49 Balancing Chemical Equations Worksheets [with Answers] - Free Printable

Let's solve each of these chemical equations by balancing them. The goal is to ensure the number of atoms of each element is the same on both sides of the equation.

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1) Li₃PO₄ + NaOH → LiOH + Na₃PO₄

- Left: Li = 3, P = 1, O = 4 + 1 = 5 (from PO₄ and OH), Na = 1, H = 1
- Right: Li = 1, O = 1 + 4 = 5, H = 1, Na = 3, P = 1

We need 3 LiOH to balance Li and H:
→ 3 LiOH → 3 Li, 3 H, 3 O

But we only have 1 NaOH → need 3 NaOH for 3 Na.

So:
- 1 Li₃PO₄ + 3 NaOH → 3 LiOH + 1 Na₃PO₄

✔ Balanced:
1 Li₃PO₄ + 3 NaOH → 3 LiOH + 1 Na₃PO₄

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2) MgF₂ + Li₂CO₃ → MgCO₃ + LiF

- Left: Mg = 1, F = 2, Li = 2, C = 1, O = 3
- Right: Mg = 1, C = 1, O = 3, Li = 1, F = 1

LiF has only 1 Li and 1 F → need 2 LiF to match 2 Li and 2 F

So:
→ 2 LiF

Now:
- MgF₂ + Li₂CO₃ → MgCO₃ + 2 LiF

✔ Balanced:
1 MgF₂ + 1 Li₂CO₃ → 1 MgCO₃ + 2 LiF

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3) P₄ + O₂ → P₂O₃

- Left: P = 4, O = 2
- Right: P = 2, O = 3

Need to make P equal: 2 P₂O₃ → 4 P, 6 O

So:
→ 2 P₂O₃

Now need 6 O → 3 O₂

P₄ already has 4 P → good

So:
→ P₄ + 3 O₂ → 2 P₂O₃

✔ Balanced:
1 P₄ + 3 O₂ → 2 P₂O₃

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4) RbNO₃ + MgF₂ → Mg(NO₃)₂ + RbF

- Left: Rb = 1, N = 1, O = 3, Mg = 1, F = 2
- Right: Mg = 1, N = 2, O = 6, Rb = 1, F = 1

Need 2 NO₃⁻ → so need 2 RbNO₃

Then Rb = 2 → need 2 RbF

F from MgF₂ = 2 → gives 2 F → matches 2 RbF

MgF₂ → 1 Mg → matches Mg(NO₃)₂

But now:
- 2 RbNO₃ + 1 MgF₂ → 1 Mg(NO₃)₂ + 2 RbF

Check:
- Left: Rb = 2, N = 2, O = 6, Mg = 1, F = 2
- Right: Mg = 1, N = 2, O = 6, Rb = 2, F = 2

✔ Balanced:
2 RbNO₃ + 1 MgF₂ → 1 Mg(NO₃)₂ + 2 RbF

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5) AgNO₃ + Cu → Cu(NO₃)₂ + Ag

- Left: Ag = 1, N = 1, O = 3, Cu = 1
- Right: Cu = 1, N = 2, O = 6, Ag = 1

Need 2 NO₃⁻ → so need 2 AgNO₃

Then Ag = 2 → need 2 Ag on right

Cu(NO₃)₂ has 1 Cu → OK

So:
→ 2 AgNO₃ + Cu → Cu(NO₃)₂ + 2 Ag

✔ Balanced:
2 AgNO₃ + 1 Cu → 1 Cu(NO₃)₂ + 2 Ag

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6) CF₄ + Br₂ → CBr₄ + F₂

- Left: C = 1, F = 4, Br = 2
- Right: C = 1, Br = 4, F = 2

Need 4 Br → 2 Br₂ → 4 Br

F: 4 F on left → 2 F₂ on right → 4 F

So:
→ CF₄ + 2 Br₂ → CBr₄ + 2 F₂

✔ Balanced:
1 CF₄ + 2 Br₂ → 1 CBr₄ + 2 F₂

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7) HCN + CuSO₄ → H₂SO₄ + Cu(CN)₂

- Left: H = 1, C = 1, N = 1, Cu = 1, S = 1, O = 4
- Right: H = 2, S = 1, O = 4, Cu = 1, C = 2, N = 2

Need 2 HCN to get 2 C, 2 N, 2 H

Then H₂SO₄ needs 2 H → good

So:
→ 2 HCN + CuSO₄ → H₂SO₄ + Cu(CN)₂

✔ Balanced:
2 HCN + 1 CuSO₄ → 1 H₂SO₄ + 1 Cu(CN)₂

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8) GaF₃ + Cs → CsF + Ga

- Left: Ga = 1, F = 3, Cs = 1
- Right: Cs = 1, F = 1, Ga = 1

Need 3 CsF → 3 F and 3 Cs

So need 3 Cs

→ GaF₃ + 3 Cs → 3 CsF + Ga

✔ Balanced:
1 GaF₃ + 3 Cs → 3 CsF + 1 Ga

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9) SrS + PtF₂ → SrF₂ + PtS

- Left: Sr = 1, S = 1, Pt = 1, F = 2
- Right: Sr = 1, F = 2, Pt = 1, S = 1

All atoms balanced as is.

