Chemistry worksheet featuring a series of unbalanced double replacement reactions for students to solve.

Explanation:
We are balancing double replacement (also called double displacement) reactions. In these reactions, two compounds swap ions to form two new compounds. The general pattern is:

AB + CD → AD + CB

To balance, we need the same number of each atom on both sides.

Let’s go one by one and balance carefully.

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35. AlI₃ + HgCl₂ → AlCl₃ + HgI₂ (ppt)

Check ions:
- Al³⁺ pairs with I⁻ → AlI₃
- Hg²⁺ pairs with Cl⁻ → HgCl₂
Swap: Al³⁺ with Cl⁻ → AlCl₃; Hg²⁺ with I⁻ → HgI₂

Now balance:
Left: Al=1, I=3, Hg=1, Cl=2
Right: Al=1, Cl=3, Hg=1, I=2 → Not balanced.

Find LCM for I and Cl:
- I: 3 on left, 2 on right → need 6 I total → multiply AlI₃ by 2 → 2 AlI₃ → 6 I
- Then HgI₂ needs 3 × HgI₂ → 6 I
So: 2 AlI₃ + ? HgCl₂ → 2 AlCl₃ + 3 HgI₂

Now check Hg: right has 3 Hg → need 3 HgCl₂ on left
Cl: 3 HgCl₂ gives 6 Cl → right: 2 AlCl₃ has 6 Cl ✔
Al: 2 on left, 2 on right ✔
I: 2×3 = 6 left, 3×2 = 6 right ✔
Hg: 3 left, 3 right ✔

→ 2 AlI₃ + 3 HgCl₂ → 2 AlCl₃ + 3 HgI₂

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36. HCl + NaOH → NaCl + H₂O
Already balanced:
H: 1+1 = 2 → right: 2 in H₂O
Cl: 1 → 1
Na: 1 → 1
O: 1 → 1
→ 1 HCl + 1 NaOH → 1 NaCl + 1 H₂O

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37. BaCl₂ + H₂SO₄ → BaSO₄ + HCl
Ba: 1 both sides
Cl: 2 left → need 2 HCl
H: 2 left → 2 in 2 HCl ✔
SO₄: 1 both sides
→ 1 BaCl₂ + 1 H₂SO₄ → 1 BaSO₄ + 2 HCl

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38. Al₂(SO₄)₃ + Ca(OH)₂ → Al(OH)₃ + CaSO₄
Left: Al=2, S=3, O=12 from SO₄ + O from OH? Wait—better count by ions.

Al₂(SO₄)₃ has: 2 Al³⁺, 3 SO₄²⁻
Ca(OH)₂ has: 1 Ca²⁺, 2 OH⁻

Products: Al(OH)₃ (Al³⁺ + 3 OH⁻), CaSO₄ (Ca²⁺ + SO₄²⁻)

Need to match charges and counts.

Let’s try:
Al₂(SO₄)₃ → gives 2 Al³⁺ and 3 SO₄²⁻
To make Al(OH)₃, each Al³⁺ needs 3 OH⁻ → 2 Al need 6 OH⁻
So need 3 Ca(OH)₂ (each gives 2 OH⁻ → 3×2 = 6 OH⁻)
Also, 3 Ca(OH)₂ gives 3 Ca²⁺ → can pair with 3 SO₄²⁻ → 3 CaSO₄
So:

Al₂(SO₄)₃ + 3 Ca(OH)₂ → 2 Al(OH)₃ + 3 CaSO₄

Check:
Al: 2 = 2
S: 3 = 3
O: left: SO₄ has 4×3=12 O, OH has 3×2=6 O → total 18 O
Right: Al(OH)₃: 2×3=6 OH → 6 O; CaSO₄: 3×4=12 O → total 18 O ✔
H: left: 3×2=6 H; right: 2×3=6 H ✔
Ca: 3 = 3
→ Balanced.

