Solved Name: Lecture 9: Chemical Formula and Equations ***DO ... - 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 the subscripts!

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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: On right, Pb(NO₃)₂ has 2 NO₃ groups → so we need 2 NaNO₃ on left.
→ 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=6 from NaNO₃ + ? from PbO → total O = 6 + x
Right: 2*3=6 from Pb(NO₃)₂ + 1 from Na₂O → 7 O
So 6 + x = 7 → x = 1 → so 1 PbO

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

✔ Balanced.

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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 + 1 Fe₂(CO₃)₃ → 2 FeI₃ + __ Ag₂CO₃

I: 2*3=6 on right → 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)
→ 6 AgI + 1 Fe₂(CO₃)₃ → 2 FeI₃ + 3 Ag₂CO₃

Check CO₃: Left: 3, Right: 3 → good
C and O also match.

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

✔ Balanced.

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

This is combustion. Let’s balance C first.

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)
→ __ C₂H₄O₂ + __ O₂ → 2 CO₂ + 2 H₂O

Now count O:

Left: from C₂H₄O₂ → 2 O; from O₂ → 2x O → total = 2 + 2x
Right: 2*2=4 from CO₂ + 2*1=2 from H₂O → total 6 O

So: 2 + 2x = 6 → 2x = 4 → x = 2

Final: 1 C₂H₄O₂ + 2 O₂ → 2 CO₂ + 2 H₂O

✔ Balanced.

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

Looks like double replacement. All ions swap partners.

Zn: 1 each side
S: 1 each side
O: lots, but let’s check Li and CO₃

Li: 2 on left → need 1 Li₂SO₄ on right → already there
CO₃: 1 on left → 1 on right

Actually, everything is 1:1.

Try: 1 ZnSO₄ + 1 Li₂CO₃ → 1 ZnCO₃ + 1 Li₂SO₄

Check all atoms:

Zn: 1=1
S: 1=1
O: 4+3=7 left; 3+4=7 right
Li: 2=2
C: 1=1

✔ Balanced.

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

V: 2 on left → 2 on right → OK
S: 1 on left, 5 on right → need 5 CaS on left
→ __ V₂O₅ + 5 CaS → __ CaO + 1 V₂S₅

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

O: 5 on left → 5 on right → perfect

Final: 1 V₂O₅ + 5 CaS → 5 CaO + 1 V₂S₅

✔ Balanced.

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

Double replacement. Swap cations.

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

All 1:1 works.

1 Mn(NO₂)₂ + 1 BeCl₂ → 1 Be(NO₂)₂ + 1 MnCl₂

✔ Balanced.

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

Ag: 1 on left, 3 on right → need 3 AgBr
→ 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 left → 1 on right → good
PO₄: 1 on left → 1 on right → good

Final: 3 AgBr + 1 GaPO₄ → 1 Ag₃PO₄ + 1 GaBr₃

✔ Balanced.

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

B: 1 on left, 2 on right → need 2 B(OH)₃
→ __ H₂SO₄ + 2 B(OH)₃ → 1 B₂(SO₄)₃ + __ H₂O

SO₄: 1 on left, 3 on right → need 3 H₂SO₄
→ 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 12 H
Right: only in H₂O → so need 6 H₂O (each has 2 H → 6*2=12)

Check O: Left: 3*4=12 from H₂SO₄ + 2*3=6 from B(OH)₃ → 18 O
Right: 3*4=12 from B₂(SO₄)₃ + 6*1=6 from H₂O → 18 O → good

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

✔ Balanced.

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9) __ S₈ + __ O₂ → __ SO₂

S: 8 on left → need 8 SO₂ on right
→ 1 S₈ + __ O₂ → 8 SO₂

O: 8*2=16 on right → need 8 O₂ on left (since each O₂ has 2 O → 8*2=16)

Final: 1 S₈ + 8 O₂ → 8 SO₂

✔ Balanced.

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

Fe: 1 each side
Ag: 1 on left, 1 on right? Wait — 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
→ 1 Fe + 2 AgNO₃ → 1 Fe(NO₃)₂ + 2 Ag

Check NO₃: 2 on left, 2 on right → good

Final: 1 Fe + 2 AgNO₃ → 1 Fe(NO₃)₂ + 2 Ag

✔ Balanced.

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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