Worksheet titled "Balancing Equations Race" containing 20 chemical equations to be balanced, including reactions involving C₃H₈, Al, Na, H₂O₂, and other compounds.

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.

We’ll start with #1 and work our way down. I’ll show you how to think about it, then give you the final balanced coefficients for all 20.

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1) C₃H₈ + O₂ → CO₂ + H₂O

Left: C=3, H=8, O=?
Right: C=1, H=2, O=3 (from CO₂ and H₂O)

→ Put 3 in front of CO₂ to match carbon:
C₃H₈ + O₂ → 3CO₂ + H₂O

→ Now hydrogen: left has 8H, right has 2H → put 4 in front of H₂O:
C₃H₈ + O₂ → 3CO₂ + 4H₂O

→ Now oxygen: right side = 3×2 + 4×1 = 6+4=10 oxygen atoms → so we need 5 O₂ on left (since each O₂ has 2 oxygens):
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O ✔

Balanced: 1, 5, 3, 4

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2) Al + Fe₃N₂ → AlN + Fe

Left: Al=1, Fe=3, N=2
Right: Al=1, N=1, Fe=1

→ Need 2 AlN to get 2 nitrogens:
Al + Fe₃N₂ → 2AlN + Fe

→ Now aluminum: right has 2 Al → put 2 in front of Al:
2Al + Fe₃N₂ → 2AlN + Fe

→ Iron: left has 3 Fe → put 3 in front of Fe:
2Al + Fe₃N₂ → 2AlN + 3Fe ✔

Balanced: 2, 1, 2, 3

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3) Na + Cl₂ → NaCl

Left: Na=1, Cl=2
Right: Na=1, Cl=1

→ Put 2 in front of NaCl:
Na + Cl₂ → 2NaCl

→ Now sodium: right has 2 Na → put 2 in front of Na:
2Na + Cl₂ → 2NaCl ✔

Balanced: 2, 1, 2

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4) H₂O₂ → H₂O + O₂

Left: H=2, O=2
Right: H=2, O=1 (from H₂O) + 2 (from O₂) = 3? Wait — let’s count properly.

Actually, if we write:
H₂O₂ → H₂O + O₂

Left: H=2, O=2
Right: H=2, O=1+2=3 → not balanced.

Try putting 2 H₂O₂:
2H₂O₂ → H₂O + O₂ → Left: H=4, O=4
Right: H=2, O=1+2=3 → still bad.

Put 2 H₂O:
2H₂O₂ → 2H₂O + O₂ → Right: H=4, O=2+2=4 ✔

Balanced: 2, 2, 1

Wait — product is H₂O and O₂, so:

2H₂O₂ → 2H₂O + O₂ → yes! That works.

Balanced: 2, 2, 1

But order is: H₂O₂ → H₂O + O₂ → so coefficients: 2, 2, 1

✔

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5) C₆H₁₂O₆ + O₂ → H₂O + CO₂

This is glucose combustion.

Left: C=6, H=12, O=6 + ? from O₂
Right: H=2, O=1; C=1, O=2

→ Put 6 CO₂ for carbon:
C₆H₁₂O₆ + O₂ → H₂O + 6CO₂

→ Put 6 H₂O for hydrogen (12 H ÷ 2 = 6):
C₆H₁₂O₆ + O₂ → 6H₂O + 6CO₂

→ Now oxygen on right: 6×1 (from H₂O) + 6×2 (from CO₂) = 6 + 12 = 18
Left: glucose has 6 O → so O₂ must supply 12 O → that’s 6 O₂ molecules.

So:
C₆H₁₂O₆ + 6O₂ → 6H₂O + 6CO₂ ✔

Balanced: 1, 6, 6, 6

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6) H₂O + CO₂ → C₇H₈ + O₂

This looks weird — usually photosynthesis is CO₂ + H₂O → glucose + O₂, but here it’s C₇H₈ (toluene?). Let’s balance as written.

Left: H=2, O=1+2=3, C=1
Right: C=7, H=8, O=2

We need to make left have 7 C, 8 H, and enough O.

