Balancing Equations Practice Worksheet with problems and solutions for chemistry students.

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) ___ NaNO₃ + ___ PbO → ___ Pb(NO₃)₂ + ___ Na₂O

Left: Na, N, O, Pb
Right: Pb, N, O, Na

Start with Pb — 1 on left, 1 on right → OK for now.
Look at NO₃ — it’s a group. On right, Pb(NO₃)₂ has 2 NO₃ groups. So we need 2 NaNO₃ on left to get 2 NO₃.
→ 2 NaNO₃ + ___ PbO → 1 Pb(NO₃)₂ + ___ Na₂O

Now Na: 2 on left → need 1 Na₂O on right (since Na₂O has 2 Na).
→ 2 NaNO₃ + ___ PbO → 1 Pb(NO₃)₂ + 1 Na₂O

Check O: Left = 2*3 (from NaNO₃) + 1 (from PbO) = 6 + 1 = 7
Right = 2*3 (from Pb(NO₃)₂) + 1 (from Na₂O) = 6 + 1 = 7 → OK
Pb: 1 on each side → OK
N: 2 on each side → OK

✔ Balanced: 2 NaNO₃ + PbO → Pb(NO₃)₂ + Na₂O

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2) ___ AgI + ___ Fe₂(CO₃)₃ → ___ FeI₃ + ___ Ag₂CO₃

Left: Ag, I, Fe, C, O
Right: Fe, I, Ag, C, O

Fe₂(CO₃)₃ has 2 Fe and 3 CO₃ groups.
On right, FeI₃ has 1 Fe → so we need 2 FeI₃ to match 2 Fe from left.
→ ___ AgI + 1 Fe₂(CO₃)₃ → 2 FeI₃ + ___ Ag₂CO₃

Now I: 2 FeI₃ has 6 I → so need 6 AgI on left.
→ 6 AgI + 1 Fe₂(CO₃)₃ → 2 FeI₃ + ___ Ag₂CO₃

Ag: 6 on left → need 3 Ag₂CO₃ on right (since each has 2 Ag → 3×2=6)
→ 6 AgI + 1 Fe₂(CO₃)₃ → 2 FeI₃ + 3 Ag₂CO₃

Check CO₃: Left has 3 (from Fe₂(CO₃)₃), right has 3 (from 3 Ag₂CO₃) → OK
C and O will also match since CO₃ is intact.

✔ Balanced: 6 AgI + Fe₂(CO₃)₃ → 2 FeI₃ + 3 Ag₂CO₃

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3) ___ C₂H₄O₂ + ___ O₂ → ___ CO₂ + ___ H₂O

This is acetic acid burning in oxygen.

Left: C=2, H=4, O=2 (from acid) + ? from O₂
Right: CO₂ and H₂O

Start with C: 2 on left → need 2 CO₂ on right
→ ___ C₂H₄O₂ + ___ O₂ → 2 CO₂ + ___ H₂O

H: 4 on left → need 2 H₂O on right (each has 2 H → 2×2=4)
→ ___ C₂H₄O₂ + ___ O₂ → 2 CO₂ + 2 H₂O

Now count O on right:
2 CO₂ → 4 O
2 H₂O → 2 O
Total = 6 O

Left: C₂H₄O₂ has 2 O → so O₂ must supply 4 more O → that’s 2 O₂ molecules (since each O₂ has 2 O → 2×2=4)

→ 1 C₂H₄O₂ + 2 O₂ → 2 CO₂ + 2 H₂O

Check:
C: 2=2
H: 4=4
O: 2 + 4 = 6; right: 4 + 2 = 6 → OK

✔ Balanced: C₂H₄O₂ + 2 O₂ → 2 CO₂ + 2 H₂O

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4) ___ ZnSO₄ + ___ Li₂CO₃ → ___ ZnCO₃ + ___ Li₂SO₄

This looks like a double replacement. All ions swap partners.

ZnSO₄ + Li₂CO₃ → ZnCO₃ + Li₂SO₄

Check atoms:

Zn: 1=1
S: 1=1
O: 4+3=7 left; 3+4=7 right → wait, let’s break it down:

Actually, SO₄ is a group, CO₃ is a group.

Left: Zn, SO₄, 2Li, CO₃
Right: Zn, CO₃, 2Li, SO₄ → already balanced!

So coefficients are all 1.

✔ Balanced: ZnSO₄ + Li₂CO₃ → ZnCO₃ + Li₂SO₄

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5) ___ V₂O₅ + ___ CaS → ___ CaO + ___ V₂S₅

Left: V=2, O=5, Ca=?, S=?
Right: Ca=?, O=?, V=2, S=5

V₂S₅ has 5 S → so need 5 CaS on left to get 5 S.
→ ___ V₂O₅ + 5 CaS → ___ CaO + 1 V₂S₅

Ca: 5 on left → need 5 CaO on right
→ ___ V₂O₅ + 5 CaS → 5 CaO + 1 V₂S₅

O: Left = 5 (from V₂O₅), Right = 5 (from 5 CaO) → OK
V: 2=2
S: 5=5

✔ Balanced: V₂O₅ + 5 CaS → 5 CaO + V₂S₅

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6) ___ Mn(NO₂)₂ + ___ BeCl₂ → ___ Be(NO₂)₂ + ___ MnCl₂

Double replacement again.

