Balancing Word Equations worksheet for chemistry students to practice converting word equations into balanced chemical equations.

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

1. Write the correct chemical formulas for reactants and products.
2. Balance the equation so that the number of atoms of each element is the same on both sides.

We’ll do this carefully, step by step.

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A. carbon + oxygen → carbon dioxide

- Carbon = C
- Oxygen (as a gas) = O₂
- Carbon dioxide = CO₂

Unbalanced:
C + O₂ → CO₂

Check atoms:
Left: 1 C, 2 O
Right: 1 C, 2 O → Already balanced!

✔ Balanced: C + O₂ → CO₂

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B. copper + silver nitrate → copper (II) nitrate + silver

- Copper = Cu
- Silver nitrate = AgNO₃
- Copper(II) nitrate = Cu(NO₃)₂ (because Cu²⁺ and NO₃⁻)
- Silver = Ag

Unbalanced:
Cu + AgNO₃ → Cu(NO₃)₂ + Ag

Count atoms:
Left: 1 Cu, 1 Ag, 1 N, 3 O
Right: 1 Cu, 2 N, 6 O, 1 Ag → Not balanced!

Need 2 AgNO₃ on left to get 2 NO₃ groups for Cu(NO₃)₂.

Try:
Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag

Now check:
Left: 1 Cu, 2 Ag, 2 N, 6 O
Right: 1 Cu, 2 N, 6 O, 2 Ag → Balanced!

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

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C. zinc + copper (II) sulfate → Zinc sulfate + copper

- Zinc = Zn
- Copper(II) sulfate = CuSO₄
- Zinc sulfate = ZnSO₄
- Copper = Cu

Unbalanced:
Zn + CuSO₄ → ZnSO₄ + Cu

Check atoms:
Left: 1 Zn, 1 Cu, 1 S, 4 O
Right: 1 Zn, 1 Cu, 1 S, 4 O → Already balanced!

✔ Balanced: Zn + CuSO₄ → ZnSO₄ + Cu

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D. mercury (II) nitrate + ammonium sulfide → mercury (II) sulfide + ammonium nitrate

- Mercury(II) nitrate = Hg(NO₃)₂
- Ammonium sulfide = (NH₄)₂S
- Mercury(II) sulfide = HgS
- Ammonium nitrate = NH₄NO₃

Unbalanced:
Hg(NO₃)₂ + (NH₄)₂S → HgS + NH₄NO₃

Count atoms:
Left: 1 Hg, 2 N (from NO₃), 6 O, 2 N (from NH₄), 8 H, 1 S → Total N=4? Wait — let’s break it down properly.

Actually:

Hg(NO₃)₂ has: 1 Hg, 2 N, 6 O
(NH₄)₂S has: 2 N, 8 H, 1 S
→ Left total: Hg=1, N=4, O=6, H=8, S=1

Right:
HgS: 1 Hg, 1 S
NH₄NO₃: 2 N, 4 H, 3 O → but we need two of them to match left?

Try:
Hg(NO₃)₂ + (NH₄)₂S → HgS + 2NH₄NO₃

Now right:
HgS: 1 Hg, 1 S
2NH₄NO₃: 4 N, 8 H, 6 O
Total right: Hg=1, S=1, N=4, H=8, O=6 → matches left!

✔ Balanced: Hg(NO₃)₂ + (NH₄)₂S → HgS + 2NH₄NO₃

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E. iron (III) hydroxide → iron (III) oxide + water

- Iron(III) hydroxide = Fe(OH)₃
- Iron(III) oxide = Fe₂O₃
- Water = H₂O

Unbalanced:
Fe(OH)₃ → Fe₂O₃ + H₂O

Left: 1 Fe, 3 O, 3 H
Right: 2 Fe, 3 O + 1 O from H₂O? Wait — Fe₂O₃ has 2 Fe and 3 O; H₂O has 2 H and 1 O.

So right: 2 Fe, 4 O, 2 H → not matching.

Need to balance Fe first. Put 2 Fe(OH)₃ on left.

2Fe(OH)₃ → Fe₂O₃ + H₂O

Left: 2 Fe, 6 O, 6 H
Right: 2 Fe, 3 O (from Fe₂O₃) + ? from H₂O

We need 3 H₂O to use up 6 H and 3 O → total O on right: 3 (from Fe₂O₃) + 3 (from 3H₂O) = 6 O → perfect.

So:
2Fe(OH)₃ → Fe₂O₃ + 3H₂O

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

✔ Balanced: 2Fe(OH)₃ → Fe₂O₃ + 3H₂O

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F. water → hydrogen + oxygen

- Water = H₂O
- Hydrogen gas = H₂
- Oxygen gas = O₂

Unbalanced:
H₂O → H₂ + O₂

Left: 2 H, 1 O
Right: 2 H, 2 O → Not balanced.

Need more H₂O or adjust coefficients.

Try 2H₂O → 2H₂ + O₂

Left: 4 H, 2 O
Right: 4 H, 2 O → Balanced!

✔ Balanced: 2H₂O → 2H₂ + O₂

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G. copper (II) nitrate + water → hydrogen nitrate + copper (II) hydroxide

Wait — “hydrogen nitrate” is another name for nitric acid = HNO₃
Copper(II) hydroxide = Cu(OH)₂
Copper(II) nitrate = Cu(NO₃)₂
Water = H₂O

Unbalanced:
Cu(NO₃)₂ + H₂O → HNO₃ + Cu(OH)₂

Left: 1 Cu, 2 N, 6 O + 2 H, 1 O → total: Cu=1, N=2, O=7, H=2
Right: HNO₃: 1 H, 1 N, 3 O; Cu(OH)₂: 1 Cu, 2 O, 2 H → total: Cu=1, N=1, O=5, H=3 → Not balanced.