✔ Balanced:
1 SrS + 1 PtF₂ → 1 SrF₂ + 1 PtS

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10) N₂ + H₂ → NH₃

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

Need 2 NH₃ → 2 N, 6 H

So need 3 H₂ → 6 H

And 1 N₂ → 2 N

→ N₂ + 3 H₂ → 2 NH₃

✔ Balanced:
1 N₂ + 3 H₂ → 2 NH₃

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11) LiF + Br₂ → LiBr + F₂

- Left: Li = 1, F = 1, Br = 2
- Right: Li = 1, Br = 1, F = 2

Need 2 LiBr → 2 Li, 2 Br

So need 2 LiF → 2 Li, 2 F

F₂ → 2 F → good

Br₂ → 2 Br → matches

So:
→ 2 LiF + Br₂ → 2 LiBr + F₂

✔ Balanced:
2 LiF + 1 Br₂ → 2 LiBr + 1 F₂

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12) Pb(OH)₂ + HCl → H₂O + PbCl₂

- Left: Pb = 1, O = 2, H = 2 + 1 = 3, Cl = 1
- Right: H = 2, O = 1, Pb = 1, Cl = 2

Need 2 HCl → 2 Cl, 2 H

Then H₂O → 2 H, 1 O

But we have 2 O from OH in Pb(OH)₂ → need 2 H₂O

So:
→ Pb(OH)₂ + 2 HCl → 2 H₂O + PbCl₂

Check:
- Left: Pb = 1, O = 2, H = 2 (from OH) + 2 (from HCl) = 4, Cl = 2
- Right: H = 4 (2×H₂O), O = 2, Pb = 1, Cl = 2

✔ Balanced:
1 Pb(OH)₂ + 2 HCl → 2 H₂O + 1 PbCl₂

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13) GaBr₃ + Na₂CO₃ → NaBr + Ga₂(CO₃)₃ (tough)

- Left: Ga = 1, Br = 3, Na = 2, C = 1, O = 3
- Right: Na = 1, Br = 1, Ga = 2, C = 3, O = 9

We need:
- Ga₂(CO₃)₃ → 2 Ga, 3 CO₃ → 3 C, 9 O
- So need 3 Na₂CO₃ → 6 Na, 3 C, 9 O

Now Na = 6 → need 6 NaBr → 6 Na, 6 Br

GaBr₃ has 3 Br → need 2 GaBr₃ → 6 Br

Ga: 2 Ga → matches Ga₂(CO₃)₃

So:
→ 2 GaBr₃ + 3 Na₂CO₃ → 6 NaBr + 1 Ga₂(CO₃)₃

✔ Balanced:
2 GaBr₃ + 3 Na₂CO₃ → 6 NaBr + 1 Ga₂(CO₃)₃

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14) CH₄ + O₂ → CO₂ + H₂O

- Left: C = 1, H = 4, O = 2
- Right: C = 1, O = 2 + 1 = 3, H = 2

Need 2 H₂O → 4 H, 2 O

Now O: CO₂ has 2 O, 2 H₂O has 2 O → total 4 O

So need 2 O₂ → 4 O

So:
→ CH₄ + 2 O₂ → CO₂ + 2 H₂O

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

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15) Li₃PO₄ + CaCl₂ → LiCl + Ca₃(PO₄)₂ (also tough)

- Left: Li = 3, P = 1, O = 4, Ca = 1, Cl = 2
- Right: Li = 1, Cl = 1, Ca = 3, P = 2, O = 8

We need Ca₃(PO₄)₂ → 3 Ca, 2 PO₄ → 2 P, 8 O

So need 2 Li₃PO₄ → 6 Li, 2 P, 8 O

Ca: need 3 Ca → 3 CaCl₂ → 3 Ca, 6 Cl

LiCl: need 6 Li → 6 LiCl → 6 Cl

So:
→ 2 Li₃PO₄ + 3 CaCl₂ → 6 LiCl + 1 Ca₃(PO₄)₂

✔ Balanced:
2 Li₃PO₄ + 3 CaCl₂ → 6 LiCl + 1 Ca₃(PO₄)₂

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16) Na + Cl₂ → NaCl

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

Need 2 NaCl → 2 Na, 2 Cl

So:
→ 2 Na + Cl₂ → 2 NaCl

✔ Balanced:
2 Na + 1 Cl₂ → 2 NaCl

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17) Ga + HCl → H₂ + GaCl₃ (last tough one)

