Types of Chemical Reactions Worksheets - Free Printable - Free Printable

Let’s solve each part step by step.

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Part A: Balance the chemical reactions

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

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

Try putting 2 in front of NaBr and 2 in front of NaOH:

→ 2NaBr + Ca(OH)₂ → CaBr₂ + 2NaOH

Check:
Left: Na=2, Br=2, Ca=1, O=2, H=2
Right: Ca=1, Br=2, Na=2, O=2, H=2 ✔ Balanced!

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

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2) N₂ + H₂ → NH₃

Left: N=2, H=2
Right: N=1, H=3

Need even H on right → try 2NH₃ → N=2, H=6
Then left needs 3H₂ → H=6

→ N₂ + 3H₂ → 2NH₃ ✔

Answer: N₂ + 3H₂ → 2NH₃

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3) NaCl + F₂ → NaF + Cl₂

Left: Na=1, Cl=1, F=2
Right: Na=1, F=1, Cl=2 → Cl not balanced

Put 2NaCl and 2NaF:

→ 2NaCl + F₂ → 2NaF + Cl₂

Check:
Left: Na=2, Cl=2, F=2
Right: Na=2, F=2, Cl=2 ✔

Answer: 2NaCl + F₂ → 2NaF + Cl₂

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4) Pb(OH)₂ + HCl → PbCl₂ + H₂O

Left: Pb=1, O=2, H=2+1=3? Wait — Pb(OH)₂ has 2O and 2H, plus HCl has 1H and 1Cl → total H=3, Cl=1
Right: Pb=1, Cl=2, H=2, O=1 → not balanced

Try 2HCl:

→ Pb(OH)₂ + 2HCl → PbCl₂ + 2H₂O

Check:
Left: Pb=1, O=2, H=2+2=4, Cl=2
Right: Pb=1, Cl=2, H=4, O=2 ✔

Answer: Pb(OH)₂ + 2HCl → PbCl₂ + 2H₂O

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5) CH₄ + O₂ → CO₂ + H₂O

Left: C=1, H=4, O=2
Right: C=1, H=2, O=3 → H and O off

Make H₂O have 4H → 2H₂O → then O on right = 2 (from CO₂) + 2 (from 2H₂O) = 4 → so need 2O₂ on left

→ CH₄ + 2O₂ → CO₂ + 2H₂O ✔

Answer: CH₄ + 2O₂ → CO₂ + 2H₂O

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6) H₂SO₄ + B(OH)₃ → B₂(SO₄)₃ + H₂O

Right has B₂ → need 2B(OH)₃ on left
Right has SO₄×3 → need 3H₂SO₄ on left

Try: 3H₂SO₄ + 2B(OH)₃ → B₂(SO₄)₃ + ?

Now count H and O for water.

Left H: from 3H₂SO₄ → 6H; from 2B(OH)₃ → 6H → total H=12
Left O: 3×4=12 from sulfate, 2×3=6 from hydroxide → 18 O

Right: B₂(SO₄)₃ has 12 O (each SO₄ has 4O ×3), no H yet → so all H must go to H₂O → 12H → 6H₂O → which adds 6O → total O on right = 12 + 6 = 18 ✔

So: 3H₂SO₄ + 2B(OH)₃ → B₂(SO₄)₃ + 6H₂O

Answer: 3H₂SO₄ + 2B(OH)₃ → B₂(SO₄)₃ + 6H₂O

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7) C₅H₉O + O₂ → CO₂ + H₂O

This looks like combustion. Let’s balance C first → 5CO₂
H: 9H → need 9/2 H₂O → better multiply everything by 2 later.

Try: C₅H₉O + ?O₂ → 5CO₂ + 4.5H₂O → not whole numbers.

Multiply entire equation by 2:

2C₅H₉O + ?O₂ → 10CO₂ + 9H₂O

Now count O:

Left: 2C₅H₉O has 2O, plus ?O₂ → let’s say x O₂ → 2x O atoms → total O = 2 + 2x
Right: 10CO₂ → 20O, 9H₂O → 9O → total 29O

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

Wait — maybe I made a mistake. Original molecule is C₅H₉O — that’s odd H count. Maybe it's correct.

Actually, standard way: set coefficients.

Let: a C₅H₉O + b O₂ → c CO₂ + d H₂O

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

Substitute c=5a, d=9a/2:

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

So: 2b = 14.5a - a = 13.5a → b = 6.75a

To eliminate decimals, let a=4:

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

Check O: left: 4 (from C₅H₉O) + 54 (from 27O₂) = 58
Right: 20×2=40 from CO₂, 18×1=18 from H₂O → 58 ✔

So: 4C₅H₉O + 27O₂ → 20CO₂ + 18H₂O

But this seems big — maybe the formula is wrong? Or perhaps it’s acceptable.

Alternatively, if we assume it’s C₅H₁₀O or something, but as written, we’ll go with this.

Actually, let me double-check common compounds — maybe it’s cyclopentanol or something, but for balancing, we proceed.

