Balancing Equations Practice Worksheet - Complete the chemical equations by adjusting coefficients.

Let’s go through each equation one by one and balance them step by step. 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
Now look at NO₃ group — appears as a unit in products. On right, Pb(NO₃)₂ has 2 NO₃ groups. So we need 2 NaNO₃ on left.

→ 2 NaNO₃ + PbO → Pb(NO₃)₂ + Na₂O

Now check Na: 2 on left, 2 on right (in Na₂O) → OK
N: 2 on left, 2 on right → OK
O: Left = 2×3 (from NaNO₃) + 1 (from PbO) = 7; Right = 6 (from Pb(NO₃)₂) + 1 (from Na₂O) = 7 → OK
Pb: 1 each side → OK

✔ Balanced: 2 NaNO₃ + 1 PbO → 1 Pb(NO₃)₂ + 1 Na₂O

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2) AgI + Fe₂(CO₃)₃ → FeI₃ + Ag₂CO₃

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

Fe: 2 on left → need 2 FeI₃ on right
→ AgI + Fe₂(CO₃)₃ → 2 FeI₃ + Ag₂CO₃

I: Now 6 on right (2×3), so need 6 AgI on left
→ 6 AgI + Fe₂(CO₃)₃ → 2 FeI₃ + Ag₂CO₃

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

C: Left = 3 (from Fe₂(CO₃)₃), Right = 3 (from 3 Ag₂CO₃) → OK
O: Left = 9 (3×3), Right = 9 (3×3) → OK
Fe: 2 each → OK
I: 6 each → OK
Ag: 6 each → OK

✔ Balanced: 6 AgI + 1 Fe₂(CO₃)₃ → 2 FeI₃ + 3 Ag₂CO₃

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3) C₂H₄O₂ + O₂ → CO₂ + H₂O

This is acetic acid burning. Let’s count atoms.

Left: C=2, H=4, O=2 (from acid) + ? from O₂
Right: C=1 per CO₂, H=2 per H₂O, O=?

Try putting 2 CO₂ to match C:
C₂H₄O₂ + O₂ → 2 CO₂ + H₂O

H: 4 on left → need 2 H₂O on right
C₂H₄O₂ + O₂ → 2 CO₂ + 2 H₂O

Now count O:

Left: 2 (from acid) + 2x (from x O₂ molecules)
Right: 2×2 = 4 from CO₂ + 2×1 = 2 from H₂O → total 6

So: 2 + 2x = 6 → 2x = 4 → x = 2

Check:
C₂H₄O₂ + 2 O₂ → 2 CO₂ + 2 H₂O

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

✔ Balanced: 1 C₂H₄O₂ + 2 O₂ → 2 CO₂ + 2 H₂O

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4) ZnSO₄ + Li₂CO₃ → ZnCO₃ + Li₂SO₄

Looks like double displacement. Check atom counts.

Zn: 1 each
S: 1 each
O: 4+3=7 left? Wait — better to treat polyatomic ions if they stay together.

SO₄ stays as SO₄, CO₃ stays as CO₃.

So: ZnSO₄ + Li₂CO₃ → ZnCO₃ + Li₂SO₄

All elements already balanced! Each side has:

Zn:1, S:1, O:4+3=7? Actually, let's count properly:

Left: Zn, S, 4O from sulfate + 2Li, C, 3O from carbonate → total O = 7
Right: Zn, C, 3O from carbonate + 2Li, S, 4O from sulfate → total O = 7

Yes, all match.

✔ Balanced: 1 ZnSO₄ + 1 Li₂CO₃ → 1 ZnCO₃ + 1 Li₂SO₄

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5) V₂O₅ + CaS → CaO + V₂S₅

V: 2 each → OK
O: 5 on left → need 5 CaO on right? But then Ca would be 5, so need 5 CaS on left.

Try: V₂O₅ + 5 CaS → 5 CaO + V₂S₅

Check S: left 5, right 5 → OK
Ca: 5 each → OK
O: 5 each → OK
V: 2 each → OK

✔ Balanced: 1 V₂O₅ + 5 CaS → 5 CaO + 1 V₂S₅

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

Double displacement. All ions swap partners.

Mn: 1 each
Be: 1 each
NO₂: 2 each
Cl: 2 each

Already balanced!

✔ Balanced: 1 Mn(NO₂)₂ + 1 BeCl₂ → 1 Be(NO₂)₂ + 1 MnCl₂

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7) AgBr + GaPO₄ → Ag₃PO₄ + GaBr₃

Ag: 1 left, 3 right → need 3 AgBr
Br: now 3 left, 3 right (GaBr₃) → OK
Ga: 1 each → OK
PO₄: 1 each → OK

So: 3 AgBr + GaPO₄ → Ag₃PO₄ + GaBr₃

Check:
Ag: 3=3
Br: 3=3
Ga:1=1
P:1=1
O:4=4

✔ Balanced: 3 AgBr + 1 GaPO₄ → 1 Ag₃PO₄ + 1 GaBr₃

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8) H₂SO₄ + B(OH)₃ → B₂(SO₄)₃ + H₂O

B: 1 left, 2 right → need 2 B(OH)₃
S: 1 left, 3 right → need 3 H₂SO₄

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

Now H: left = 3×2 + 2×3 = 6 + 6 = 12
Right: only in H₂O → need 6 H₂O to get 12 H

O: Let’s verify later.

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

Check O:

Left:
From 3 H₂SO₄: 3×4 = 12 O
From 2 B(OH)₃: 2×3 = 6 O → total 18 O

Right:
B₂(SO₄)₃: 3×4 = 12 O
6 H₂O: 6×1 = 6 O → total 18 O → OK

H: left 6+6=12, right 12 → OK
S: 3=3
B:2=2

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

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9) S₈ + O₂ → SO₂

S: 8 on left → need 8 SO₂ on right
Then O: 8×2 = 16 on right → need 8 O₂ on left (since each O₂ gives 2 O)

So: S₈ + 8 O₂ → 8 SO₂

Check:
S:8=8
O:16=16

✔ Balanced: 1 S₈ + 8 O₂ → 8 SO₂

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10) Fe + AgNO₃ → Fe(NO₃)₂ + Ag

Fe: 1 each
Ag: 1 left, 1 right → but wait, Fe becomes Fe²⁺, so it takes two NO₃⁻ → so need 2 AgNO₃

Try: Fe + 2 AgNO₃ → Fe(NO₃)₂ + 2 Ag

Check:
Fe:1=1
Ag:2=2
N:2=2
O:6=6

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

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