Assume coefficient a for H₂O, b for CO₂, c for C₇H₈, d for O₂.

But easier to trial:

To get 7 C on right → need 7 CO₂ on left.

To get 8 H on right → need 4 H₂O on left (since each has 2 H).

So try:
4H₂O + 7CO₂ → C₇H₈ + O₂

Now check oxygen:

Left: 4×1 + 7×2 = 4 + 14 = 18 O
Right: C₇H₈ has no O, so all O must be in O₂ → 18 O atoms → 9 O₂ molecules.

So:
4H₂O + 7CO₂ → C₇H₈ + 9O₂ ✔

Check atoms:

Left: H=8, C=7, O=4+14=18
Right: C=7, H=8, O=18 → perfect.

Balanced: 4, 7, 1, 9

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7) NaClO₃ → NaCl + O₂

Decomposition.

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

Need even oxygen on right → multiply O₂ by 3 → 6 O → so left needs 2 NaClO₃ (gives 6 O)

Try:
2NaClO₃ → NaCl + O₂ → Left: Na=2, Cl=2, O=6
Right: need 2 NaCl and 3 O₂ → 2NaCl + 3O₂ → O=6 ✔

So:
2NaClO₃ → 2NaCl + 3O₂

Balanced: 2, 2, 3

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8) (NH₄)₃PO₄ + Pb(NO₃)₄ → Pb₃(PO₄)₄ + NH₄NO₃

This is double displacement.

Left: N from ammonium and nitrate, etc.

Let’s assign variables or balance step by step.

First, look at phosphate: PO₄³⁻

On left: 1 PO₄ per (NH₄)₃PO₄
On right: 4 PO₄ in Pb₃(PO₄)₄ → so need 4 (NH₄)₃PO₄

Similarly, lead: Pb₃(PO₄)₄ has 3 Pb → so need 3 Pb(NO₃)₄? But Pb(NO₃)₄ has 1 Pb → so 3 Pb(NO₃)₄

Try:

4(NH₄)₃PO₄ + 3Pb(NO₃)₄ → Pb(PO₄)₄ + ?

Now ammonium: left: 4 × 3 = 12 NH₄⁺ → so right should have 12 NH₄NO₃

Nitrate: left: 3 Pb(NO₃)₄ → 3×4=12 NO₃⁻ → right: 12 NH₄NO₃ → 12 NO₃⁻ ✔

Phosphate: left: 4 PO₄, right: 4 PO₄ ✔

Lead: left: 3 Pb, right: 3 Pb ✔

So:
4(NH₄)₃PO₄ + 3Pb(NO₃)₄ → Pb₃(PO₄)₄ + 12NH₄NO₃

Balanced: 4, 3, 1, 12

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9) BF₃ + Li₂SO₃ → B₂(SO₃)₃ + LiF

Left: B=1, F=3, Li=2, S=1, O=3
Right: B=2, S=3, O=9, Li=1, F=1

Need 2 B on right → so 2 BF₃ on left.

Need 3 SO₃ on right → so 3 Li₂SO₃ on left.

Then lithium: left: 3×2=6 Li → right: need 6 LiF

Fluorine: left: 2×3=6 F → right: 6 LiF → 6 F ✔

Sulfur: left: 3 S, right: 3 S ✔

Oxygen: left: 3×3=9, right: 3×3=9 ✔

Boron: left: 2, right: 2 ✔

So:
2BF₃ + 3Li₂SO₃ → B₂(SO₃)₃ + 6LiF

Balanced: 2, 3, 1, 6

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10) C₇H₁₇ + O₂ → CO₂ + H₂O

Wait — C₇H₁₇? That’s not standard. Usually alkanes are CₙH₂ₙ₊₂ → C₇H₁₆. Maybe typo? But we’ll balance as given.

Left: C=7, H=17, O=?

Right: C=1, H=2, O=3

→ Put 7 CO₂ for carbon:
C₇H₁₇ + O₂ → 7CO₂ + H₂O

→ Hydrogen: 17 H on left → right: each H₂O has 2 H → so 17/2 = 8.5 → not integer.

