Balancing Equations Race worksheet with 20 chemical equations to balance.

Let's solve each of these balanced chemical equations step by step. The goal is to balance the number of atoms of each element on both sides of the equation using coefficients (whole numbers placed in front of the formulas).

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1) C₄H₈ + O₂ → CO₂ + H₂O

Unbalanced:
C₄H₈ + O₂ → CO₂ + H₂O

- Carbon: 4 on left, 1 on right → need 4 CO₂
- Hydrogen: 8 on left, 2 on right → need 4 H₂O
- Oxygen: Right has 4×2 + 4×1 = 12; Left has 2 per O₂ → need 6 O₂

✔ Balanced:
1 C₄H₈ + 6 O₂ → 4 CO₂ + 4 H₂O

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2) Al + Fe₃N₂ → AlN + Fe

Unbalanced:
Al + Fe₃N₂ → AlN + Fe

- Fe: 3 on left, 1 on right → need 3 Fe
- N: 2 on left, 1 on right → need 2 AlN
- Al: 2 on right, 1 on left → need 2 Al

✔ Balanced:
2 Al + Fe₃N₂ → 2 AlN + 3 Fe

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3) Na + Cl₂ → NaCl

Unbalanced:
Na + Cl₂ → NaCl

- Cl: 2 on left, 1 on right → need 2 NaCl
- Na: 2 on right, 1 on left → need 2 Na

✔ Balanced:
2 Na + Cl₂ → 2 NaCl

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4) H₂O₂ → H₂O + O₂

Unbalanced:
H₂O₂ → H₂O + O₂

- H: 2 on left, 2 on right → OK
- O: 2 on left, 1+2=3 on right → not balanced

Try 2 H₂O₂ → 2 H₂O + O₂

- H: 4 → 4 → OK
- O: 4 → 2 + 2 = 4 → OK

✔ Balanced:
2 H₂O₂ → 2 H₂O + O₂

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5) C₆H₁₂O₆ + O₂ → H₂O + CO₂

This is glucose combustion.

Unbalanced:
C₆H₁₂O₆ + O₂ → H₂O + CO₂

- C: 6 → need 6 CO₂
- H: 12 → need 6 H₂O
- O: Left: 6 (from glucose) + 2×? (from O₂); Right: 6×2 + 6×1 = 12 + 6 = 18

So oxygen needed: 18 - 6 = 12 → 6 O₂ molecules

✔ Balanced:
1 C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O

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6) H₂O + CO₂ → C₇H₁₆ + O₂

Wait — this seems chemically incorrect. Water and CO₂ don't form heptane (C₇H₁₆) and O₂. But let’s assume it's a hypothetical or typo.

But looking at it:
Left: H₂O + CO₂ → elements: H, O, C
Right: C₇H₁₆ + O₂ → C, H, O

We can try balancing:

- C: 7 on right → need 7 CO₂
- H: 16 on right → need 8 H₂O
- O: Left: 8 (from H₂O) + 14 (from CO₂) = 22; Right: 16 (in C₇H₁₆) + 2 (in O₂)? No — O₂ is separate.

Wait: C₇H₁₆ has no O, so all O must go to O₂.

Left O: from 8 H₂O → 8 O, 7 CO₂ → 14 O → total 22 O
Right: O₂ → must have 11 O₂ → 22 O

But that gives:
8 H₂O + 7 CO₂ → C₇H₁₆ + 11 O₂

Check atoms:
- H: 16 → 16 → OK
- C: 7 → 7 → OK
- O: 8 + 14 = 22 → 22 → OK

✔ Balanced:
8 H₂O + 7 CO₂ → C₇H₁₆ + 11 O₂

But this is not realistic (it's reverse of combustion), but mathematically correct.

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7) NaClO₃ → NaCl + O₂

Decomposition of sodium chlorate.

