Free Printable Word Equations Worksheets - Free Printable

Let's solve each of these chemical reactions by writing balanced chemical equations. We'll go step-by-step for each one, identifying the correct formulas for reactants and products, then balancing the atoms on both sides.

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a) aluminum bromide + chlorine → aluminum chloride + bromine

Step 1: Write formulas
- Aluminum bromide: AlBr₃
- Chlorine: Cl₂ (diatomic)
- Aluminum chloride: AlCl₃
- Bromine: Br₂ (diatomic)

Unbalanced equation:
AlBr₃ + Cl₂ → AlCl₃ + Br₂

Balance:
We need to balance Al, Br, and Cl.

- Al is already balanced (1 on each side).
- Br: 3 on left, 2 on right → LCM = 6 → use 2 AlBr₃ and 3 Br₂
- Cl: 2 on left (Cl₂), 3 on right → LCM = 6 → use 3 Cl₂ and 2 AlCl₃

Now:
2AlBr₃ + 3Cl₂ → 2AlCl₃ + 3Br₂

✔ Balanced.

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b) calcium sulfite → calcium oxide + sulfur dioxide

Step 1: Formulas
- Calcium sulfite: CaSO₃
- Calcium oxide: CaO
- Sulfur dioxide: SO₂

Unbalanced:
CaSO₃ → CaO + SO₂

Already balanced:
- Ca: 1 = 1
- S: 1 = 1
- O: 3 = 1 (in CaO) + 2 (in SO₂) = 3

✔ Balanced as is.

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c) phosphoric acid + calcium hydroxide → calcium phosphate + water

Step 1: Formulas
- Phosphoric acid: H₃PO₄
- Calcium hydroxide: Ca(OH)₂
- Calcium phosphate: Ca₃(PO₄)₂ (calcium is +2, phosphate is -3)
- Water: H₂O

Unbalanced:
H₃PO₄ + Ca(OH)₂ → Ca₃(PO₄)₂ + H₂O

Balance:
- PO₄³⁻: 2 in product → need 2 H₃PO₄
- Ca²⁺: 3 in product → need 3 Ca(OH)₂
- Now count H and O

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

Right:
- Ca₃(PO₄)₂: 8O (from PO₄), plus H₂O: let’s say x H₂O → 2x H and x O

So:
- H: 2x = 12 → x = 6
- O: 8 (from phosphate) + 6 (from water) = 14 → matches

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

✔ Balanced.

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d) silver nitrate + ammonium carbonate → silver carbonate + ammonium nitrate

Formulas:
- AgNO₃
- (NH₄)₂CO₃
- Ag₂CO₃ (silver is +1, carbonate -2)
- NH₄NO₃

Unbalanced:
AgNO₃ + (NH₄)₂CO₃ → Ag₂CO₃ + NH₄NO₃

Balance:
- Ag: 2 on right → need 2 AgNO₃
- CO₃²⁻: 1 on each side → OK
- NH₄⁺: 2 on left → need 2 NH₄NO₃ on right

So:
2AgNO₃ + (NH₄)₂CO₃ → Ag₂CO₃ + 2NH₄NO₃

✔ Balanced.

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e) lithium hydrogen carbonate → lithium carbonate + water + carbon dioxide

Formulas:
- Lithium hydrogen carbonate: LiHCO₃
- Lithium carbonate: Li₂CO₃
- Water: H₂O
- CO₂

Unbalanced:
LiHCO₃ → Li₂CO₃ + H₂O + CO₂

Balance:
- Li: 1 on left, 2 on right → need 2 LiHCO₃
- C: 2 on left → 1 in Li₂CO₃, 1 in CO₂ → OK
- H: 2 on left → need 1 H₂O → H is balanced
- O: check later

Left: 2 LiHCO₃ → 2 Li, 2 H, 2 C, 6 O

Right:
- Li₂CO₃: 2 Li, 1 C, 3 O
- H₂O: 2 H, 1 O
- CO₂: 1 C, 2 O
→ Total: 2 Li, 2 H, 2 C, 6 O

✔ Balanced.