Problem: H=17 is odd → can’t balance with whole numbers unless we use fractions.

Multiply entire equation by 2 to eliminate fraction later.

Set:
a C₇H₁₇ + b O₂ → c CO₂ + d H₂O

Carbon: 7a = c
Hydrogen: 17a = 2d → d = (17a)/2
Oxygen: 2b = 2c + d

Substitute c=7a, d=17a/2

2b = 2*(7a) + 17a/2 = 14a + 8.5a = 22.5a → b = 11.25a

To make integers, set a=4 → then c=28, d=34, b=45

Check:

4 C₇H₁₇ + 45 O₂ → 28 CO₂ + 34 H₂O

Carbon: 4×7=28, right 28 ✔
Hydrogen: 4×17=68, right 34×2=68 ✔
Oxygen: left 45×2=90, right: 28×2 + 34×1 = 56+34=90 ✔

So balanced: 4, 45, 28, 34

But this is unusual. Perhaps it was meant to be C₇H₁₆? But we follow what’s written.

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11) CaCO₃ + H₃PO₄ → Ca₃(PO₄)₂ + H₂CO₃

Left: Ca=1, C=1, O=3+4=7, H=3, P=1
Right: Ca=3, P=2, O=8+3=11? Wait H₂CO₃ has 3 O, Ca₃(PO₄)₂ has 8 O → total 11 O? Not matching.

Balance step by step.

Calcium: right has 3 Ca → so 3 CaCO₃ on left.

Phosphate: right has 2 PO₄ → so 2 H₃PO₄ on left.

Now:
3CaCO₃ + 2H₃PO₄ → Ca₃(PO₄)₂ + ?

Carbon: left: 3 C → so 3 H₂CO₃ on right.

Hydrogen: left: 2×3=6 H → right: 3 H₂CO₃ has 6 H ✔

Oxygen: let’s verify.

Left: 3 CaCO₃ → 3×3=9 O; 2 H₃PO₄ → 2×4=8 O → total 17 O
Right: Ca₃(PO₄)₂ → 8 O; 3 H₂CO₃ → 3×3=9 O → total 17 O ✔

Perfect.

Balanced: 3, 2, 1, 3

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12) Ag₂S → Ag + S₈

Decomposition.

Left: Ag=2, S=1
Right: Ag=1, S=8

Need 8 S on left → so 8 Ag₂S → gives 16 Ag and 8 S

Right: S₈ is already 8 S → good.

Silver: 16 Ag on left → so 16 Ag on right.

So:
8Ag₂S → 16Ag + S₈

Balanced: 8, 16, 1

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13) KBr + Fe(OH)₃ → KOH + FeBr₃

Double displacement.

Left: K=1, Br=1, Fe=1, O=3, H=3
Right: K=1, O=1, H=1, Fe=1, Br=3

Bromine: right has 3 Br → so 3 KBr on left.

Potassium: left 3 K → so 3 KOH on right.

Iron: 1 on each side.

Hydroxide: left: Fe(OH)₃ has 3 OH → right: 3 KOH has 3 OH ✔

So:
3KBr + Fe(OH)₃ → 3KOH + FeBr₃

Balanced: 3, 1, 3, 1

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14) KNO₃ + H₂CO₃ → K₂CO₃ + HNO₃

Left: K=1, N=1, O=3+3=6, H=2, C=1
Right: K=2, C=1, O=3+3=6, H=1, N=1

Potassium: right has 2 K → so 2 KNO₃ on left.

Nitrate: left 2 NO₃ → right need 2 HNO₃.

Carbonate: left H₂CO₃ has 1 CO₃ → right K₂CO₃ has 1 CO₃ → good.

Hydrogen: left H₂CO₃ has 2 H → right 2 HNO₃ has 2 H ✔

So:
2KNO₃ + H₂CO₃ → K₂CO₃ + 2HNO₃

Balanced: 2, 1, 1, 2

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15) Pb(OH)₄ + Cu₂O → PbO₂ + CuOH

Left: Pb=1, O=4+1=5, H=4, Cu=2
Right: Pb=1, O=2+1=3, H=1, Cu=1

Copper: left 2 Cu → right need 2 CuOH.