Mn(NO₂)₂ + BeCl₂ → Be(NO₂)₂ + MnCl₂

Check:

Mn: 1=1
NO₂: 2=2
Be: 1=1
Cl: 2=2

Already balanced!

✔ Balanced: Mn(NO₂)₂ + BeCl₂ → Be(NO₂)₂ + MnCl₂

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7) ___ AgBr + ___ GaPO₄ → ___ Ag₃PO₄ + ___ GaBr₃

Left: Ag, Br, Ga, P, O
Right: Ag, P, O, Ga, Br

Ag₃PO₄ has 3 Ag → so need 3 AgBr on left.
→ 3 AgBr + ___ GaPO₄ → 1 Ag₃PO₄ + ___ GaBr₃

Br: 3 on left → need 1 GaBr₃ on right (has 3 Br)
→ 3 AgBr + ___ GaPO₄ → 1 Ag₃PO₄ + 1 GaBr₃

Ga: 1 on right → need 1 GaPO₄ on left
→ 3 AgBr + 1 GaPO₄ → 1 Ag₃PO₄ + 1 GaBr₃

Check PO₄: 1 on each side → OK

✔ Balanced: 3 AgBr + GaPO₄ → Ag₃PO₄ + GaBr₃

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8) ___ H₂SO₄ + ___ B(OH)₃ → ___ B₂(SO₄)₃ + ___ H₂O

Left: H, S, O, B
Right: B, S, O, H

B₂(SO₄)₃ has 2 B and 3 SO₄ → so need 2 B(OH)₃ and 3 H₂SO₄

Try:
3 H₂SO₄ + 2 B(OH)₃ → 1 B₂(SO₄)₃ + ___ H₂O

Now H:
Left: 3×2 = 6 from H₂SO₄; 2×3 = 6 from B(OH)₃ → total H = 12
Right: H only in H₂O → so need 6 H₂O (6×2=12 H)

→ 3 H₂SO₄ + 2 B(OH)₃ → 1 B₂(SO₄)₃ + 6 H₂O

Check O:
Left: 3×4 = 12 from H₂SO₄; 2×3 = 6 from B(OH)₃ → total 18 O
Right: B₂(SO₄)₃ has 3×4=12 O; 6 H₂O has 6 O → total 18 → OK
S: 3=3
B: 2=2

✔ Balanced: 3 H₂SO₄ + 2 B(OH)₃ → B₂(SO₄)₃ + 6 H₂O

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9) ___ S₈ + ___ O₂ → ___ SO₂

S₈ has 8 S → so need 8 SO₂ on right.
→ 1 S₈ + ___ O₂ → 8 SO₂

O: 8×2=16 on right → need 8 O₂ on left (8×2=16)

→ 1 S₈ + 8 O₂ → 8 SO₂

✔ Balanced: S₈ + 8 O₂ → 8 SO₂

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10) ___ Fe + ___ AgNO₃ → ___ Fe(NO₃)₂ + ___ Ag

Single replacement: Fe replaces Ag.

Fe + AgNO₃ → Fe(NO₃)₂ + Ag

Fe(NO₃)₂ has 2 NO₃ → so need 2 AgNO₃ on left.
→ ___ Fe + 2 AgNO₃ → 1 Fe(NO₃)₂ + ___ Ag

Ag: 2 on left → need 2 Ag on right
→ ___ Fe + 2 AgNO₃ → 1 Fe(NO₃)₂ + 2 Ag

Fe: 1=1

✔ Balanced: Fe + 2 AgNO₃ → Fe(NO₃)₂ + 2 Ag

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Final Answer:
1) 2 NaNO₃ + PbO → Pb(NO₃)₂ + Na₂O
2) 6 AgI + Fe₂(CO₃)₃ → 2 FeI₃ + 3 Ag₂CO₃
3) C₂H₄O₂ + 2 O₂ → 2 CO₂ + 2 H₂O
4) ZnSO₄ + Li₂CO₃ → ZnCO₃ + Li₂SO₄
5) V₂O₅ + 5 CaS → 5 CaO + V₂S₅
6) Mn(NO₂)₂ + BeCl₂ → Be(NO₂)₂ + MnCl₂
7) 3 AgBr + GaPO₄ → Ag₃PO₄ + GaBr₃
8) 3 H₂SO₄ + 2 B(OH)₃ → B₂(SO₄)₃ + 6 H₂O
9) S₈ + 8 O₂ → 8 SO₂
10) Fe + 2 AgNO₃ → Fe(NO₃)₂ + 2 Ag