Types of Chemical Reactions Worksheet Balance the reactions and ... - Free Printable

Let’s go through each reaction one by one. We’ll balance the equation first (make sure atoms on left = atoms on right), then figure out what type of reaction it is.

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Reaction 1:
NaBr + Ca(OH)₂ → CaBr₂ + NaOH

Left: Na=1, Br=1, Ca=1, O=2, H=2
Right: Ca=1, Br=2, Na=1, O=1, H=1 → Not balanced!

We need 2 NaOH on right to match O and H from Ca(OH)₂. That gives us 2 Na on right → so we need 2 NaBr on left.

Balanced:
2 NaBr + Ca(OH)₂ → CaBr₂ + 2 NaOH

Type: Two compounds swap partners → double-displacement

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Reaction 2:
NH₃ + H₂SO₄ → (NH₄)₂SO₄

Left: N=1, H=3+2=5, S=1, O=4
Right: N=2, H=8, S=1, O=4 → Need more NH₃ and H?

Actually, (NH₄)₂SO₄ has 2 NH₄⁺ ions → needs 2 NH₃ and 2 H⁺ → but H₂SO₄ gives 2 H⁺.

So:
2 NH₃ + H₂SO₄ → (NH₄)₂SO₄

Check: Left: N=2, H=6+2=8, S=1, O=4 → Right: same → Balanced!

Type: Two things combine into one → synthesis

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Reaction 3:
C₅H₉O + O₂ → CO₂ + H₂O

This looks like combustion (hydrocarbon + oxygen → CO₂ + H₂O). But C₅H₉O isn’t a standard hydrocarbon — still, if it burns, it’s combustion.

Balance:

Carbon: 5 on left → 5 CO₂ on right
Hydrogen: 9 on left → need 4.5 H₂O on right? Wait, can’t have half molecules.

Multiply everything by 2 to avoid fractions:

Start with:
C₅H₉O + O₂ → 5 CO₂ + ? H₂O

H: 9 → need 4.5 H₂O → multiply entire equation by 2:

2 C₅H₉O + ? O₂ → 10 CO₂ + 9 H₂O

Now count O:

Left: 2×1 (from C₅H₉O) + 2×? (from O₂) = 2 + 2x
Right: 10×2 + 9×1 = 20 + 9 = 29

So: 2 + 2x = 29 → 2x = 27 → x = 13.5 → again fraction.

Wait — maybe I made a mistake. Let me try balancing properly.

Set coefficients:

a C₅H₉O + b O₂ → c CO₂ + d H₂O

C: 5a = c
H: 9a = 2d → d = 9a/2
O: a + 2b = 2c + d

Plug in c=5a, d=9a/2:

a + 2b = 2(5a) + 9a/2 = 10a + 4.5a = 14.5a
→ 2b = 13.5a → b = 6.75a

To eliminate decimals, set a=4:

Then c=20, d=18, b=27

So:
4 C₅H₉O + 27 O₂ → 20 CO₂ + 18 H₂O

Check O: Left: 4×1 + 27×2 = 4 + 54 = 58
Right: 20×2 + 18×1 = 40 + 18 = 58 → Good.

Type: Fuel + O₂ → CO₂ + H₂O → combustion

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Reaction 4:
Pb + H₃PO₄ → H₂ + Pb₃(PO₄)₂

Left: Pb=1, H=3, P=1, O=4
Right: H=2, Pb=3, P=2, O=8 → Not balanced.

Need 3 Pb on left → put 3 before Pb
Need 2 PO₄ on right → so need 2 H₃PO₄ on left? But that gives 6 H → need 3 H₂ on right.

Try:

3 Pb + 2 H₃PO₄ → 3 H₂ + Pb₃(PO₄)₂

Check:
Left: Pb=3, H=6, P=2, O=8
Right: H=6, Pb=3, P=2, O=8 → Balanced!

Type: Element replaces another in compound → Pb takes place of H in acid → single-displacement

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Reaction 5:
Li₃N + NH₄NO₃ → LiNO₃ + (NH₄)N

Left: Li=3, N=1+2=3, H=4, O=3
Right: Li=1, N=1+3=4, H=12, O=3 → Way off.

Notice: (NH₄)₃N means 3 NH₄ and 1 N → total N=4, H=12

Li₃N has 3 Li, 1 N
NH₄NO₃ has 2 N, 4 H, 3 O

Product: LiNO₃ has 1 Li, 1 N, 3 O
(NH₄)₃N has 3 N, 12 H, 1 N? Wait no — (NH₄)₃N is triammonium nitride? Actually, it's likely meant as ammonium nitride, which is unstable, but let’s balance as written.

Assume formula is correct.

Set:

a Li₃N + b NH₄NO₃ → c LiNO₃ + d (NH₄)₃N

Li: 3a = c
N: a + 2b = c + 4d [since (NH₄)₃N has 3N from NH₄ + 1N = 4N? Wait no — actually, (NH₄)₃N contains 3 nitrogen atoms from ammonium and 1 from nitride? No — chemical formula (NH₄)₃N implies 3 NH₄ groups and 1 N atom → total N atoms = 3 (from NH₄) + 1 (central N) = 4? Actually, no — in ionic terms, it’s 3 NH₄⁺ and N³⁻, so total N atoms = 3 + 1 = 4.