Hydrogen: right 2 CuOH has 2 H → left has 4 H → not enough.

Oxygen also off.

Try balancing.

Assume:
a Pb(OH)₄ + b Cu₂O → c PbO₂ + d CuOH

Pb: a = c
Cu: 2b = d
O: 4a + b = 2c + d
H: 4a = d

From H: d = 4a
From Cu: 2b = d = 4a → b = 2a
From Pb: c = a
From O: 4a + b = 2c + d → 4a + 2a = 2a + 4a → 6a = 6a ✔

So set a=1 → b=2, c=1, d=4

Thus:
Pb(OH)₄ + 2Cu₂O → PbO₂ + 4CuOH

Check atoms:

Left: Pb=1, O=4+2=6, H=4, Cu=4
Right: Pb=1, O=2+4=6, H=4, Cu=4 ✔

Balanced: 1, 2, 1, 4

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16) Cr(NO₂)₂ + (NH₄)₂SO₄ → CrSO₄ + NH₄NO₂

Left: Cr=1, N=2+2=4? Wait Cr(NO₂)₂ has 2 N from NO₂, (NH₄)₂SO₄ has 2 N from NH₄ → total N=4
O: from NO₂: 4 O, from SO₄: 4 O → total 8 O? Let's list:

Cr(NO₂)₂: Cr, 2N, 4O
(NH₄)₂SO₄: 2N, 8H, S, 4O
Total left: Cr=1, N=4, O=8, H=8, S=1

Right: CrSO₄: Cr, S, 4O
NH₄NO₂: N=2, H=4, O=2

So for one NH₄NO₂: N=2, H=4, O=2

But left has N=4, H=8 → so need 2 NH₄NO₂ on right.

Then:
Cr(NO₂)₂ + (NH₄)₂SO₄ → CrSO₄ + 2NH₄NO₂

Check:

Left: Cr=1, N=2 (from Cr(NO₂)₂) + 2 (from (NH₄)₂SO₄) =4, O=4+4=8, H=8, S=1
Right: Cr=1, S=1, O=4 (from CrSO₄) + 2×2=4 (from two NH₄NO₂) =8, N=2×2=4, H=2×4=8 ✔

Balanced: 1, 1, 1, 2

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17) KOH + Co₃(PO₄)₂ → K₃PO₄ + Co(OH)₂

Left: K=1, O=1+8=9? Co₃(PO₄)₂ has 8 O, KOH has 1 O → total 9 O? Better to count per compound.

Co₃(PO₄)₂: Co=3, P=2, O=8
KOH: K=1, O=1, H=1

Right: K₃PO₄: K=3, P=1, O=4
Co(OH)₂: Co=1, O=2, H=2

Phosphate: right has 1 P per K₃PO₄, left has 2 P → so 2 K₃PO₄ on right.

Then potassium: right 2×3=6 K → so 6 KOH on left.

Cobalt: left 3 Co → right need 3 Co(OH)₂.

Hydrogen: left 6 KOH → 6 H → right 3 Co(OH)₂ → 6 H ✔

Oxygen: left: 6 KOH → 6 O; Co₃(PO₄)₂ → 8 O → total 14 O
Right: 2 K₃PO₄ → 2×4=8 O; 3 Co(OH)₂ → 3×2=6 O → total 14 O ✔

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

Balanced: 6, 1, 2, 3

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18) Sn(NO₂)₄ + Pt₃N₄ → Sn₃N₄ + Pt(NO₂)₄

Left: Sn=1, N=4+4=8? Sn(NO₂)₄ has 4 N, Pt₃N₄ has 4 N → total 8 N
O: 8 O from Sn(NO₂)₄
Pt=3

Right: Sn₃N₄: Sn=3, N=4
Pt(NO₂)₄: Pt=1, N=4, O=8

So to balance tin: left 1 Sn, right 3 Sn → so 3 Sn(NO₂)₄ on left.

