Free Printable Balancing and Classifying Chemical Equations Worksheets - Free Printable

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

1. Identify the type of reaction (e.g., synthesis, decomposition, single displacement, double displacement, combustion, etc.).
2. Balance the chemical equation.

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(i)

Unbalanced:
$$ \text{Na}_2\text{CO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{CO}_3 $$

- Type of Reaction: Double Displacement (also called metathesis). Two compounds exchange ions.
- Balancing:
- Na: 2 on left → need 2 NaCl on right
- Cl: 2 on right → need 2 HCl on left
- H and CO₃: H₂CO₃ already matches with 2 H from 2 HCl

Balanced:
$$
\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{CO}_3
$$

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(ii)

Unbalanced:
$$ \text{NH}_4\text{NO}_2 \rightarrow \text{N}_2 + \text{H}_2\text{O} $$

- Type of Reaction: Decomposition — one compound breaks down into two or more substances.
- Balancing:
- N: 2 on left → 2 on right in N₂ → OK
- H: 4 on left → need 2 H₂O to have 4 H → OK
- O: 2 on left → 2 on right in 2 H₂O → OK

Balanced:
$$
\text{NH}_4\text{NO}_2 \rightarrow \text{N}_2 + 2\text{H}_2\text{O}
$$

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(iii)

Unbalanced:
$$ \text{N}_2 + \text{O}_2 \rightarrow \text{N}_2\text{O}_5 $$

- Type of Reaction: Synthesis (Combination) — two elements combine to form a compound.
- Balancing:
- N: 2 on both sides → OK
- O: 2 on left, 5 on right → need LCM of 2 and 5 = 10
- Multiply N₂O₅ by 2 → 2N₂O₅ (10 O, 4 N)
- So need 2 N₂ (4 N) and 5 O₂ (10 O)

Balanced:
$$
2\text{N}_2 + 5\text{O}_2 \rightarrow 2\text{N}_2\text{O}_5
$$

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(iv)

Unbalanced:
$$ \text{MgCO}_3 \rightarrow \text{MgO} + \text{CO}_2 $$

- Type of Reaction: Decomposition — carbonate breaks down into metal oxide and CO₂.
- Balancing:
- Already balanced as written.

Balanced:
$$
\text{MgCO}_3 \rightarrow \text{MgO} + \text{CO}_2
$$

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(v)

Unbalanced:
$$ \text{KBr} + \text{Cl}_2 \rightarrow \text{KCl} + \text{Br}_2 $$

- Type of Reaction: Single Displacement — more reactive halogen (Cl₂) displaces Br⁻ from KBr.
- Balancing:
- Cl₂ is diatomic → needs 2 KCl
- So 2 KBr needed → gives 2 Br⁻ → forms Br₂
- Br₂ is diatomic → OK

Balanced:
$$
2\text{KBr} + \text{Cl}_2 \rightarrow 2\text{KCl} + \text{Br}_2
$$

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(vi)

Unbalanced:
$$ \text{Zn} + \text{CuSO}_4 \rightarrow \text{Cu} + \text{ZnSO}_4 $$

- Type of Reaction: Single Displacement — Zn displaces Cu²⁺ because Zn is more reactive.
- Balancing:
- Already balanced.

Balanced:
$$
\text{Zn} + \text{CuSO}_4 \rightarrow \text{Cu} + \text{ZnSO}_4
$$

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(vii)

Unbalanced:
$$ \text{P} + \text{O}_2 \rightarrow \text{P}_4\text{O}_6 $$

- Type of Reaction: Synthesis — phosphorus combines with oxygen.
- Balancing:
- P₄O₆ has 4 P atoms → need 4 P on left
- O₂ → 6 O atoms → need 3 O₂ molecules

Balanced:
$$
4\text{P} + 3\text{O}_2 \rightarrow \text{P}_4\text{O}_6
$$

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(viii)

Unbalanced:
$$ \text{SrBr}_2 + (\text{NH}_4)_2\text{CO}_3 \rightarrow \text{SrCO}_3 + \text{NH}_4\text{Br} $$

- Type of Reaction: Double Displacement — ions swap partners.
- Balancing:
- Sr: 1 each side → OK
- CO₃: 1 each side → OK
- NH₄: 2 on left → need 2 NH₄Br on right
- Br: 2 on left → 2 on right → OK

Balanced:
$$
\text{SrBr}_2 + (\text{NH}_4)_2\text{CO}_3 \rightarrow \text{SrCO}_3 + 2\text{NH}_4\text{Br}
$$

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(ix)

Unbalanced:
$$ \text{AgNO}_3 + (\text{NH}_4)_2\text{CrO}_4 \rightarrow \text{Ag}_2\text{CrO}_4 + \text{NH}_4\text{NO}_3 $$

- Type of Reaction: Double Displacement — precipitation of Ag₂CrO₄.
- Balancing:
- Ag: 1 on left → 2 on right → need 2 AgNO₃
- NO₃: 2 on left → need 2 NH₄NO₃ on right
- NH₄: 2 on left → 2 on right → OK
- CrO₄: 1 each side → OK

