balancing-equations-practice-worksheet - wikiDownload - Free Printable

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 use coefficients (numbers in front) to balance — never change subscripts!

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1) N₂ + H₂ → NH₃

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

Try putting 2 in front of NH₃ → now right has 2 N, 6 H
So left needs 3 H₂ (to get 6 H) and 1 N₂ (already 2 N)

✔ 1 N₂ + 3 H₂ → 2 NH₃

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2) KClO₃ → KCl + O₂

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

Need even oxygen on right → try 2 KClO₃ → gives 6 O
Then need 3 O₂ on right (3×2=6 O)
Also need 2 KCl to match K and Cl

✔ 2 KClO₃ → 2 KCl + 3 O₂

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3) NaCl + F₂ → NaF + Cl₂

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

Fluorine: need 2 NaF to use up F₂ → then 2 Na on right → so 2 NaCl on left
Chlorine: 2 Cl on left → need 1 Cl₂ on right

✔ 2 NaCl + 1 F₂ → 2 NaF + 1 Cl₂

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4) H₂ + O₂ → H₂O

Left: 2 H, 2 O
Right: 2 H, 1 O

Need 2 H₂O → 4 H, 2 O → so need 2 H₂ on left

✔ 2 H₂ + 1 O₂ → 2 H₂O

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5) Pb(OH)₂ + HCl → H₂O + PbCl₂

Left: Pb, 2 O, 2 H from OH, plus H and Cl from HCl
Better to count total:

Pb(OH)₂ has: 1 Pb, 2 O, 2 H
HCl has: 1 H, 1 Cl per molecule

Right: H₂O → 2 H, 1 O; PbCl₂ → 1 Pb, 2 Cl

Try 2 HCl → gives 2 H, 2 Cl → matches PbCl₂
Now H: left = 2 (from Pb(OH)₂) + 2 (from 2 HCl) = 4 H
Right: need 2 H₂O → 4 H, 2 O → matches O from Pb(OH)₂

✔ 1 Pb(OH)₂ + 2 HCl → 2 H₂O + 1 PbCl₂

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6) AlBr₃ + K₂SO₄ → KBr + Al₂(SO₄)₃

Right has Al₂ → need 2 AlBr₃ on left
Right has 3 SO₄ → need 3 K₂SO₄ on left → that gives 6 K → so 6 KBr on right
Check Br: 2 AlBr₃ = 6 Br → 6 KBr = 6 Br ✔️

✔ 2 AlBr₃ + 3 K₂SO₄ → 6 KBr + 1 Al₂(SO₄)₃

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7) CH₄ + O₂ → CO₂ + H₂O

C: 1=1 ✔️
H: 4 on left → need 2 H₂O on right → 4 H
O: right = 2 (CO₂) + 1 (each H₂O) ×2 = 4 O → so need 2 O₂ on left

✔ 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O

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8) C₃H₈ + O₂ → CO₂ + H₂O

C: 3 → need 3 CO₂
H: 8 → need 4 H₂O
O: right = 3×2 + 4×1 = 6+4=10 → so 5 O₂ on left

✔ 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

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9) C₆H₁₈ + O₂ → CO₂ + H₂O

Wait — C₆H₁? That’s unusual. Usually it’s C₆H₁₄ or C₈H₁₈. But let’s assume it’s correct.

C: 6 → 6 CO₂
H: 18 → 9 H₂O
O: right = 6×2 + 9×1 = 12+9=21 → so 10.5 O₂ → multiply all by 2

→ 2 C₆H₁ + 21 O₂ → 12 CO₂ + 18 H₂O

But maybe typo? If it’s C₈H₁ (octane), common problem.

Assuming it’s C₈H₁ (probably typo):

C₈H₁ + O₂ → CO₂ + H₂O

C: 8 → 8 CO₂
H: 18 → 9 H₂O
O: 8×2 + 9 = 25 → so 12.5 O₂ → ×2 → 2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O

But since written as C₆H₁₈, we’ll do that:

C₆H₁₈ → 6 CO₂, 9 H₂O → O = 12 + 9 = 21 → 10.5 O₂ → ×2:

✔ 2 C₆H₁₈ + 21 O₂ → 12 CO₂ + 18 H₂O

*(Note: This compound isn’t standard — likely meant C₈H₁₈, but we follow what’s written.)*

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10) FeCl₃ + NaOH → Fe(OH)₃ + NaCl

Fe: 1=1
Cl: 3 → need 3 NaCl
Na: 3 → need 3 NaOH
OH: 3 on right → 3 NaOH gives 3 OH ✔️

✔ 1 FeCl₃ + 3 NaOH → 1 Fe(OH)₃ + 3 NaCl

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11) P + O₂ → P₂O₅

Right: 2 P, 5 O
Left: P and O₂

Need 2 P on left → 2 P
Need 5 O → 2.5 O₂ → ×2 → 4 P + 5 O₂ → 2 P₂O₅

✔ 4 P + 5 O₂ → 2 P₂O₅

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12) Na + H₂O → NaOH + H₂

Na: 1=1
H: 2 on left → right: 1 in NaOH + 2 in H₂? Wait no — H₂ has 2 H, NaOH has 1 H → total 3 H? No.

Actually: H₂O has 2 H, 1 O
NaOH has 1 Na, 1 O, 1 H
H₂ has 2 H

So if 1 Na + 1 H₂O → 1 NaOH + 0.5 H₂ → ×2:

2 Na + 2 H₂O → 2 NaOH + 1 H₂

Check: Na:2=2, H:4=2+2, O:2=2 ✔️

✔ 2 Na + 2 H₂O → 2 NaOH + 1 H₂

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13) Ag₂O → Ag + O₂

Left: 2 Ag, 1 O
Right: Ag and O₂

Need 2 Ag on right → 2 Ag
O: 1 on left → need 0.5 O₂ → ×2:

2 Ag₂O → 4 Ag + 1 O₂

✔ 2 Ag₂O → 4 Ag + 1 O₂

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14) S₈ + O₂ → SO₃

S: 8 → need 8 SO₃
O: 8×3=24 → need 12 O₂

✔ 1 S₈ + 12 O₂ → 8 SO₃

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15) CO₂ + H₂O → C₆H₁₂O₆ + O₂

This is photosynthesis.

