Free Printable Writing and Balancing Chemical Equations Worksheets - Free Printable

Let’s go through each word equation one by one. We’ll write the correct chemical formulas for each substance, then balance the equation so that the number of atoms of each element is the same on both sides.

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(i) Aluminum metal reacts with iron (II) oxide powder to produce aluminum oxide solid and iron metal.

- Aluminum metal = Al
- Iron (II) oxide = FeO (since iron is +2, oxygen is -2)
- Aluminum oxide = Al₂O₃ (Al is +3, O is -2 → need 2 Al and 3 O)
- Iron metal = Fe

Unbalanced:
Al + FeO → Al₂O₃ + Fe

Balance step-by-step:

Look at oxygen: left has 1 O in FeO, right has 3 O in Al₂O₃ → multiply FeO by 3
→ Al + 3FeO → Al₂O₃ + Fe

Now iron: left has 3 Fe, right has 1 Fe → multiply Fe by 3
→ Al + 3FeO → Al₂O₃ + 3Fe

Now aluminum: left has 1 Al, right has 2 Al → multiply Al by 2
→ 2Al + 3FeO → Al₂O₃ + 3Fe

Check:
Left: 2 Al, 3 Fe, 3 O
Right: 2 Al, 3 O, 3 Fe → Balanced!

✔ Final: 2Al + 3FeO → Al₂O₃ + 3Fe

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(ii) Aluminum sulfate solution and calcium hydroxide solution produce a precipitate of aluminum hydroxide and solid calcium sulfate.

- Aluminum sulfate = Al₂(SO₄)₃ (Al³⁺, SO₄²⁻ → need 2 Al, 3 SO₄)
- Calcium hydroxide = Ca(OH)₂ (Ca²⁺, OH⁻ → need 2 OH)
- Aluminum hydroxide = Al(OH)₃ (Al³⁺, OH⁻ → need 3 OH)
- Calcium sulfate = CaSO₄ (Ca²⁺, SO₄²⁻)

Unbalanced:
Al₂(SO₄)₃ + Ca(OH)₂ → Al(OH)₃ + CaSO₄

Balance step-by-step:

Start with sulfate: left has 3 SO₄, right has 1 → multiply CaSO₄ by 3
→ Al₂(SO₄)₃ + Ca(OH)₂ → Al(OH)₃ + 3CaSO₄

Now calcium: right has 3 Ca, left has 1 → multiply Ca(OH)₂ by 3
→ Al₂(SO₄)₃ + 3Ca(OH)₂ → Al(OH)₃ + 3CaSO₄

Now aluminum: left has 2 Al, right has 1 → multiply Al(OH)₃ by 2
→ Al₂(SO₄)₃ + 3Ca(OH)₂ → 2Al(OH)₃ + 3CaSO₄

Check hydroxide: left has 3 × 2 = 6 OH, right has 2 × 3 = 6 OH → good!

Check all:
Left: 2 Al, 3 S, 12 O from sulfate + 6 O and 6 H from hydroxide → total O: 18? Wait — better to count per element.

Actually, let’s list atoms:

Left:
Al: 2
S: 3
O from SO₄: 3×4=12; from OH: 3×2=6 → total O: 18? But wait — Ca(OH)₂ contributes O and H separately.

Better breakdown:

Left:
Al₂(SO₄)₃ → 2 Al, 3 S, 12 O
3Ca(OH)₂ → 3 Ca, 6 O, 6 H
Total left: Al=2, S=3, Ca=3, O=18, H=6

Right:
2Al(OH)₃ → 2 Al, 6 O, 6 H
3CaSO₄ → 3 Ca, 3 S, 12 O
Total right: Al=2, S=3, Ca=3, O=18, H=6 → Perfect!

✔ Final: Al₂(SO₄)₃ + 3Ca(OH)₂ → 2Al(OH)₃ + 3CaSO₄

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(iii) Ammonia and oxygen gas yield nitrogen (II) oxide gas plus water vapor.

- Ammonia = NH₃
- Oxygen gas = O₂
- Nitrogen (II) oxide = NO (N is +2, O is -2)
- Water vapor = H₂O

Unbalanced:
NH₃ + O₂ → NO + H₂O

Balance step-by-step:

Start with N: 1 on each side → ok for now.

H: left 3, right 2 → find LCM of 3 and 2 = 6 → multiply NH₃ by 2, H₂O by 3
→ 2NH₃ + O₂ → NO + 3H₂O

Now N: left 2, right 1 → multiply NO by 2
→ 2NH₃ + O₂ → 2NO + 3H₂O

Now O: right has 2 (from 2NO) + 3 (from 3H₂O) = 5 O atoms → left has O₂ → need 5/2 O₂ → use fraction temporarily
→ 2NH₃ + ⁵/₂O₂ → 2NO + 3H₂O

Multiply entire equation by 2 to eliminate fraction:
→ 4NH₃ + 5O₂ → 4NO + 6H₂O

Check:
Left: N=4, H=12, O=10
Right: N=4, O=4+6=10, H=12 → Balanced!

✔ Final: 4NH₃ + 5O₂ → 4NO + 6H₂O

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(iv) Calcium hydroxide solution and carbon dioxide gas yield solid calcium carbonate and liquid water.

- Calcium hydroxide = Ca(OH)₂
- Carbon dioxide = CO₂
- Calcium carbonate = CaCO₃
- Water = H₂O

Unbalanced:
Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

Check atoms:

Left: Ca=1, O=2+2=4, H=2, C=1
Right: Ca=1, C=1, O=3+1=4, H=2 → Already balanced!