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39. AgNO₃ + K₃PO₄ → Ag₃PO₄ + KNO₃
Ag⁺ and NO₃⁻ from first; K⁺ and PO₄³⁻ from second.

PO₄³⁻ needs 3 Ag⁺ → Ag₃PO₄
So need 3 AgNO₃
That gives 3 NO₃⁻ → need 3 KNO₃
K₃PO₄ gives 3 K⁺ → matches 3 KNO₃

→ 3 AgNO₃ + 1 K₃PO₄ → 1 Ag₃PO₄ + 3 KNO₃

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40. CuBr₂ + AlCl₃ → CuCl₂ + AlBr₃
Cu²⁺, Br⁻; Al³⁺, Cl⁻
Swap: Cu²⁺ + Cl⁻ → CuCl₂; Al³⁺ + Br⁻ → AlBr₃

LCM of charges: need 3 Cu²⁺ (6+) and 2 Al³⁺ (6+) to balance total charge.

Try: 3 CuBr₂ + 2 AlCl₃ → 3 CuCl₂ + 2 AlBr₃

Check:
Cu: 3 = 3
Br: 3×2 = 6 → 2×3 = 6 ✔
Al: 2 = 2
Cl: 2×3 = 6 → 3×2 = 6 ✔
→ Balanced.

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41. Ca(C₂H₃O₂)₂ + Na₂CO₃ → CaCO₃ + NaC₂H₃O₂
Note: C₂H₃O₂ is acetate, often written as CH₃COO⁻, charge −1.

Ca(C₂H₃O₂)₂: Ca²⁺ + 2 acetate
Na₂CO₃: 2 Na⁺ + CO₃²⁻
Products: CaCO₃ (Ca²⁺ + CO₃²⁻), NaC₂H₃O₂ (Na⁺ + acetate)

So 1 Ca(C₂H₃O₂)₂ + 1 Na₂CO₃ → 1 CaCO₃ + 2 NaC₂H₃O₂

Check:
Ca: 1 = 1
C₂H₃O₂: 2 left → 2 right
Na: 2 left → 2 right
CO₃: 1 both sides
→ Balanced.

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42. NH₄Cl + Hg₂(C₂H₃O₂)₂ → NH₄C₂H₃O₂ + Hg₂Cl₂
Note: Hg₂²⁺ is a *diatomic ion* (two Hg⁺ bonded), so Hg₂(C₂H₃O₂)₂ has Hg₂²⁺ and 2 acetate.

NH₄⁺ + Cl⁻ → NH₄Cl
Swap: NH₄⁺ with acetate → NH₄C₂H₃O₂
Hg₂²⁺ with Cl⁻ → Hg₂Cl₂ (since Hg₂²⁺ needs 2 Cl⁻)

So: 2 NH₄Cl + 1 Hg₂(C₂H₃O₂)₂ → 2 NH₄C₂H₃O₂ + 1 Hg₂Cl₂

Check:
NH₄: 2 left → 2 right
Cl: 2 left → 2 in Hg₂Cl₂ ✔
Hg₂: 1 both sides
C₂H₃O₂: 2 left → 2 right
→ Balanced.

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43. Ca(NO₃)₂ + HCl → CaCl₂ + HNO₃
Ca²⁺, 2 NO₃⁻; H⁺, Cl⁻
Products: CaCl₂ (Ca²⁺ + 2 Cl⁻), HNO₃ (H⁺ + NO₃⁻)

Need 2 HNO₃ to use 2 NO₃⁻ → need 2 HCl
Then CaCl₂ uses 2 Cl⁻ ✔

→ 1 Ca(NO₃)₂ + 2 HCl → 1 CaCl₂ + 2 HNO₃

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44. FeS + HCl → FeCl₂ + H₂S
Fe²⁺, S²⁻; H⁺, Cl⁻
Products: FeCl₂ (Fe²⁺ + 2 Cl⁻), H₂S (2 H⁺ + S²⁻)