- Left: Ga = 1, H = 1, Cl = 1
- Right: H = 2, Ga = 1, Cl = 3

GaCl₃ → 3 Cl → need 3 HCl

H: 3 H → but H₂ has 2 H → need 3/2 H₂ → not integer

Multiply all by 2:

→ 2 Ga + 6 HCl → 3 H₂ + 2 GaCl₃

Check:
- Left: Ga = 2, H = 6, Cl = 6
- Right: H = 6, Ga = 2, Cl = 6

✔ Balanced:
2 Ga + 6 HCl → 3 H₂ + 2 GaCl₃

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18) N₂ + F₂ → NF₃

- Left: N = 2, F = 2
- Right: N = 1, F = 3

Need 2 NF₃ → 2 N, 6 F

So need 3 F₂ → 6 F

N₂ → 2 N → good

→ N₂ + 3 F₂ → 2 NF₃

✔ Balanced:
1 N₂ + 3 F₂ → 2 NF₃

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19) SO₂ + Li₂Se → SSe₂ + Li₂O

- Left: S = 1, O = 2, Li = 2, Se = 1
- Right: S = 1, Se = 2, Li = 2, O = 1

Need 2 Se → 2 Li₂Se → 4 Li, 2 Se

But Li₂O needs 2 Li → need 2 Li₂O → 4 Li, 2 O

SO₂ has 2 O → good

S: 1 SO₂ → 1 S → good

SSe₂ has 1 S, 2 Se → good

So:
→ SO₂ + 2 Li₂Se → SSe₂ + 2 Li₂O

✔ Balanced:
1 SO₂ + 2 Li₂Se → 1 SSe₂ + 2 Li₂O

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20) NH₃ + H₂SO₄ → (NH₄)₂SO₄

- Left: N = 1, H = 3 + 2 = 5, S = 1, O = 4
- Right: N = 2, H = 8, S = 1, O = 4

Need 2 N → 2 NH₃

H: 2 NH₃ → 6 H, H₂SO₄ → 2 H → total 8 H → matches (NH₄)₂SO₄

So:
→ 2 NH₃ + H₂SO₄ → (NH₄)₂SO₄

✔ Balanced:
2 NH₃ + 1 H₂SO₄ → 1 (NH₄)₂SO₄

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## ✔ Final Answers (Balanced Equations):

1) 1 Li₃PO₄ + 3 NaOH → 3 LiOH + 1 Na₃PO₄
2) 1 MgF₂ + 1 Li₂CO₃ → 1 MgCO₃ + 2 LiF
3) 1 P₄ + 3 O₂ → 2 P₂O₃
4) 2 RbNO₃ + 1 MgF₂ → 1 Mg(NO₃)₂ + 2 RbF
5) 2 AgNO₃ + 1 Cu → 1 Cu(NO₃)₂ + 2 Ag
6) 1 CF₄ + 2 Br₂ → 1 CBr₄ + 2 F₂
7) 2 HCN + 1 CuSO₄ → 1 H₂SO₄ + 1 Cu(CN)₂
8) 1 GaF₃ + 3 Cs → 3 CsF + 1 Ga
9) 1 SrS + 1 PtF₂ → 1 SrF₂ + 1 PtS
10) 1 N₂ + 3 H₂ → 2 NH₃
11) 2 LiF + 1 Br₂ → 2 LiBr + 1 F₂
12) 1 Pb(OH)₂ + 2 HCl → 2 H₂O + 1 PbCl₂
13) 2 GaBr₃ + 3 Na₂CO₃ → 6 NaBr + 1 Ga₂(CO₃)₃
14) 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O
15) 2 Li₃PO₄ + 3 CaCl₂ → 6 LiCl + 1 Ca₃(PO₄)₂
16) 2 Na + 1 Cl₂ → 2 NaCl
17) 2 Ga + 6 HCl → 3 H₂ + 2 GaCl₃
18) 1 N₂ + 3 F₂ → 2 NF₃
19) 1 SO₂ + 2 Li₂Se → 1 SSe₂ + 2 Li₂O
20) 2 NH₃ + 1 H₂SO₄ → 1 (NH₄)₂SO₄

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