2LiHCO₃ → Li₂CO₃ + H₂O + CO₂

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f) magnesium + silver acetate → magnesium acetate + silver

Formulas:
- Mg (elemental)
- Silver acetate: AgCH₃COO or AgC₂H₃O₂
- Magnesium acetate: Mg(C₂H₃O₂)₂
- Silver: Ag

Unbalanced:
Mg + AgC₂H₃O₂ → Mg(C₂H₃O₂)₂ + Ag

Balance:
- Mg: 1 on each side → OK
- Ag: 1 on left, 1 on right → OK
- Acetate: 1 on left, 2 on right → need 2 AgC₂H₃O₂
- Then Ag: 2 on left → need 2 Ag on right

So:
Mg + 2AgC₂H₃O₂ → Mg(C₂H₃O₂)₂ + 2Ag

✔ Balanced.

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g) aluminum bromide + potassium sulfate → potassium bromide + aluminum sulfate

Formulas:
- AlBr₃
- K₂SO₄
- KBr
- Al₂(SO₄)₃

Unbalanced:
AlBr₃ + K₂SO₄ → KBr + Al₂(SO₄)₃

Balance:
- Al: 1 on left, 2 on right → need 2 AlBr₃
- SO₄: 1 on left, 3 on right → need 3 K₂SO₄
- K: 6 on left (3×2) → need 6 KBr
- Br: 6 on left (2×3) → 6 on right → OK

So:
2AlBr₃ + 3K₂SO₄ → 6KBr + Al₂(SO₄)₃

✔ Balanced.

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h) sodium chloride + fluorine → sodium fluoride + chlorine

Formulas:
- NaCl
- F₂ (diatomic)
- NaF
- Cl₂ (diatomic)

Unbalanced:
NaCl + F₂ → NaF + Cl₂

Balance:
- Na: 1 on each side → OK
- Cl: 1 on left, 2 on right → need 2 NaCl
- F: 2 on left → need 2 NaF

So:
2NaCl + F₂ → 2NaF + Cl₂

✔ Balanced.

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i) copper + sulfuric acid → copper sulfate + water + oxygen

Wait: This reaction is not standard. Copper does not react with dilute sulfuric acid because it's below hydrogen in reactivity series. But if it's concentrated H₂SO₄ and heated, it can produce SO₂, not O₂.

But the problem says: → copper sulfate + water + oxygen

That suggests oxidation of water? But that doesn't make sense.

Actually, this is likely a mistake. The correct reaction of copper with concentrated H₂SO₄ is:

Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O

But here it says "oxygen" — which is unusual.

However, let's assume the problem intends a redox reaction where copper displaces oxygen? That’s chemically invalid.

Alternatively, perhaps it's a typo, but let's try to write based on given products.

Assume:
Cu + H₂SO₄ → CuSO₄ + H₂O + O₂

Try balancing:

Left: Cu, H₂, S, O₄
Right: Cu, S, O₄ (in CuSO₄), H₂O, O₂ → too many O

Let’s suppose:
2Cu + 2H₂SO₄ → 2CuSO₄ + 2H₂O + O₂

Check:
- Cu: 2 = 2
- S: 2 = 2
- H: 4 = 4 (in 2H₂O)
- O: left: 8 (2×4), right: 8 (from 2CuSO₄: 8O) + 2 (from H₂O) + 2 (from O₂) = 12 → too many

Not working.

Alternatively, maybe it's a decomposition?

No, copper is unreactive.

But if we must follow the given products, it may be incorrect.

Wait — perhaps it's a misstatement. A known reaction is:

2H₂SO₄ + Cu → CuSO₄ + SO₂ + 2H₂O — no O₂

But the problem says oxygen, so maybe it's:

2Cu + 2H₂SO₄ + O₂ → 2CuSO₄ + 2H₂O — but that's not what’s written.

Given the instruction, perhaps it's a typo. But let's try to balance as per given.

Suppose:
Cu + H₂SO₄ → CuSO₄ + H₂O + O₂

Try:
2Cu + 2H₂SO₄ → 2CuSO₄ + 2H₂O + O₂

Atoms:
- Cu: 2 = 2
- S: 2 = 2
- H: 4 = 4
- O: left: 8 (2×4), right: 8 (in 2CuSO₄) + 2 (in 2H₂O) + 2 (in O₂) = 12 → imbalance

Too many O on right.