Notice: We have 2 NO₃ on left, so we need 2 HNO₃ on right.

Try:
Cu(NO₃)₂ + H₂O → 2HNO₃ + Cu(OH)₂

Now right: 2HNO₃ → 2 H, 2 N, 6 O; Cu(OH)₂ → 1 Cu, 2 O, 2 H → total: Cu=1, N=2, O=8, H=4

Left: Cu(NO₃)₂ → 1 Cu, 2 N, 6 O; H₂O → 2 H, 1 O → total: Cu=1, N=2, O=7, H=2 → Still not matching.

We need more water. Try 2H₂O on left.

Cu(NO₃)₂ + 2H₂O → 2HNO₃ + Cu(OH)₂

Left: Cu=1, N=2, O=6+2=8, H=4
Right: 2HNO₃ → 2H, 2N, 6O; Cu(OH)₂ → 1Cu, 2O, 2H → total: Cu=1, N=2, O=8, H=4 → Balanced!

✔ Balanced: Cu(NO₃)₂ + 2H₂O → 2HNO₃ + Cu(OH)₂

*(Note: This reaction doesn’t actually happen easily in real life, but for balancing practice, it’s fine.)*

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H. phosphorus + oxygen → diphosphorous pentoxide

Phosphorus = P (usually written as P₄, but often simplified to P in basic problems — however, diphosphorus pentoxide is P₂O₅)

But standard formula for diphosphorus pentoxide is P₂O₅, and phosphorus burns as P₄.

However, many textbooks simplify and use P.

Let’s follow common school approach: Use P and P₂O₅.

Unbalanced:
P + O₂ → P₂O₅

Left: 1 P, 2 O
Right: 2 P, 5 O → Not balanced.

Find LCM for P: 2 on right → put 2P on left.

2P + O₂ → P₂O₅

Now O: right has 5, left has 2 → need 5/2 O₂ → multiply all by 2.

4P + 5O₂ → 2P₂O₅

Check:
Left: 4 P, 10 O
Right: 4 P, 10 O → Balanced!

✔ Balanced: 4P + 5O₂ → 2P₂O₅

*(If your teacher uses P₄, then: P₄ + 5O₂ → 2P₂O₅ — also balanced. But since the problem says "phosphorus", using P is acceptable at this level.)*

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I. iron + hydrochloric acid → iron (II) chloride + hydrogen

Iron = Fe
Hydrochloric acid = HCl
Iron(II) chloride = FeCl₂
Hydrogen gas = H₂

Unbalanced:
Fe + HCl → FeCl₂ + H₂

Left: 1 Fe, 1 H, 1 Cl
Right: 1 Fe, 2 Cl, 2 H → Not balanced.

Need 2 HCl on left.

Fe + 2HCl → FeCl₂ + H₂

Check:
Left: 1 Fe, 2 H, 2 Cl
Right: 1 Fe, 2 Cl, 2 H → Balanced!

✔ Balanced: Fe + 2HCl → FeCl₂ + H₂

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J. sodium hydroxide + phosphoric acid → sodium phosphate + water

Sodium hydroxide = NaOH
Phosphoric acid = H₃PO₄
Sodium phosphate = Na₃PO₄ (since PO₄³⁻ needs 3 Na⁺)
Water = H₂O

Unbalanced:
NaOH + H₃PO₄ → Na₃PO₄ + H₂O

Left: 1 Na, 1 O, 1 H + 3 H, 1 P, 4 O → total: Na=1, H=4, P=1, O=5
Right: Na₃PO₄: 3 Na, 1 P, 4 O; H₂O: 2 H, 1 O → total: Na=3, P=1, O=5, H=2 → Not balanced.

Need 3 NaOH to get 3 Na on left.

3NaOH + H₃PO₄ → NaPO₄ + H₂O

Now left: Na=3, O=3+4=7, H=3+3=6, P=1
Right: Na₃PO₄: 3 Na, 1 P, 4 O; H₂O: 2 H, 1 O → total: Na=3, P=1, O=5, H=2 → Still not balanced.

We have 6 H on left, only 2 on right → need 3 H₂O on right (which gives 6 H and 3 O).

Try:
3NaOH + H₃PO₄ → Na₃PO₄ + 3H₂O

Left: Na=3, O=3 (from NaOH) + 4 (from H₃PO₄) = 7 O; H=3+3=6; P=1
Right: Na₃PO₄: 3 Na, 1 P, 4 O; 3H₂O: 6 H, 3 O → total: Na=3, P=1, O=7, H=6 → Perfect!

✔ Balanced: 3NaOH + H₃PO₄ → NaPO₄ + 3H₂O

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Final Answer:

A. C + O₂ → CO₂
B. Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag
C. Zn + CuSO₄ → ZnSO₄ + Cu
D. Hg(NO₃)₂ + (NH₄)₂S → HgS + 2NH₄NO₃
E. 2Fe(OH)₃ → Fe₂O₃ + 3H₂O
F. 2H₂O → 2H₂ + O₂
G. Cu(NO₃)₂ + 2H₂O → 2HNO₃ + Cu(OH)₂
H. 4P + 5O₂ → 2P₂O₅
I. Fe + 2HCl → FeCl₂ + H₂
J. 3NaOH + H₃PO₄ → Na₃PO₄ + 3H₂O