Unbalanced:
NaClO₃ → NaCl + O₂

- Na: 1 → 1 → OK
- Cl: 1 → 1 → OK
- O: 3 → 2 → not balanced

Try 2 NaClO₃ → 2 NaCl + 3 O₂

- O: 6 → 6 → OK
- Na: 2 → 2 → OK
- Cl: 2 → 2 → OK

✔ Balanced:
2 NaClO₃ → 2 NaCl + 3 O₂

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8) (NH₄)₃PO₄ + Pb(NO₃)₂ → Pb₃(PO₄)₂ + NH₄NO₃

Double displacement.

Unbalanced:
(NH₄)₃PO₄ + Pb(NO₃)₂ → Pb₃(PO₄)₂ + NH₄NO₃

- Pb: 3 on right → need 3 Pb(NO₃)₂
- PO₄: 2 on right → need 2 (NH₄)₃PO₄
- NH₄: 2×3 = 6 → need 6 NH₄NO₃
- NO₃: 3×2 = 6 → matches 6 NH₄NO₃

✔ Balanced:
2 (NH₄)₃PO₄ + 3 Pb(NO₃)₂ → Pb₃(PO₄)₂ + 6 NH₄NO₃

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9) BF₃ + Li₂SO₄ → B₂(SO₄)₃ + LiF

Unbalanced:
BF₃ + Li₂SO₄ → B₂(SO₄)₃ + LiF

- B: 2 on right → need 2 BF₃
- SO₄: 3 on right → need 3 Li₂SO₄
- Li: 3×2 = 6 → need 6 LiF
- F: 2×3 = 6 → 6 LiF → OK

✔ Balanced:
2 BF₃ + 3 Li₂SO₄ → B₂(SO₄)₃ + 6 LiF

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10) C₇H₁₇ + O₂ → CO₂ + H₂O

Combustion of heptane.

Unbalanced:
C₇H₁₇ + O₂ → CO₂ + H₂O

- C: 7 → 7 CO₂
- H: 17 → need 17/2 H₂O → better use even number

Multiply by 2:
2 C₇H₁₇ → 14 CO₂ + 17 H₂O

Now balance O:

Right: 14×2 + 17×1 = 28 + 17 = 45 O atoms → need 45/2 = 22.5 O₂ → multiply all by 2

Final:
4 C₇H₁₇ + 45 O₂ → 28 CO₂ + 34 H₂O

Check:
- C: 4×7 = 28 → 28 → OK
- H: 4×17 = 68 → 34×2 = 68 → OK
- O: 45×2 = 90 → 28×2 = 56 + 34×1 = 34 → 56+34=90 → OK

✔ Balanced:
4 C₇H₁₇ + 45 O₂ → 28 CO₂ + 34 H₂O

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11) CaCO₃ + H₃PO₄ → Ca₃(PO₄)₂ + H₂CO₃

Unbalanced:
CaCO₃ + H₃PO₄ → Ca₃(PO₄)₂ + H₂CO₃

- Ca: 3 on right → need 3 CaCO₃
- PO₄: 2 on right → need 2 H₃PO₄
- H: 2×3 = 6 → need 3 H₂CO₃
- C: 3 on left → 3 on right → OK

But H₂CO₃: 3 H₂CO₃ → 6 H, 3 C, 3 O

Left: H: 2×3 = 6 → OK

So:
3 CaCO₃ + 2 H₃PO₄ → Ca₃(PO₄)₂ + 3 H₂CO₃

✔ Balanced:
3 CaCO₃ + 2 H₃PO₄ → Ca₃(PO₄)₂ + 3 H₂CO₃

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12) Ag₂S → Ag + S₈

Unbalanced:
Ag₂S → Ag + S₈

- S: 1 on left, 8 on right → need 8 Ag₂S → 8 S
- Ag: 8×2 = 16 → need 16 Ag

✔ Balanced:
8 Ag₂S → 16 Ag + S₈

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13) KBr + Fe(OH)₃ → KOH + FeBr₃

Unbalanced:
KBr + Fe(OH)₃ → KOH + FeBr₃

- Fe: 1 → 1 → OK
- Br: 3 on right → need 3 KBr
- K: 3 → need 3 KOH
- OH: 3 on left → 3 on right → OK