But let’s look at actual atoms:

In (NH₄)₃N: N from NH₄: 3 atoms, plus central N: 1 atom → total 4 N atoms? Actually, no — each NH₄ has one N, so 3 NH₄ = 3 N, and the “N” in (NH₄)₃N is an additional N? That doesn't make sense chemically. Probably it's a typo or misnomer. In reality, ammonium nitride is not stable, but for worksheet purposes, let’s assume (NH₄)₃N means 3 NH₄ and 1 N, so total N=4, H=12.

Similarly, NH₄NO₃ has 2 N (one in NH₄, one in NO₃), 4 H, 3 O.

Li₃N: 3 Li, 1 N

LiNO₃: 1 Li, 1 N, 3 O

So equations:

Li: 3a = c
H: 4b = 12d → b = 3d
N: a + 2b = c + 4d
O: 3b = 3c → b = c

From b = c and 3a = c → b = 3a
From b = 3d → 3a = 3d → a = d

Set a=1 → then d=1, b=3, c=3

Check N: left: a + 2b = 1 + 6 = 7
right: c + 4d = 3 + 4 = 7 → good

H: left: 4b=12, right: 12d=12 → good

O: left: 3b=9, right: 3c=9 → good

Li: 3a=3, c=3 → good

So:
Li₃N + 3 NH₄NO₃ → 3 LiNO₃ + (NH₄)₃N

Type: Looks like double displacement — ions swapping. Li pairs with NO₃, NH₄ pairs with N. So double-displacement

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Reaction 6:
HBr + Al(OH)₃ → H₂O + AlBr₃

Acid + base → water + salt → classic neutralization, which is double displacement.

Balance:

Al: 1 on both sides
Br: 1 on left, 3 on right → need 3 HBr
H: left: 3 (from HBr) + 3 (from Al(OH)₃) = 6 H → right: H₂O has 2H per molecule → need 3 H₂O
O: left: 3 from Al(OH)₃ → right: 3 from 3 H₂O → good

So:
3 HBr + Al(OH)₃ → 3 H₂O + AlBr₃

Type: double-displacement (also called neutralization)

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Now reactions 7 to 20 — they are already balanced, just identify type.

7. Na₃PO₄ + 3 KOH → 3 NaOH + K₃PO₄
Swap partners → double-displacement

8. MgCl₂ + Li₂CO₃ → MgCO₃ + 2 LiCl
Swap → double-displacement

9. C₆H₁₂ + 9 O₂ → 6 CO₂ + 6 H₂O
Hydrocarbon + O₂ → CO₂ + H₂O → combustion

10. Pb + FeSO₄ → PbSO₄ + Fe
Element replaces another → single-displacement

11. CaCO₃ → CaO + CO₂
One breaks into two → decomposition

12. P₄ + 3 O₂ → 2 P₂O₃
Two elements combine → synthesis

13. 2 RbNO₃ + BeF₂ → Be(NO₃)₂ + 2 RbF
Swap → double-displacement

14. 2 AgNO₃ + Cu → Cu(NO₃)₂ + 2 Ag
Cu replaces Ag → single-displacement

15. C₃H₆O + 4 O₂ → 3 CO₂ + 3 H₂O
Organic + O₂ → CO₂ + H₂O → combustion

16. 2 C₅H₅ + Fe → Fe(C₅H₅)₂
Two things combine → synthesis

17. SeCl₆ + O₂ → SeO₂ + 3 Cl₂
Not obvious — SeCl₆ loses Cl, gains O? Actually, this might be decomposition or replacement. But O₂ is reacting, and Cl₂ is produced. It’s not synthesis or decomposition. Possibly single displacement? Oxygen displaces chlorine? Unusual. Alternatively, it could be considered a redox reaction, but based on common types, since one element (O) is replacing part of compound, but it’s messy. Looking at products: SeO₂ and Cl₂ — so SeCl₆ is breaking down with help of O₂. Some might call it decomposition, but O₂ is reactant. Actually, this is likely single-displacement if we think O displaces Cl, but typically single displacement is metal replacing metal or H. Another way: it’s not fitting perfectly, but in many curricula, if an element reacts with a compound to form new compound and element, it’s single displacement. Here, O₂ (element) reacts with SeCl₆ to give SeO₂ and Cl₂ (element). So yes, single-displacement

Wait — O₂ is diatomic element, and it’s producing Cl₂ (element). So oxygen is displacing chlorine from selenium. So single-displacement

18. 2 MgI₂ + Mn(SO₃)₂ → 2 MgSO₃ + MnI₄
Swap ions → double-displacement

19. O₃ → O + O₂
One molecule breaks into two → decomposition

20. 2 NO₂ → 2 O₂ + N₂
One compound breaks into elements → decomposition

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Final Answers:

1. 2 NaBr + Ca(OH)₂ → CaBr₂ + 2 NaOH — double-displacement
2. 2 NH₃ + H₂SO₄ → (NH₄)₂SO₄ — synthesis
3. 4 C₅H₉O + 27 O₂ → 20 CO₂ + 18 H₂O — combustion
4. 3 Pb + 2 H₃PO₄ → 3 H₂ + Pb₃(PO₄)₂ — single-displacement
5. Li₃N + 3 NH₄NO₃ → 3 LiNO₃ + (NH₄)₃N — double-displacement
6. 3 HBr + Al(OH)₃ → 3 H₂O + AlBr₃ — double-displacement
7. double-displacement
8. double-displacement
9. combustion
10. single-displacement
11. decomposition
12. synthesis
13. double-displacement
14. single-displacement
15. combustion
16. synthesis
17. single-displacement
18. double-displacement
19. decomposition
20. decomposition