Chemical reactions worksheet featuring exercises on balancing equations and identifying reaction types.

Let's solve this step by step.

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A. Balance the following chemical reactions:

We will balance each equation by ensuring the number of atoms of each element is equal on both sides.

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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

Balance Br: 2 NaBr
Then Na becomes 2 → need 2 NaOH
Now H and O: 2 NaOH has 2 H and 2 O → matches left side (Ca(OH)₂ has 2 O and 2 H)

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

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2) N₂ + H₂ → NH₃

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

Balance N: 2 NH₃ → now N=2, H=6
So H₂ must be 3 → 3 H₂

✔ Balanced:
N₂ + 3 H₂ → 2 NH₃

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3) NaCl + F₂ → NaF + Cl₂

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

Need 2 NaF → then F=2, Na=2 → so 2 NaCl
Then Cl=2 → Cl₂ is correct

✔ Balanced:
2 NaCl + F₂ → 2 NaF + Cl₂

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4) Pb(OH)₂ + HCl → PbCl₂ + H₂O

- Left: Pb=1, O=2, H=2+1=3, Cl=1
- Right: Pb=1, Cl=2, H=2, O=1 → not balanced

PbCl₂ needs 2 Cl → so 2 HCl
Now H = 2 from HCl + 2 from OH → total H = 4? Wait:

Pb(OH)₂ has 2 H and 2 O
2 HCl has 2 H and 2 Cl
Total left: Pb=1, O=2, H=4, Cl=2

Right: PbCl₂ → Pb=1, Cl=2; H₂O → H=2, O=1 → only one water?

Need 2 H₂O → H=4, O=2

✔ Balanced:
Pb(OH)₂ + 2 HCl → PbCl₂ + 2 H₂O

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5) CH₄ + O₂ → CO₂ + H₂O

- Left: C=1, H=4, O=2
- Right: C=1, O=2+1=3, H=2 → not balanced

Balance H: 2 H₂O → H=4, O=2 from H₂O + 2 from CO₂ = 4 O → so O₂ must be 2

✔ Balanced:
CH₄ + 2 O₂ → CO₂ + 2 H₂O

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6) H₂SO₄ + B(OH)₃ → B₂(SO₄)₃ + H₂O

- Right: B=2, S=3, O=12 from SO₄ + 3 from B(OH)₃? Let’s count carefully.

B₂(SO₄)₃ → B=2, S=3, O=12
H₂O → H=2, O=1

Left: H₂SO₄ → H=2, S=1, O=4
B(OH)₃ → B=1, O=3, H=3

We need 2 B → so 2 B(OH)₃
We need 3 S → so 3 H₂SO₄

Now left:
- H: 3×2 = 6 from H₂SO₄, 2×3 = 6 from B(OH)₃ → total H=12
- O: 3×4 = 12 from H₂SO₄, 2×3 = 6 from B(OH)₃ → total O=18
- B=2, S=3

Right: B₂(SO₄)₃ → B=2, S=3, O=12
Water: need to account for remaining H and O

H: 12 → so 6 H₂O
O: 6 H₂O → 6 O, plus 12 from sulfate → 18 O → matches!

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

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7) C₅H₉O + O₂ → CO₂ + H₂O

This is combustion. General form: hydrocarbon + O₂ → CO₂ + H₂O

C₅H₉O → 5 C, 9 H, 1 O

Products: CO₂ and H₂O

- 5 CO₂ → 5 C, 10 O
- 9/2 H₂O → 4.5 H₂O → better use whole numbers

Multiply entire equation by 2 to avoid fractions.

Start with:
C₅H₉O + O₂ → 5 CO₂ + (9/2) H₂O

Multiply by 2:
2 C₅H₉O + ? O₂ → 10 CO₂ + 9 H₂O

Now count O atoms:

Left: 2 C₅H₉O → 2 O
Right: 10 CO₂ → 20 O, 9 H₂O → 9 O → total 29 O

So O₂ must supply 29 - 2 = 27 O → so 27/2 O₂ → multiply all by 2 again?

Wait, better:

From above:
2 C₅H₉O → provides 2 O
Need 29 O on right → so O₂ must provide 27 O → 27/2 O₂

So:
2 C₅H₉O + 27/2 O₂ → 10 CO₂ + 9 H₂O

Multiply entire equation by 2:

4 C₅H₉O + 27 O₂ → 20 CO₂ + 18 H₂O

✔ Balanced.