✔ Final: Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

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(v) Aqueous iron (III) chloride and sodium carbonate solution yield aqueous sodium chloride and a precipitate of iron (III) carbonate.

- Iron (III) chloride = FeCl₃ (Fe³⁺, Cl⁻ → need 3 Cl)
- Sodium carbonate = Na₂CO₃ (Na⁺, CO₃²⁻ → need 2 Na)
- Sodium chloride = NaCl
- Iron (III) carbonate = Fe₂(CO₃)₃ (Fe³⁺, CO₃²⁻ → need 2 Fe, 3 CO₃)

Unbalanced:
FeCl₃ + Na₂CO₃ → NaCl + Fe₂(CO₃)₃

Balance step-by-step:

Start with Fe: left 1, right 2 → multiply FeCl₃ by 2
→ 2FeCl₃ + Na₂CO₃ → NaCl + Fe₂(CO₃)₃

Now Cl: left 6, right 1 → multiply NaCl by 6
→ 2FeCl₃ + Na₂CO₃ → 6NaCl + Fe₂(CO₃)₃

Now Na: right 6, left 2 → multiply Na₂CO₃ by 3
→ 2FeCl₃ + 3Na₂CO₃ → 6NaCl + Fe₂(CO₃)₃

Check CO₃: left 3, right 3 → good
Check Fe: 2=2
Check Cl: 6=6
Check Na: 6=6

✔ Final: 2FeCl₃ + 3Na₂CO₃ → 6NaCl + Fe₂(CO₃)₃

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(vi) Solid iron (III) oxide and carbon monoxide gas yield iron metal and carbon dioxide gas.

- Iron (III) oxide = Fe₂O₃
- Carbon monoxide = CO
- Iron metal = Fe
- Carbon dioxide = CO₂

Unbalanced:
Fe₂O₃ + CO → Fe + CO₂

Balance step-by-step:

Fe: left 2, right 1 → multiply Fe by 2
→ Fe₂O₃ + CO → 2Fe + CO₂

O: left 3 + ? from CO, right 2 from CO₂ → tricky because CO and CO₂ both have O.

Try balancing O by adjusting CO and CO₂.

Note: Each CO takes one O to become CO₂. Fe₂O₃ has 3 O atoms → need 3 CO to take them → makes 3 CO₂.

So try:
Fe₂O₃ + 3CO → 2Fe + 3CO₂

Check:
Left: Fe=2, O=3+3=6, C=3
Right: Fe=2, C=3, O=6 → Balanced!

✔ Final: Fe₂O₃ + 3CO → 2Fe + 3CO₂

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(vii) Magnesium carbonate solution plus aqueous hydrochloric acid yield magnesium chloride solution plus liquid water and carbon dioxide gas.

- Magnesium carbonate = MgCO₃
- Hydrochloric acid = HCl
- Magnesium chloride = MgCl₂
- Water = H₂O
- Carbon dioxide = CO₂

Unbalanced:
MgCO₃ + HCl → MgCl₂ + H₂O + CO₂

Balance step-by-step:

Mg: 1=1 → ok
C: 1=1 → ok
O: 3 on left, 1 in H₂O + 2 in CO₂ = 3 → ok
H: left 1 (in HCl), right 2 (in H₂O) → need 2 HCl
Cl: left 2 (if 2 HCl), right 2 (in MgCl₂) → good

So:
MgCO₃ + 2HCl → MgCl₂ + H₂O + CO₂

Check all:
Left: Mg=1, C=1, O=3, H=2, Cl=2
Right: Mg=1, Cl=2, H=2, O=1+2=3, C=1 → Balanced!

✔ Final: MgCO₃ + 2HCl → MgCl₂ + H₂O + CO₂

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(viii) Silicon dioxide solid plus aqueous hydrofluoric acid yield solid silicon tetrafluoride plus.

Wait — it says “plus.” at the end. That seems incomplete. Probably meant to say “plus water” or something else. Let me think.

Standard reaction: SiO₂ + HF → SiF₄ + H₂O

Yes, that’s common.

So:

- Silicon dioxide = SiO₂
- Hydrofluoric acid = HF
- Silicon tetrafluoride = SiF₄
- Water = H₂O (assumed, since not specified but needed for balance)

Unbalanced:
SiO₂ + HF → SiF₄ + H₂O

Balance step-by-step:

Si: 1=1 → ok
O: 2 on left, 1 on right → multiply H₂O by 2
→ SiO₂ + HF → SiF₄ + 2H₂O

H: right 4, left 1 → multiply HF by 4
→ SiO₂ + 4HF → SiF₄ + 2H₂O

F: left 4, right 4 → good

Check:
Left: Si=1, O=2, H=4, F=4
Right: Si=1, F=4, H=4, O=2 → Balanced!

✔ Final: SiO₂ + 4HF → SiF₄ + 2H₂O

(Note: The original said “plus.” — we assumed water is the other product, which is standard.)

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Final Answer:

(i) 2Al + 3FeO → Al₂O₃ + 3Fe
(ii) Al₂(SO₄)₃ + 3Ca(OH)₂ → 2Al(OH)₃ + 3CaSO₄
(iii) 4NH₃ + 5O₂ → 4NO + 6H₂O
(iv) Ca(OH)₂ + CO₂ → CaCO₃ + H₂O
(v) 2FeCl₃ + 3Na₂CO₃ → 6NaCl + Fe₂(CO₃)₃
(vi) Fe₂O₃ + 3CO → 2Fe + 3CO₂
(vii) MgCO₃ + 2HCl → MgCl₂ + H₂O + CO₂
(viii) SiO₂ + 4HF → SiF₄ + 2H₂O