Then nitrogen from Sn(NO₂)₄: 3×4=12 N
Plus Pt₃N₄ has 4 N → total 16 N on left.

Right: Sn₃N₄ has 4 N, and Pt(NO₂)₄ has 4 N per molecule.

Also platinum: left 3 Pt → so 3 Pt(NO₂)₄ on right → which has 3×4=12 N

Plus Sn₃N₄ has 4 N → total 16 N ✔

Oxygen: left: 3 Sn(NO₂)₄ → 3×8=24 O? Each NO₂ has 2 O, so Sn(NO₂)₄ has 8 O → 3×8=24 O
Right: 3 Pt(NO₂)₄ → 3×8=24 O ✔

Tin: left 3, right 3 ✔

Platinum: left 3, right 3 ✔

Nitrogen: left: 3×4 (from Sn) + 4 (from Pt₃N₄) = 12+4=16
Right: Sn₃N₄ has 4 N, 3 Pt(NO₂)₄ has 12 N → 16 ✔

So:
3Sn(NO₂)₄ + Pt₃N₄ → Sn₃N₄ + 3Pt(NO₂)₄

Balanced: 3, 1, 1, 3

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19) B₂Br₆ + HNO₃ → B(NO₃)₃ + HBr

Left: B=2, Br=6, H=1, N=1, O=3
Right: B=1, N=3, O=9, H=1, Br=1

Boron: left 2 B → right need 2 B(NO₃)₃

Then nitrogen: right 2×3=6 N → so 6 HNO₃ on left.

Hydrogen: left 6 H → right need 6 HBr.

Bromine: left 6 Br → right 6 HBr → 6 Br ✔

Oxygen: left: 6 HNO₃ → 6×3=18 O
Right: 2 B(NO₃)₃ → 2×9=18 O ✔

So:
B₂Br₆ + 6HNO₃ → 2B(NO₃)₃ + 6HBr

Balanced: 1, 6, 2, 6

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20) ZnS + AlP → Zn₃P₂ + Al₂S₃

Left: Zn=1, S=1, Al=1, P=1
Right: Zn=3, P=2, Al=2, S=3

Zinc: right 3 Zn → so 3 ZnS on left.

Sulfur: left 3 S → right Al₂S₃ has 3 S → good.

Aluminum: right 2 Al → so 2 AlP on left.

Phosphorus: left 2 P → right Zn₃P₂ has 2 P ✔

So:
3ZnS + 2AlP → Zn₃P₂ + Al₂S₃

Check:

Left: Zn=3, S=3, Al=2, P=2
Right: Zn=3, P=2, Al=2, S=3 ✔

Balanced: 3, 2, 1, 1

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Now, compiling all answers in order:

1) 1, 5, 3, 4
2) 2, 1, 2, 3
3) 2, 1, 2
4) 2, 2, 1
5) 1, 6, 6, 6
6) 4, 7, 1, 9
7) 2, 2, 3
8) 4, 3, 1, 12
9) 2, 3, 1, 6
10) 4, 45, 28, 34
11) 3, 2, 1, 3
12) 8, 16, 1
13) 3, 1, 3, 1
14) 2, 1, 1, 2
15) 1, 2, 1, 4
16) 1, 1, 1, 2
17) 6, 1, 2, 3
18) 3, 1, 1, 3
19) 1, 6, 2, 6
20) 3, 2, 1, 1

Final Answer:
1) 1, 5, 3, 4
2) 2, 1, 2, 3
3) 2, 1, 2
4) 2, 2, 1
5) 1, 6, 6, 6
6) 4, 7, 1, 9
7) 2, 2, 3
8) 4, 3, 1, 12
9) 2, 3, 1, 6
10) 4, 45, 28, 34
11) 3, 2, 1, 3
12) 8, 16, 1
13) 3, 1, 3, 1
14) 2, 1, 1, 2
15) 1, 2, 1, 4
16) 1, 1, 1, 2
17) 6, 1, 2, 3
18) 3, 1, 1, 3
19) 1, 6, 2, 6
20) 3, 2, 1, 1