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8) Li₃N + NH₄NO₃ → LiNO₃ + (NH₄)₃N

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

Right: LiNO₃ → Li=1, N=1, O=3
(NH₄)₃N → N=4 (3 from NH₄, 1 central), H=12

So total right: Li=1, N=5, H=12, O=3

Left: Li=3, N=1 (from Li₃N) + 2 (from NH₄NO₃) = 3 N, H=4, O=3

Not matching.

Try balancing:

We want LiNO₃ and (NH₄)₃N

Let’s suppose we have 3 LiNO₃ → then Li=3 → so 1 Li₃N
Then N: 3 from LiNO₃ → but (NH₄)₃N has 4 N → total N needed = 3 + 4 = 7

But left: Li₃N → 1 N, NH₄NO₃ → 2 N → total 3 N per NH₄NO₃

So need more NH₄NO₃.

Let’s try:

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

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

Now from b = c and 3a = c → 3a = b
Also b = 3d → d = b/3

Now plug into N balance:

a + 2b = b + 4(b/3)
a + 2b = b + 4b/3
a = -b + 4b/3 = (-3b + 4b)/3 = b/3

But earlier: 3a = b → a = b/3 → consistent!

So let b = 3 → then a = 1, c = 3, d = 1

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

Check atoms:

Left:
- Li: 3
- N: 1 (Li₃N) + 3×2 = 6 → total 7
- H: 3×4 = 12
- O: 3×3 = 9

Right:
- Li: 3
- N: 3 (LiNO₃) + 4 ((NH₄)₃N) = 7
- H: 12
- O: 3×3 = 9

✔ Balanced:
Li₃N + 3 NH₄NO₃ → 3 LiNO₃ + (NH₄)₃N

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9) HBr + Al(OH)₃ → AlBr₃ + H₂O

Acid-base reaction.

Al(OH)₃ has 3 OH → needs 3 H⁺ → so 3 HBr

Then AlBr₃ → Al=1, Br=3 → so 3 HBr

H₂O: 3 HBr → 3 H, 3 OH from Al(OH)₃ → makes 3 H₂O

✔ Balanced:
3 HBr + Al(OH)₃ → AlBr₃ + 3 H₂O

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10) Pb + H₃PO₄ → Pb₃(PO₄)₂ + H₂

Pb₃(PO₄)₂ → Pb=3, PO₄=2 → so P=2, O=8

H₃PO₄ → H=3, P=1, O=4

So need 2 H₃PO₄ → P=2, H=6, O=8

Pb: need 3 Pb

H₂: H=6 → so 3 H₂

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

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B. Identify the type of reaction

Types:
- Synthesis: A + B → AB
- Decomposition: AB → A + B
- Single-replacement: A + BC → AC + B
- Double-replacement: AB + CD → AD + CB
- Combustion: fuel + O₂ → CO₂ + H₂O

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1) Na₃PO₄ + 3 KOH → 3 NaOH + K₃PO₄

Two compounds exchanging ions → double replacement
✔ Double-replacement

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2) Pb + FeSO₄ → PbSO₄ + Fe

Pb replaces Fe → single replacement
✔ Single-replacement

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3) 2 BF₃ + 3 H₂O → B₂O₃ + 6 HF

BF₃ reacts with water to form oxide and acid → appears like a hydrolysis, but no element replaced.
Actually, it's a double-replacement or decomposition-like, but more accurately, it's a combination of two reactants forming two products — not typical synthesis.

But notice: BF₃ is a Lewis acid, reacting with water → not standard types.

However, since it involves exchange of groups, and forms HF and B₂O₃, it can be seen as a double-replacement or hydrolysis.

But in basic classification: Double-replacement is best fit.

Alternatively, some might say decomposition, but both reactants are consumed.

Actually, it's not a simple category — but in many curricula, such reactions are considered double-replacement.

✔ Double-replacement

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4) 2 Al + 6 HCl → 2 AlCl₃ + 3 H₂

Aluminum replaces hydrogen → single-replacement

✔ Single-replacement

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5) 2 Fe + O₂ + 2 H₂O → 2 Fe(OH)₃

Iron reacts with oxygen and water to form rust → oxidation, but also includes combination.

Fe and O₂ and H₂O combine to form Fe(OH)₃ → synthesis (multiple reactants forming one product)

Even though three reactants, it's a synthesis reaction.

✔ Synthesis

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✔ Final Answers:

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#### A. Balanced Equations:

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

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#### B. Reaction Types:

1) Double-replacement
2) Single-replacement
3) Double-replacement
4) Single-replacement
5) Synthesis

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