Free Printable Balancing Equations Worksheets - Free Printable

Let’s go step by step to solve all the balancing and writing problems.

---

Part 1: Balance the following chemical equations

We need to make sure the number of atoms of each element is the same on both sides of the arrow.

---

1) CaC₂ + ___ H₂O → C₂H₂ + Ca(OH)₂

Left: Ca=1, C=2, H=?, O=?
Right: Ca=1, C=2, H=2+2=4 (from C₂H₂ and Ca(OH)₂), O=2

Try 2 H₂O on left:
H = 4, O = 2 → matches right side.

✔ Balanced:
CaC₂ + 2 H₂O → C₂H₂ + Ca(OH)₂

---

2) ___ Ca(OH)₂ + ___ H₃PO₄ → Ca₃(PO₄)₂ + ___ H₂O

Right has Ca=3, P=2, O=8+? from water. Left: Ca from Ca(OH)₂, P from H₃PO₄.

Need 3 Ca on left → 3 Ca(OH)₂
Need 2 P on left → 2 H₃PO₄

Now check H and O:

Left:
Ca=3, O from Ca(OH)₂: 3×2=6, from H₃PO₄: 2×4=8 → total O=14
H: from Ca(OH)₂: 3×2=6, from H₃PO₄: 2×3=6 → total H=12

Right:
Ca₃(PO₄)₂ has O=8, so remaining O must be in H₂O → 14 - 8 = 6 O → 6 H₂O? But 6 H₂O has 12 H → matches!

✔ Balanced:
3 Ca(OH)₂ + 2 H₃PO₄ → Ca₃(PO₄)₂ + 6 H₂O

---

3) SrCO₃ + ___ NH₄Br → SrBr₂ + (NH₄)₂CO₃

Left: Sr=1, C=1, O=3, N=?, H=?, Br=?
Right: Sr=1, Br=2, N=2, H=8, C=1, O=3

So we need 2 Br on left → 2 NH₄Br
Then N=2, H=8 → matches (NH₄)₂CO₃

✔ Balanced:
SrCO₃ + 2 NH₄Br → SrBr₂ + (NH₄)₂CO₃

---

4) C₇H₁₆ + ___ O₂ → ___ CO₂ + ___ H₂O

Left: C=7, H=16
Right: need 7 CO₂ for carbon, and 8 H₂O for hydrogen (since 8×2=16 H)

Now count O on right:
CO₂: 7×2=14 O
H₂O: 8×1=8 O → total O=22 → so O₂ needed = 22/2 = 11

✔ Balanced:
C₇H₁₆ + 11 O₂ → 7 CO₂ + 8 H₂O

---

5) ___ C₄H₁₀ + ___ O₂ → ___ CO₂ + ___ H₂O

Left: C=4, H=10
Right: need 4 CO₂, and 5 H₂O (for 10 H)

O on right:
CO₂: 4×2=8
H₂O: 5×1=5 → total O=13 → O₂ = 13/2 → not whole number.

Multiply entire equation by 2 to eliminate fraction:

2 C₄H₁₀ + ? O₂ → 8 CO₂ + 10 H₂O

O on right: 8×2 + 10×1 = 16 + 10 = 26 → O₂ = 13

✔ Balanced:
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

---

6) ___ Fe + ___ O₂ → ___ Fe₂O₃

Right: Fe=2, O=3
Left: O₂ → need even O, so multiply Fe₂O₃ by 2 → Fe=4, O=6

Then O₂ = 3 (since 3×2=6 O)
Fe = 4

✔ Balanced:
4 Fe + 3 O₂ → 2 Fe₂O₃

---

Part 2: Write and balance the following chemical equations

---

a) Hydrogen plus oxygen yields water.

Hydrogen gas is H₂, oxygen gas is O₂, water is H₂O.

Unbalanced: H₂ + O₂ → H₂O

Balance:
Need 2 H₂O to get even O → then 2 H₂O has 4 H → need 2 H₂

✔ Balanced:
2 H₂ + O₂ → 2 H₂O

---

b) Sodium plus water yields sodium hydroxide plus hydrogen.