✔ Balanced:
3 KBr + Fe(OH)₃ → 3 KOH + FeBr₃

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14) KNO₃ + H₂CO₃ → K₂CO₃ + HNO₃

Unbalanced:
KNO₃ + H₂CO₃ → K₂CO₃ + HNO₃

- K: 2 on right → need 2 KNO₃
- NO₃: 2 → need 2 HNO₃
- H: 2 → 2 → OK
- CO₃: 1 → 1 → OK

✔ Balanced:
2 KNO₃ + H₂CO₃ → K₂CO₃ + 2 HNO₃

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15) Pb(OH)₄ + Cu₂O → PbO₂ + CuOH

Unbalanced:
Pb(OH)₄ + Cu₂O → PbO₂ + CuOH

- Pb: 1 → 1 → OK
- O: left: 4 + 1 = 5; right: 2 + 1 = 3 → not balanced
- Cu: 2 on left → need 2 CuOH
- OH: 4 on left → 2 CuOH → 2 OH → need more?

Wait: Pb(OH)₄ → PbO₂ + 2 H₂O (common decomposition)

But here: Pb(OH)₄ → PbO₂ + 2 H₂O

Then H₂O + Cu₂O → 2 CuOH

But we can try direct balancing.

Try:
Pb(OH)₄ + Cu₂O → PbO₂ + 2 CuOH

Atoms:
- Pb: 1 → 1 → OK
- Cu: 2 → 2 → OK
- O: left: 4 (OH) + 1 (Cu₂O) = 5; right: 2 (PbO₂) + 2 (CuOH) = 4 → not enough
- H: 4 → 2 → not balanced

Try:
Pb(OH)₄ + Cu₂O → PbO₂ + 2 CuOH + H₂O

Now:
- H: 4 → 2 (in CuOH) + 2 (in H₂O) = 4 → OK
- O: 4 + 1 = 5 → 2 (PbO₂) + 2 (CuOH) + 1 (H₂O) = 5 → OK

✔ Balanced:
1 Pb(OH)₄ + 1 Cu₂O → 1 PbO₂ + 2 CuOH + 1 H₂O

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16) Cr(NO₃)₂ + (NH₄)₂SO₄ → CrSO₄ + NH₄NO₃

Unbalanced:
Cr(NO₃)₂ + (NH₄)₂SO₄ → CrSO₄ + NH₄NO₃

- Cr: 1 → 1 → OK
- SO₄: 1 → 1 → OK
- NO₃: 2 on left → need 2 NH₄NO₃
- NH₄: 2 → 2 → OK

✔ Balanced:
1 Cr(NO₃)₂ + 1 (NH₄)₂SO₄ → 1 CrSO₄ + 2 NH₄NO₃

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17) KOH + Co₂(PO₄)₃ → K₃PO₄ + Co(OH)₂

Wait: Co₂(PO₄)₃ is correct for cobalt(III) phosphate.

But Co(OH)₂ suggests Co²⁺ — inconsistency.

Assuming it's Co₂(PO₄)₃ → 2 Co³⁺ and 3 PO₄³⁻

And KOH → K⁺ and OH⁻

Products: K₃PO₄ and Co(OH)₃? But it says Co(OH)₂ → problem.

Possibly typo. Let's assume it should be Co(OH)₃, since Co is +3.

But as written: Co(OH)₂ implies Co²⁺.

Alternatively, maybe it's Co₃(PO₄)₂? But given as Co₂(PO₄)₃.

Let’s proceed with what’s written.

Assume:
Co₂(PO₄)₃ → 2 Co³⁺ + 3 PO₄³⁻

To make Co(OH)₂, Co must be +2 — contradiction.

So likely, the formula is wrong. Perhaps it's Co₃(PO₄)₂ for Co²⁺.

But let's suppose it's Co₂(PO₄)₃ and product is Co(OH)₃.

But problem says Co(OH)₂ — so probably mistake.

Alternatively, maybe it's CoPO₄ and Co(OH)₂ — but not matching.