Balanced:
$$
2\text{AgNO}_3 + (\text{NH}_4)_2\text{CrO}_4 \rightarrow \text{Ag}_2\text{CrO}_4 + 2\text{NH}_4\text{NO}_3
$$

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(x)

Unbalanced:
$$ \text{K} + \text{H}_2\text{O} \rightarrow \text{KOH} + \text{H}_2 $$

- Type of Reaction: Single Displacement — K displaces H from water.
- Balancing:
- K: 1 on left → 1 on right → OK
- H: 2 on left → KOH has 1 H, H₂ has 2 → total 3 H on right? No!
- Actually: H₂O has 2 H → products: KOH has 1 H, H₂ has 2 H → total 3 H → not balanced
- Better: Try 2K → 2KOH → needs 2H from water → so 2H₂O → produces H₂

Let’s balance:
- 2K → 2KOH → need 2H from water → so 2H₂O → gives 2H₂O → 4H → 2H in 2KOH, 2H in H₂ → OK
- O: 2 on left → 2 in 2KOH → OK

Balanced:
$$
2\text{K} + 2\text{H}_2\text{O} \rightarrow 2\text{KOH} + \text{H}_2
$$

But wait: H₂ has only 2H → so 2H₂O provides 4H → 2H in 2KOH, 2H in H₂ → yes.

Alternatively, can simplify:
$$
2\text{K} + 2\text{H}_2\text{O} \rightarrow 2\text{KOH} + \text{H}_2
$$

This is correct.

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(xi)

Unbalanced:
$$ \text{Al} + \text{Pb(NO}_3)_2 \rightarrow \text{Al(NO}_3)_3 + \text{Pb} $$

- Type of Reaction: Single Displacement — Al displaces Pb²⁺ (Al is more reactive).
- Balancing:
- Al: 1 on left → 1 on right → OK
- Pb: 1 each → OK
- NO₃: 2 on left, 3 on right → LCM = 6
- Multiply Pb(NO₃)₂ by 3 → 3Pb(NO₃)₂ → 6 NO₃
- Multiply Al(NO₃)₃ by 2 → 2Al(NO₃)₃ → 6 NO₃
- Now Al: 2 on right → need 2 Al on left
- Pb: 3 on left → need 3 Pb on right

Balanced:
$$
2\text{Al} + 3\text{Pb(NO}_3)_2 \rightarrow 2\text{Al(NO}_3)_3 + 3\text{Pb}
$$

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(xii)

Unbalanced:
$$ \text{Fe} + \text{O}_2 \rightarrow \text{Fe}_3\text{O}_4 $$

- Type of Reaction: Synthesis — iron reacts with oxygen to form magnetite.
- Balancing:
- Fe: 3 on right → need 3 Fe on left
- O: 4 on right → O₂ has 2 O → need 2 O₂

Balanced:
$$
3\text{Fe} + 2\text{O}_2 \rightarrow \text{Fe}_3\text{O}_4
$$

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✔ Final Answers Summary:

| Equation | Type of Reaction | Balanced Equation |
|--------|------------------|-------------------|
| (i) | Double Displacement | $\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{CO}_3$ |
| (ii) | Decomposition | $\text{NH}_4\text{NO}_2 \rightarrow \text{N}_2 + 2\text{H}_2\text{O}$ |
| (iii) | Synthesis | $2\text{N}_2 + 5\text{O}_2 \rightarrow 2\text{N}_2\text{O}_5$ |
| (iv) | Decomposition | $\text{MgCO}_3 \rightarrow \text{MgO} + \text{CO}_2$ |
| (v) | Single Displacement | $2\text{KBr} + \text{Cl}_2 \rightarrow 2\text{KCl} + \text{Br}_2$ |
| (vi) | Single Displacement | $\text{Zn} + \text{CuSO}_4 \rightarrow \text{Cu} + \text{ZnSO}_4$ |
| (vii) | Synthesis | $4\text{P} + 3\text{O}_2 \rightarrow \text{P}_4\text{O}_6$ |
| (viii) | Double Displacement | $\text{SrBr}_2 + (\text{NH}_4)_2\text{CO}_3 \rightarrow \text{SrCO}_3 + 2\text{NH}_4\text{Br}$ |
| (ix) | Double Displacement | $2\text{AgNO}_3 + (\text{NH}_4)_2\text{CrO}_4 \rightarrow \text{Ag}_2\text{CrO}_4 + 2\text{NH}_4\text{NO}_3$ |
| (x) | Single Displacement | $2\text{K} + 2\text{H}_2\text{O} \rightarrow 2\text{KOH} + \text{H}_2$ |
| (xi) | Single Displacement | $2\text{Al} + 3\text{Pb(NO}_3)_2 \rightarrow 2\text{Al(NO}_3)_3 + 3\text{Pb}$ |
| (xii) | Synthesis | $3\text{Fe} + 2\text{O}_2 \rightarrow \text{Fe}_3\text{O}_4$ |

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✔ All equations are now balanced and types identified. Let me know if you'd like this formatted for printing!