Sodium = Na, water = H₂O, sodium hydroxide = NaOH, hydrogen = H₂

Unbalanced: Na + H₂O → NaOH + H₂

Left: Na=1, H=2, O=1
Right: Na=1, O=1, H=1+2=3 → not balanced.

Try 2 Na + 2 H₂O → 2 NaOH + H₂

Left: Na=2, H=4, O=2
Right: Na=2, O=2, H=2+2=4 → yes!

✔ Balanced:
2 Na + 2 H₂O → 2 NaOH + H₂

---

c) Hydrochloric acid plus iron (III) sulfide yields iron (III) chloride plus hydrogen sulfide.

Hydrochloric acid = HCl
Iron (III) sulfide = Fe₂S₃ (Fe³⁺, S²⁻ → Fe₂S₃)
Iron (III) chloride = FeCl₃
Hydrogen sulfide = H₂S

Unbalanced: HCl + Fe₂S₃ → FeCl₃ + H₂S

Fe: 2 on left → 2 FeCl₃ on right
S: 3 on left → 3 H₂S on right
H: 3×2=6 → need 6 HCl
Cl: 6 on left → 2×3=6 on right → good

✔ Balanced:
6 HCl + Fe₂S₃ → 2 FeCl₃ + 3 H₂S

---

d) Aluminum sulfate plus calcium hydroxide yields aluminum hydroxide plus calcium sulfate.

Aluminum sulfate = Al₂(SO₄)₃
Calcium hydroxide = Ca(OH)₂
Aluminum hydroxide = Al(OH)₃
Calcium sulfate = CaSO₄

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

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

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

---

e) Zinc and lead (II) nitrate react to form zinc nitrate and lead.

Zinc = Zn
Lead (II) nitrate = Pb(NO₃)₂
Zinc nitrate = Zn(NO₃)₂
Lead = Pb

This is single replacement: Zn + Pb(NO₃)₂ → Zn(NO₃)₂ + Pb

Already balanced: 1 Zn, 1 Pb, 2 NO₃ on each side.

✔ Balanced:
Zn + Pb(NO₃)₂ → Zn(NO₃)₂ + Pb

---

f) Aluminum bromide and chlorine gas react to form aluminum chloride and bromine gas.

Aluminum bromide = AlBr₃
Chlorine gas = Cl₂
Aluminum chloride = AlCl₃
Bromine gas = Br₂

Unbalanced: AlBr₃ + Cl₂ → AlCl₃ + Br₂

Al: 1 each side
Br: 3 → need 3/2 Br₂ → multiply by 2

2 AlBr₃ + 3 Cl₂ → 2 AlCl₃ + 3 Br₂

Check:
Al: 2, Br: 6, Cl: 6 → right: Al:2, Cl:6, Br:6 → good

✔ Balanced:
2 AlBr₃ + 3 Cl₂ → 2 AlCl₃ + 3 Br₂

---

Final Answer:

1) CaC₂ + 2 H₂O → C₂H₂ + Ca(OH)₂
2) 3 Ca(OH)₂ + 2 H₃PO₄ → Ca₃(PO₄)₂ + 6 H₂O
3) SrCO₃ + 2 NH₄Br → SrBr₂ + (NH₄)₂CO₃
4) C₇H₁₆ + 11 O₂ → 7 CO₂ + 8 H₂O
5) 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
6) 4 Fe + 3 O₂ → 2 Fe₂O₃

a) 2 H₂ + O₂ → 2 H₂O
b) 2 Na + 2 H₂O → 2 NaOH + H₂
c) 6 HCl + Fe₂S₃ → 2 FeCl₃ + 3 H₂S
d) Al₂(SO₄)₃ + 3 Ca(OH)₂ → 2 Al(OH)₃ + 3 CaSO₄
e) Zn + Pb(NO₃)₂ → Zn(NO₃)₂ + Pb
f) 2 AlBr₃ + 3 Cl₂ → 2 AlCl₃ + 3 Br₂