait:

Actually:
If we do:
2N₂ + 1O₂ → 2N₂O
Check: Left: N=4, O=2 → Right: N=4, O=2 ✔

Wait — that works! But let me double-check standard balancing.

Actually, common way:
N₂ + O₂ → N₂O → not balanced.

Multiply N₂O by 2:
N₂ + O₂ → 2N₂O → now N: 2 vs 4 → no.

So try:
2N₂ + 1O₂ → 2N₂O → yes! N: 4=4, O:2=2 ✔

But wait — actually, this is correct. Some might write it as:

Alternatively, think: To get even oxygen, make 2 N₂O → requires 4 N and 2 O → so 2 N₂ and 1 O₂.

✔ Balanced: 2 N₂ + 1 O₂ → 2 N₂O

*(Note: Sometimes people prefer smallest whole numbers — but 2,1,2 is already simplest.)*

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4) N₂ + H₂ → NH₃

Left: N=2, H=2
Right: N=1, H=3 → need to match.

Put 2 in front of NH₃:
N₂ + H₂ → 2NH₃ → Now N=2, H=6 → so need 3 H₂ on left (since 3×2=6 H)

✔ 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 → I mismatch.

Put 2 in front of KI:
2KI + Cl₂ → KCl + I₂ → Now K=2, I=2 → so need 2 KCl on right.

→ 2KI + Cl₂ → 2KCl + I₂
Check: K=2, I=2, Cl=2 → all good.

✔ 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 has 1 H and 1 Cl.

Better count per atom:

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 → total: Ca=1, Cl=2, H=2, O=1 → not matching.

Need more HCl and more H₂O.

Try:
2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O

Left: H=2 (from 2HCl) + 2 (from Ca(OH)₂) = 4 H; Cl=2; Ca=1; O=2
Right: Ca=1, Cl=2, H=4 (from 2H₂O), O=2 → perfect!

✔ 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 → O mismatch.

Need even oxygen on right. Multiply KClO₃ by 2 → O=6 → then O₂ should be 3 (since 3×2=6)

Try:
2KClO₃ → 2KCl + 3O₂

Check: Left: K=2, Cl=2, O=6
Right: K=2, Cl=2, O=6 ✔

✔ 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 → K and H mismatch.

Need 3 KCl on right → so 3 HCl on left.

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 ✔

✔ 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 → O mismatch.

Need even oxygen. Multiply SO₃ by 2 → O=6 → so O₂ should be 3 (3×2=6)

Then S must be 2 on left.

Try:
2S + 3O₂ → 2SO₃

Check: Left: S=2, O=6 → Right: S=2, O=6 ✔

✔ 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 → I and N/O mismatch.

PbI₂ needs 2 I → so 2 KI on left.

Then K=2 → so 2 KNO₃ on right.

Then NO₃: 2 on right → so Pb(NO₃)₂ already gives 2 NO₃ → good.

Try:
2KI + 1Pb(NO₃)₂ → 2KNO₃ + 1PbI₂

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

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

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11) CaSO₄ + AlBr₃ → CaBr₂ + Al₂(SO₄)₃

This one looks tricky. Let’s look at sulfate and aluminum.

Right: Al₂(SO₄)₃ → Al=2, SO₄=3
Left: CaSO₄ has 1 SO₄, AlBr₃ has 1 Al.

So to get 3 SO₄ on right, need 3 CaSO₄ on left.

To get 2 Al on right, need 2 AlBr₃ on left.

Now check bromine: 2 AlBr₃ → 6 Br → so need 3 CaBr₂ on right (since each has 2 Br → 3×2=6)

Calcium: 3 CaSO₄ → 3 Ca → 3 CaBr₂ → matches.

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

Check atoms:

Left:
Ca=3, S=3, O=12 (3×4), Al=2, Br=6 (2×3)
Right:
Ca=3, Br=6, Al=2, S=3, O=12 ✔

✔ 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 → too many O.

Need to balance oxygen.

Try:
2H₂O₂ → 2H₂O + 1O₂

Left: H=4, O=4
Right: H=4, O=2 (from 2H₂O) + 2 (from O₂) = 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 mismatch.

Need even hydrogen. Try 2 NaOH → then H from NaOH=2, plus H₂=2 → total H=4 on right.

So left: need 2 H₂O → H=4, O=2 → then right: 2 NaOH uses 2 O and 2 H, leaving 2 H for H₂ → which is 1 H₂ molecule.

Also Na: 2 on right → so 2 Na on left.

Try:
2Na + 2H₂O → 2NaOH + 1H₂

Check:
Left: Na=2, H=4, O=2
Right: Na=2, O=2, H=2 (from 2NaOH) + 2 (from H₂) = 4 ✔

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

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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