Answer: 4C₅H₉O + 27O₂ → 20CO₂ + 18H₂O

*(Note: This is unusual, but mathematically correct based on given formula.)*

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8) Li₃N + NH₄NO₃ → LiNO₃ + (NH₄)₃N

Left: Li=3, N=1+2=3? Wait — NH₄NO₃ has two N: one in NH₄, one in NO₃ → so N=1 (from Li₃N) + 2 (from NH₄NO₃) = 3N
H=4, O=3

Right: LiNO₃ → Li=1, N=1, O=3; (NH₄)₃N → N=3+1=4? Wait — (NH₄)₃N means 3 NH₄ groups and one N³⁻ → so N total = 3 (from NH₄) + 1 (central N) = 4N, H=12

Not matching.

Perhaps typo? Common reaction might be different, but let’s balance as is.

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

Li: 3a = c
N: a + 2b = c + 4d [since (NH₄)₃N has 4N]
H: 4b = 12d → b = 3d
O: 3b = 3c → b = c

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

Set a=1 → d=1, b=3, c=3

Check N: left: 1 + 2*3 = 7; right: 3 + 4*1 = 7 ✔
H: left: 4*3=12; right: 12*1=12 ✔
O: left: 3*3=9; right: 3*3=9 ✔
Li: 3=3 ✔

So: Li₃N + 3NH₄NO₃ → 3LiNO₃ + (NH₄)₃N

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

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9) HBr + Al(OH)₃ → AlBr₃ + H₂O

Left: H=1+3=4? HBr has 1H, Al(OH)₃ has 3H → total H=4, Br=1, Al=1, O=3
Right: Al=1, Br=3, H=2, O=1 → not balanced

Need 3HBr for Br, and 3H₂O for H and O.

Try: 3HBr + Al(OH)₃ → AlBr₃ + 3H₂O

Check:
Left: H=3+3=6, Br=3, Al=1, O=3
Right: Al=1, Br=3, H=6, O=3 ✔

Answer: 3HBr + Al(OH)₃ → AlBr₃ + 3H₂O

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10) Pb + H₃PO₄ → Pb(PO₄)₂ + H₂

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

Need 3Pb on left → 3Pb
Need 2H₃PO₄ for P and O → 2H₃PO₄ → H=6, P=2, O=8

Right: Pb₃(PO₄)₂ has Pb=3, P=2, O=8 → good
H₂: need 3H₂ to get 6H

So: 3Pb + 2H₃PO₄ → Pb₃(PO₄)₂ + 3H₂

Check:
Left: Pb=3, H=6, P=2, O=8
Right: Pb=3, P=2, O=8, H=6 ✔

Answer: 3Pb + 2H₃PO₄ → Pb₃(PO₄)₂ + 3H₂

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Part B: Identify reaction type

Types: synthesis, decomposition, single-replacement, double-replacement, combustion

1) Na₃PO₄ + 3KOH → 3NaOH + K₃PO₄

Two compounds swap partners: Na with K → double replacement ✔

Answer: double-replacement

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2) Pb + FeSO₄ → PbSO₄ + Fe

One element replaces another in compound → single replacement ✔

Answer: single-replacement

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3) 2BF₃ + 3H₂O → B₂O₃ + 6HF

Two compounds react to form two new compounds — looks like double replacement? But BF₃ and H₂O forming oxide and acid — actually, it’s a hydrolysis, but in basic classification, since no element is replacing, and not synthesis/decomposition, likely double replacement.

In double replacement, ions swap: here B and H are swapping with F and OH? Not exactly, but often classified as such.

Alternatively, some might call it acid-base, but per options, double-replacement fits best.

Answer: double-replacement

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4) 2Al + 6HCl → 2AlCl₃ + 3H₂

Element + compound → new compound + element → single replacement ✔

Answer: single-replacement

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5) 2Fe + O₂ + 2H₂O → 2Fe(OH)₃

Multiple reactants forming one product? Actually, 3 reactants → one product type, but it’s rusting — combination of elements and compound to form a compound. Technically, it’s a synthesis (combination) reaction, though complex.

Synthesis: A + B → AB, but here multiple inputs → still considered synthesis if forming one main product.

Some might argue it’s not pure synthesis, but among choices, synthesis is closest.

Combustion usually involves fuel + O₂ → CO₂ + H₂O — not here.

So: synthesis

Answer: synthesis

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Final Answer:

Part A:
1) 2NaBr + Ca(OH)₂ → CaBr₂ + 2NaOH
2) N₂ + 3H₂ → 2NH₃
3) 2NaCl + F₂ → 2NaF + Cl₂
4) Pb(OH)₂ + 2HCl → PbCl₂ + 2H₂O
5) CH₄ + 2O₂ → CO₂ + 2H₂O
6) 3H₂SO₄ + 2B(OH)₃ → B₂(SO₄)₃ + 6H₂O
7) 4C₅H₉O + 27O₂ → 20CO₂ + 18H₂O
8) Li₃N + 3NH₄NO₃ → 3LiNO₃ + (NH₄)₃N
9) 3HBr + Al(OH)₃ → AlBr₃ + 3H₂O
10) 3Pb + 2H₃PO₄ → Pb₃(PO₄)₂ + 3H₂

Part B:
1) double-replacement
2) single-replacement
3) double-replacement
4) single-replacement
5) synthesis