Left: C, 2O from CO₂; 2H, 1O from H₂O
Right: 6C, 12H, 6O from glucose + 2O from O₂? Wait.

Standard balanced: 6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂

Check: C:6=6, H:12=12, O: 12+6=18 vs 6+12=18 ✔️

✔ 6 CO₂ + 6 H₂O → 1 C₆H₁₂O₆ + 6 O₂

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16) K + MgBr → KBr + Mg

Wait — MgBr? Should be MgBr₂ probably. Assuming typo → MgBr₂

Then: K + MgBr₂ → KBr + Mg

Mg:1=1, Br:2 → need 2 KBr → so 2 K on left

✔ 2 K + 1 MgBr₂ → 2 KBr + 1 Mg

*(If really MgBr, then: K + MgBr → KBr + Mg → already balanced? But MgBr not stable. We’ll assume MgBr₂.)*

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17) HCl + CaCO₃ → CaCl₂ + H₂O + CO₂

Ca:1=1, C:1=1, O:3 → right: 1 in H₂O + 2 in CO₂ = 3 ✔️
H:1 on left → need 2 HCl for 2 H in H₂O? Wait.

CaCO₃ + 2 HCl → CaCl₂ + H₂O + CO₂

Check: H:2=2, Cl:2=2, Ca:1=1, C:1=1, O:3=1+2 ✔️

✔ 2 HCl + 1 CaCO₃ → 1 CaCl₂ + 1 H₂O + 1 CO₂

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18) HNO₃ + NaHCO₃ → NaNO₃ + H₂O + CO₂

Already looks balanced? Let’s check:

H:1+1=2 → right: 2 in H₂O? No — H₂O has 2 H, but left has H from HNO₃ and NaHCO₃ → 1+1=2 H → right: H₂O has 2 H ✔️
N:1=1, Na:1=1, C:1=1, O:3+3=6 → right: 3 in NaNO₃ + 1 in H₂O + 2 in CO₂ = 6 ✔️

✔ 1 HNO₃ + 1 NaHCO₃ → 1 NaNO₃ + 1 H₂O + 1 CO₂

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19) H₂O + O₂ → H₂O₂

Left: 2H, 3O? Wait — H₂O has 2H,1O; O₂ has 2O → total 2H,3O
Right: H₂O₂ has 2H,2O → not balanced.

Actually, this reaction doesn’t normally happen, but to balance:

Suppose: 2 H₂O + O₂ → 2 H₂O₂

Left: 4H, 4O
Right: 4H, 4O ✔️

✔ 2 H₂O + 1 O₂ → 2 H₂O₂

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20) NaBr + CaF₂ → NaF + CaBr₂

Na:1, Br:1, Ca:1, F:2
Right: Na:1, F:1, Ca:1, Br:2

Need 2 NaBr → 2 Na, 2 Br
Then 2 NaF → 2 Na, 2 F
CaF₂ provides 2 F → good
CaBr₂ needs 2 Br → from 2 NaBr ✔️

✔ 2 NaBr + 1 CaF₂ → 2 NaF + 1 CaBr₂

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21) H₂SO₄ + NaNO₂ → HNO₂ + Na₂SO₄

Na:1 on left → need 2 NaNO₂ → 2 Na
Then Na₂SO₄ → 2 Na ✔️
SO₄:1=1
H:2 on left → need 2 HNO₂ → 2 H ✔️
N:2=2

✔ 1 H₂SO₄ + 2 NaNO₂ → 2 HNO₂ + 1 Na₂SO₄

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Final Answer:

1) 1 N₂ + 3 H₂ → 2 NH₃
2) 2 KClO₃ → 2 KCl + 3 O₂
3) 2 NaCl + 1 F₂ → 2 NaF + 1 Cl₂
4) 2 H₂ + 1 O₂ → 2 H₂O
5) 1 Pb(OH)₂ + 2 HCl → 2 H₂O + 1 PbCl₂
6) 2 AlBr₃ + 3 K₂SO₄ → 6 KBr + 1 Al₂(SO₄)₃
7) 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O
8) 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
9) 2 C₆H₁₈ + 21 O₂ → 12 CO₂ + 18 H₂O
10) 1 FeCl₃ + 3 NaOH → 1 Fe(OH)₃ + 3 NaCl
11) 4 P + 5 O₂ → 2 P₂O₅
12) 2 Na + 2 H₂O → 2 NaOH + 1 H₂
13) 2 Ag₂O → 4 Ag + 1 O₂
14) 1 S₈ + 12 O₂ → 8 SO₃
15) 6 CO₂ + 6 H₂O → 1 C₆H₁₂O₆ + 6 O₂
16) 2 K + 1 MgBr₂ → 2 KBr + 1 Mg
17) 2 HCl + 1 CaCO₃ → 1 CaCl₂ + 1 H₂O + 1 CO₂
18) 1 HNO₃ + 1 NaHCO₃ → 1 NaNO₃ + 1 H₂O + 1 CO₂
19) 2 H₂O + 1 O₂ → 2 H₂O₂
20) 2 NaBr + 1 CaF₂ → 2 NaF + 1 CaBr₂
21) 1 H₂SO₄ + 2 NaNO₂ → 2 HNO₂ + 1 Na₂SO₄