Free Printable Writing and Balancing Chemical Equations Worksheets - Free Printable

Let’s go through each reaction one by one. For each, we’ll:

1. Figure out what products form (based on reaction type).
2. Write the correct chemical formulas with states (s, l, g, aq).
3. Balance the equation so atoms are equal on both sides.

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a) Lead(II) nitrate solution + sodium iodide solution

This is a double displacement reaction. The ions swap partners.

Lead(II) = Pb²⁺
Nitrate = NO₃⁻ → so lead(II) nitrate = Pb(NO₃)₂(aq)
Sodium = Na⁺
Iodide = I⁻ → so sodium iodide = NaI(aq)

Products: Lead(II) iodide and sodium nitrate.

Lead(II) iodide = PbI₂ — it’s insoluble (yellow precipitate), so (s)
Sodium nitrate = NaNO₃ — soluble, so (aq)

Unbalanced:
Pb(NO₃)₂(aq) + NaI(aq) → PbI₂(s) + NaNO₃(aq)

Balance: Need 2 NaI to get 2 I for PbI₂, and that gives 2 Na, so need 2 NaNO₃.

✔ Balanced:
Pb(NO₃)₂(aq) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)

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b) Solid zinc sulfide reacts with oxygen in air

This is combustion/oxidation. Zinc sulfide burns to make zinc oxide and sulfur dioxide.

ZnS(s) + O₂(g) → ZnO(s) + SO₂(g)

Check atoms:

Left: Zn=1, S=1, O=2
Right: Zn=1, S=1, O=1+2=3 → not balanced

Try putting 2 in front of ZnS and 2 in front of ZnO:

2ZnS + O₂ → 2ZnO + 2SO₂? Let’s check O: right has 2 + 4 = 6 O → need 3 O₂ on left.

✔ Balanced:
2ZnS(s) + 3O₂(g) → 2ZnO(s) + 2SO₂(g)

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c) Liquid butane used as fuel to ignite lighter

Butane = C₄H₁₀(l) — burns in oxygen to make CO₂ and H₂O (combustion).

C₄H₁₀(l) + O₂(g) → CO₂(g) + H₂O(g)

Balance C first: 4 on left → 4 CO₂
Balance H: 10 H → 5 H₂O (since each has 2 H)
Now O: right = 4×2 + 5×1 = 8 + 5 = 13 O atoms → need 13/2 O₂

Multiply whole equation by 2 to eliminate fraction:

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

✔ Balanced:
2C₄H₁₀(l) + 13O₂(g) → 8CO₂(g) + 10H₂O(g)

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d) Barium hydroxide solution neutralized by hydrochloric acid

Acid-base neutralization → salt + water

Ba(OH)₂(aq) + HCl(aq) → BaCl₂(aq) + H₂O(l)

Balance: Ba is good. Cl: need 2 HCl → then 2 H on left, so need 2 H₂O? Wait:

Ba(OH)₂ has 2 OH → needs 2 H⁺ → so 2 HCl → makes 2 H₂O

✔ Balanced:
Ba(OH)₂(aq) + 2HCl(aq) → BaCl₂(aq) + 2H₂O(l)

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e) Copper metal placed in silver nitrate solution

Single displacement: Cu replaces Ag because Cu is more reactive than Ag.

Cu(s) + AgNO₃(aq) → Cu(NO₃)₂(aq) + Ag(s)

Cu goes from 0 to +2, Ag from +1 to 0.

Need 2 AgNO₃ to balance charge and atoms.

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

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f) Sulfur burns in oxygen to make sulfur dioxide gas

Simple combination:

S(s) + O₂(g) → SO₂(g)

Already balanced!

✔ Balanced:
S(s) + O₂(g) → SO₂(g)

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g) Aluminum sulfate solution mixed with calcium hydroxide solution

Double displacement:

Al₂(SO₄)₃(aq) + Ca(OH)₂(aq) → Al(OH)₃(s) + CaSO₄(s)

Wait — aluminum hydroxide is insoluble (gelatinous precipitate), calcium sulfate is slightly soluble but often written as solid in these reactions.

Balance:

Al: 2 on left → need 2 Al(OH)₃
SO₄: 3 on left → need 3 CaSO₄ → so 3 Ca on right → need 3 Ca(OH)₂ on left
OH: 3×2=6 on left → 2×3=6 on right → good

✔ Balanced:
Al₂(SO₄)₃(aq) + 3Ca(OH)₂(aq) → 2Al(OH)₃(s) + 3CaSO₄(s)

*(Note: Some sources may write CaSO₄ as aqueous if dilute, but typically in such problems, it's considered precipitate. We'll follow common textbook practice here.)*

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h) Zinc metal placed in sulfuric acid

Single displacement: Zn replaces H in acid → hydrogen gas + zinc sulfate

Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)

Already balanced!

✔ Balanced:
Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)

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i) Aluminum powder placed in chlorine gas

Combination reaction: metal + nonmetal → ionic compound

Al(s) + Cl₂(g) → AlCl₃(s)

Balance: Al is 1, Cl is 2 on left, 3 on right → LCM of 2 and 3 is 6.

So 2 Al and 3 Cl₂ → 2 AlCl₃

✔ Balanced:
2Al(s) + 3Cl₂(g) → 2AlCl₃(s)

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j) Sucrose undergoes cellular respiration

Cellular respiration: sugar + oxygen → carbon dioxide + water + energy

Sucrose = C₁₂H₂₂O₁₁(s) — though in body it’s dissolved, we can write (aq) or (s). Usually in bio contexts, it’s aqueous.

C₁₂H₂₂O₁₁(aq) + O₂(g) → CO₂(g) + H₂O(l)

Balance:

C: 12 → 12 CO₂
H: 22 → 11 H₂O
O: left = 11 + ?; right = 12×2 + 11×1 = 24 + 11 = 35 → so O₂ must supply 35 - 11 = 24 O atoms → 12 O₂ molecules

✔ Balanced:
C₁₂H₂₂O₁₁(aq) + 12O₂(g) → 12CO₂(g) + 11H₂O(l)

*(Energy is released but not included in balancing equations unless specified.)*

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Final Answer:

a) Pb(NO₃)₂(aq) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)
b) 2ZnS(s) + 3O₂(g) → 2ZnO(s) + 2SO₂(g)
c) 2C₄H₁₀(l) + 13O₂(g) → 8CO₂(g) + 10H₂O(g)
d) Ba(OH)₂(aq) + 2HCl(aq) → BaCl₂(aq) + 2H₂O(l)
e) Cu(s) + 2AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)
f) S(s) + O₂(g) → SO₂(g)
g) Al₂(SO₄)₃(aq) + 3Ca(OH)₂(aq) → 2Al(OH)₃(s) + 3CaSO₄(s)
h) Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)
i) 2Al(s) + 3Cl₂(g) → 2AlCl₃(s)
j) C₁₂H₂₂O₁₁(aq) + 12O₂(g) → 12CO₂(g) + 11H₂O(l)