Let’s try balancing as is.

Assume:
KOH + Co₂(PO₄)₃ → K₃PO₄ + Co(OH)₂

But oxidation states don’t match.

Perhaps it's a double displacement with Co³⁺ forming Co(OH)₃.

Let’s change to Co(OH)₃.

So:
KOH + Co₂(PO₄)₃ → K₃PO₄ + Co(OH)₃

Now:
- Co: 2 → need 2 Co(OH)₃
- PO₄: 3 → need 3 K₃PO₄ → 9 K
- K: 9 → need 9 KOH
- OH: 9 → 2×3 = 6 from Co(OH)₃ → 3 extra OH → need 3 H₂O?

Better:
3 K₃PO₄ → 9 K, 3 PO₄

So:
9 KOH + Co₂(PO₄)₃ → 3 K₃PO₄ + 2 Co(OH)₃

Now check:
- K: 9 → 9 → OK
- PO₄: 3 → 3 → OK
- Co: 2 → 2 → OK
- O and H: left: 9 H, 9 O (from KOH) + 12 O (from PO₄) = 21 O? Wait, better count:

Actually, better to just verify atoms.

Left:
- K: 9
- O: 9 (from KOH) + 12 (from PO₄ in Co₂(PO₄)₃) = 21
- H: 9
- Co: 2
- P: 3

Right:
- K₃PO₄: 3 K, 3 P, 12 O
- 2 Co(OH)₃: 2 Co, 6 O, 6 H

Total right: K:3, P:3, O:12+6=18, H:6 → mismatch

Need more H and O.

Wait: KOH has 1 O and 1 H per molecule.

We have 9 KOH → 9 H, 9 O

Co₂(PO₄)₃ → Co₂, P₃, O₁₂

Total left: O: 9 + 12 = 21, H: 9

Right:
- 3 K₃PO₄ → 3 K, 3 P, 12 O
- 2 Co(OH)₃ → 2 Co, 6 O, 6 H

Total: K:3, P:3, O:18, H:6 → missing 3 H and 3 O

Add 3 H₂O on right?

But then not balanced.

Alternative: perhaps coefficient adjustment.

Try:
6 KOH + Co₂(PO₄)₃ → 2 K₃PO₄ + 2 Co(OH)₃

Left: K:6, O:6+12=18, H:6, Co:2, P:3

Right: 2 K₃PO₄ → 6 K, 2 P, 8 O → wait, only 2 P → not enough

We need 3 P → 3 K₃PO₄ → 9 K

So need 9 KOH

Then:
9 KOH + Co₂(PO₄)₃ → 3 K₃PO₄ + 2 Co(OH)₃

Now:
- K: 9 → 9 → OK
- P: 3 → 3 → OK
- Co: 2 → 2 → OK
- O: left: 9 (KOH) + 12 (PO₄) = 21
- Right: 3 K₃PO₄ → 3×4 = 12 O; 2 Co(OH)₃ → 2×3 = 6 O → total 18 O → missing 3 O
- H: 9 → 2×3 = 6 → missing 3 H

So add 3 H₂O on right? But that would add 3 H and 3 O → total H: 6+3=9, O:18+3=21 → OK

But now reaction:
9 KOH + Co₂(PO₄)₃ → 3 K₃PO₄ + 2 Co(OH)₃ + 3 H₂O

But that’s not typical.

Alternatively, if the product is Co(OH)₃, and we accept water formed.

But the original says Co(OH)₂ — which is inconsistent.

So likely, the reactant is Co₃(PO₄)₂ for Co²⁺.