Try:
2Cu + 2H₂SO₄ → 2CuSO₄ + 2H₂O → O₂ not produced

So unless there's an external oxidant, this isn’t feasible.

But since the problem says "copper + sulfuric acid → copper sulfate + water + oxygen", it's likely incorrect.

However, for educational purposes, let's assume it's a decomposition of something else — but it's not.

Perhaps it's a misinterpretation of the reaction.

Alternatively, consider:

4H₂SO₄ + 2Cu → 2CuSO₄ + 2H₂O + 2SO₂ + O₂ — but that’s not standard.

After research, no such reaction produces O₂.

So likely, this reaction is not possible as written.

But let's suppose it's a hypothetical or error.

Maybe it's meant to be:

Copper(II) hydroxide decomposes? No.

Wait — perhaps the intended reaction is thermal decomposition of copper sulfate? But that’s not what’s said.

Given the ambiguity, and since copper does not react with sulfuric acid to produce oxygen, this reaction is not chemically valid.

But if we force a balance assuming the equation is:

Cu + H₂SO₄ → CuSO₄ + H₂O + O₂

Then we need to balance:

Let’s try:
2Cu + 2H₂SO₄ → 2CuSO₄ + 2H₂O + O₂

Atoms:
- Cu: 2=2
- H: 4=4
- S: 2=2
- O: left: 8, right: 8 (from CuSO₄) + 2 (from H₂O) + 2 (from O₂) = 12 → excess

Still off.

To balance oxygen, we’d need fewer O₂ or more input.

But no way.

Thus, this reaction cannot be balanced as written. It’s chemically incorrect.

Conclusion: Likely a typo. Perhaps it should be sulfur dioxide instead of oxygen.

But since the problem says "oxygen", and we must answer, let’s assume it’s a hypothetical reaction.

Alternatively, maybe it's:

2H₂O → 2H₂ + O₂, but that’s not involved.

Given all, I suspect the intended reaction might be:

2Cu + O₂ + 2H₂SO₄ → 2CuSO₄ + 2H₂O

But that’s not what’s written.

So for now, we’ll skip or note it as invalid.

But let’s proceed with the rest.

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j) zinc + iron(III) chloride → zinc chloride + iron

Formulas:
- Zn
- FeCl₃
- ZnCl₂
- Fe

Unbalanced:
Zn + FeCl₃ → ZnCl₂ + Fe

Balance:
- Zn: 1 = 1
- Fe: 1 = 1
- Cl: 3 on left, 2 on right → LCM = 6 → need 2 FeCl₃ and 3 ZnCl₂

So:
3Zn + 2FeCl₃ → 3ZnCl₂ + 2Fe

✔ Balanced.

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Final Answers:

Here are the balanced equations:

(a)
2AlBr₃ + 3Cl₂ → 2AlCl₃ + 3Br₂

(b)
CaSO₃ → CaO + SO₂

(c)
2H₃PO₄ + 3Ca(OH)₂ → Ca₃(PO₄)₂ + 6H₂O

(d)
2AgNO₃ + (NH₄)₂CO₃ → Ag₂CO₃ + 2NH₄NO₃

(e)
2LiHCO₃ → Li₂CO₃ + H₂O + CO₂

(f)
Mg + 2AgC₂H₃O₂ → Mg(C₂H₃O₂)₂ + 2Ag

(g)
2AlBr₃ + 3K₂SO₄ → 6KBr + Al₂(SO₄)₃

(h)
2NaCl + F₂ → 2NaF + Cl₂

(i)
⚠️ This reaction is not chemically valid. Copper does not react with sulfuric acid to produce oxygen. The correct reaction with concentrated H₂SO₄ produces SO₂, not O₂.
If forced: no balanced equation exists under normal conditions.

(j)
3Zn + 2FeCl₃ → 3ZnCl₂ + 2Fe

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Summary:

Most are straightforward double displacement or redox reactions. For (i), the reaction as stated is not feasible due to thermodynamic and kinetic reasons.

Let me know if you'd like the corrected version of (i).