Let’s assume it's Co₃(PO₄)₂ and Co(OH)₂

Then:
KOH + Co₃(PO₄)₂ → K₃PO₄ + Co(OH)₂

- Co: 3 → need 3 Co(OH)₂
- PO₄: 2 → need 2 K₃PO₄ → 6 K
- K: 6 → need 6 KOH
- OH: 6 → 3×2 = 6 → OK

✔ Balanced:
6 KOH + Co₃(PO₄)₂ → 2 K₃PO₄ + 3 Co(OH)₂

But since problem says Co₂(PO₄)₃, we'll assume it's a typo and go with:

6 KOH + Co₃(PO₄)₂ → 2 K₃PO₄ + 3 Co(OH)₂

But since it says Co₂(PO₄)₃, perhaps it's meant to be:

Let’s try:
KOH + Co₂(PO₄)₃ → K₃PO₄ + Co(OH)₃

As before, requires 9 KOH → 3 K₃PO₄ + 2 Co(OH)₃ + 3 H₂O

But not clean.

Given ambiguity, and since Co(OH)₂ implies Co²⁺, likely the reactant should be Co₃(PO₄)₂.

So I’ll assume typo and use:

✔ 6 KOH + Co₃(PO₄)₂ → 2 K₃PO₄ + 3 Co(OH)₂

But since the problem says Co₂(PO₄)₃, perhaps it's:

Try:
3 KOH + Co₂(PO₄)₃ → K₃PO₄ + 2 Co(OH)₃

No, not balanced.

Best guess: problem has error — but for now, we'll skip and come back.

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18) Sn(NO₃)₄ + Pt₄N₄ → Sn₃N₄ + Pt(NO₃)₄

Unbalanced:
Sn(NO₃)₄ + Pt₄N₄ → Sn₃N₄ + Pt(NO₃)₄

- Sn: 3 on right → need 3 Sn(NO₃)₄
- Pt: 4 on left → need 4 Pt(NO₃)₄
- N: left: 3×4 = 12 from Sn(NO₃)₄, plus 4 from Pt₄N₄ → 16 N
- Right: Sn₃N₄ → 4 N, 4 Pt(NO₃)₄ → 4×4 = 16 N → total 20 → too much

Wait: Pt₄N₄ → 4 Pt, 4 N

Sn₃N₄ → 3 Sn, 4 N

Pt(NO₃)₄ → Pt, 4 NO₃ → 4 N

So right: Sn₃N₄ → 4 N, 4 Pt(NO₃)₄ → 16 N → total 20 N

Left: 3 Sn(NO₃)₄ → 12 N, Pt₄N₄ → 4 N → total 16 N → not enough

Try 3 Sn(NO₃)₄ + Pt₄N₄ → Sn₃N₄ + 4 Pt(NO₃)₄

Now:
- Sn: 3 → 3 → OK
- Pt: 4 → 4 → OK
- N: left: 3×4 = 12 + 4 = 16; right: Sn₃N₄ → 4 N, 4 Pt(NO₃)₄ → 16 N → total 20 → still off

Difference: 4 N missing

Maybe Pt₄N₄ provides 4 N, but Sn₃N₄ needs 4 N, so net N used: 4

But Pt(NO₃)₄ adds 16 N

So left: Sn(NO₃)₄ provides 4 N per molecule

So 3 Sn(NO₃)₄ → 12 N, Pt₄N₄ → 4 N → total 16 N

Right: Sn₃N₄ → 4 N, 4 Pt(NO₃)₄ → 16 N → total 20 N → excess

Not working.

Perhaps it's a redox reaction.

But let's try:

Suppose:
3 Sn(NO₃)₄ + Pt₄N₄ → Sn₃N₄ + 4 Pt(NO₃)₄

Atoms:
- Sn: 3 → 3 → OK
- Pt: 4 → 4 → OK
- N: left: 3×4 = 12 (from Sn) + 4 (from Pt₄N₄) = 16
- Right: Sn₃N₄ → 4 N, 4 Pt(NO₃)₄ → 4×4 = 16 N → total 20 → not equal

So impossible.

Unless Pt₄N₄ is not providing 4 N — but it is.

Perhaps the formula is wrong.

Maybe it's Pt₄N₄ as a compound, but likely it's a typo.

Alternatively, perhaps it's Pt₄N₄ meaning 4 Pt and 4 N, but the reaction is:

3 Sn(NO₃)₄ + Pt₄N₄ → Sn₃N₄ + 4 Pt(NO₃)₄

But N doesn't balance.

Difference: right has 20 N, left has 16 → need 4 more N on left.

So perhaps 4 Sn(NO₃)₄?

Try:
4 Sn(NO₃)₄ + Pt₄N₄ → Sn₃N₄ + 4 Pt(NO₃)₄ + Sn(NO₃)₄ → not good.

Alternatively, maybe Sn₃N₄ comes from Sn and N, and Pt takes NO₃.

But it's messy.

Perhaps the intended reaction is:

3 Sn(NO₃)₄ + Pt₄N₄ → Sn₃N₄ + 4 Pt(NO₃)₄

But N imbalance.

Unless Pt₄N₄ is Pt₄N₄, but N is shared.

But it's not possible.

Maybe it's a typo: perhaps Pt₄N₄ should be Pt₄N₄ but means something else.

I think there's an error in the problem.

Skip for now.

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19) B₂Br₆ + HNO₃ → B(NO₃)₃ + HBr

Unbalanced:
B₂Br₆ + HNO₃ → B(NO₃)₃ + HBr

- B: 2 → need 2 B(NO₃)₃
- Br: 6 → need 6 HBr
- H: 6 → 6 → OK
- NO₃: 2×3 = 6 → need 6 HNO₃

✔ Balanced:
1 B₂Br₆ + 6 HNO₃ → 2 B(NO₃)₃ + 6 HBr

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20) ZnS + AlP → Zn₃P₂ + Al₂S₃

Unbalanced:
ZnS + AlP → Zn₃P₂ + Al₂S₃

- Zn: 3 on right → need 3 ZnS
- S: 3 on right → need 3 ZnS → 3 S
- Al: 2 on right → need 2 AlP
- P: 2 on right → need 2 AlP → 2 P

So:
3 ZnS + 2 AlP → Zn₃P₂ + Al₂S₃

Check:
- Zn: 3 → 3 → OK
- S: 3 → 3 → OK
- Al: 2 → 2 → OK
- P: 2 → 2 → OK

✔ Balanced:
3 ZnS + 2 AlP → Zn₃P₂ + Al₂S₃

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## ✔ Final Answers (Balanced Equations):

1) 1 C₄H₈ + 6 O₂ → 4 CO₂ + 4 H₂O
2) 2 Al + Fe₃N₂ → 2 AlN + 3 Fe
3) 2 Na + Cl₂ → 2 NaCl
4) 2 H₂O₂ → 2 H₂O + O₂
5) 1 C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O
6) 8 H₂O + 7 CO₂ → C₇H₁₆ + 11 O₂
7) 2 NaClO₃ → 2 NaCl + 3 O₂
8) 2 (NH₄)₃PO₄ + 3 Pb(NO₃)₂ → Pb₃(PO₄)₂ + 6 NH₄NO₃
9) 2 BF₃ + 3 Li₂SO₄ → B₂(SO₄)₃ + 6 LiF
10) 4 C₇H₁₇ + 45 O₂ → 28 CO₂ + 34 H₂O
11) 3 CaCO₃ + 2 H₃PO₄ → Ca₃(PO₄)₂ + 3 H₂CO₃
12) 8 Ag₂S → 16 Ag + S₈
13) 3 KBr + Fe(OH)₃ → 3 KOH + FeBr₃
14) 2 KNO₃ + H₂CO₃ → K₂CO₃ + 2 HNO₃
15) Pb(OH)₄ + Cu₂O → PbO₂ + 2 CuOH + H₂O
16) Cr(NO₃)₂ + (NH₄)₂SO₄ → CrSO₄ + 2 NH₄NO₃
17) 6 KOH + Co₃(PO₄)₂ → 2 K₃PO₄ + 3 Co(OH)₂ *(assuming typo)*
18) Problem may have error – consider revising formulas
19) B₂Br₆ + 6 HNO₃ → 2 B(NO₃)₃ + 6 HBr
20) 3 ZnS + 2 AlP → Zn₃P₂ + Al₂S₃

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Let me know if you'd like help with #17 or #18 corrected!