So need 2 HCl
→ 1 FeS + 2 HCl → 1 FeCl₂ + 1 H₂S

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45. Cu(OH)₂ + HC₂H₃O₂ → Cu(C₂H₃O₂)₂ + H₂O
Cu²⁺, 2 OH⁻; H⁺, C₂H₃O₂⁻
Product: Cu(C₂H₃O₂)₂ needs 2 acetate; H₂O from 2 H⁺ + 2 OH⁻

So need 2 HC₂H₃O₂
→ 1 Cu(OH)₂ + 2 HC₂H₃O₂ → 1 Cu(C₂H₃O₂)₂ + 2 H₂O

Check:
Cu:1=1
O: left: OH has 2 O, acetate has 2×2=4 O? Wait — better atom count:

Cu(OH)₂: Cu, 2O, 2H
2 HC₂H₃O₂: 2H, 4C, 6H, 4O → total left: Cu, 4C, 8H, 6O
Right: Cu(C₂H₃O₂)₂ = Cu + 4C + 6H + 4O
2 H₂O = 4H + 2O
Total right: Cu, 4C, 10H? Wait — mistake.

Actually, HC₂H₃O₂ = CH₃COOH = C₂H₄O₂ (common formula). Yes! Acetic acid is C₂H₄O₂, not C₂H₃O₂ (the ion is C₂H₃O₂⁻, but acid is neutral: adds one H).

So HC₂H₃O₂ = H⁺ + C₂H₃O₂⁻ → molecular formula: C₂H₄O₂.

Thus:
Cu(OH)₂ = Cu, 2O, 2H
2 HC₂H₄O₂ = 4C, 8H, 4O
Total left: Cu, 4C, 10H, 6O

Right:
Cu(C₂H₃O₂)₂ = Cu + 4C + 6H + 4O (since two acetate ions: each C₂H₃O₂ has 2C,3H,2O)
2 H₂O = 4H, 2O
Total right: Cu, 4C, 10H, 6O ✔

So coefficients: 1, 2, 1, 2

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46. Ca(OH)₂ + H₃PO₄ → Ca₃(PO₄)₂ + H₂O
Phosphate is PO₄³⁻, Ca²⁺.

Ca₃(PO₄)₂ requires 3 Ca²⁺ and 2 PO₄³⁻
So need 3 Ca(OH)₂ and 2 H₃PO₄

Left:
Ca: 3
OH: 3×2 = 6 OH → 6 H and 6 O
H₃PO₄ ×2: 6 H, 2 P, 8 O

Total left: Ca=3, P=2, H=6+6=12, O=6+8=14

Right:
Ca₃(PO₄)₂: Ca=3, P=2, O=8
H₂O: need enough to use 12 H and remaining O: 14−8=6 O → 6 H₂O has 12 H and 6 O ✔

So: 3 Ca(OH)₂ + 2 H₃PO₄ → 1 Ca₃(PO₄)₂ + 6 H₂O

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47. CaBr₂ + KOH → Ca(OH)₂ + KBr
Ca²⁺, 2 Br⁻; K⁺, OH⁻
Products: Ca(OH)₂ (Ca²⁺ + 2 OH⁻), KBr (K⁺ + Br⁻)

Need 2 KOH to give 2 OH⁻ and 2 K⁺
Then 2 KBr
CaBr₂ gives 2 Br⁻ → matches 2 KBr

→ 1 CaBr₂ + 2 KOH → 1 Ca(OH)₂ + 2 KBr

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Now compile all final coefficients:

35. 2, 3, 2, 3
36. 1, 1, 1, 1
37. 1, 1, 1, 2
38. 1, 3, 2, 3
39. 3, 1, 1, 3
40. 3, 2, 3, 2
41. 1, 1, 1, 2
42. 2, 1, 2, 1
43. 1, 2, 1, 2
44. 1, 2, 1, 1
45. 1, 2, 1, 2
46. 3, 2, 1, 6
47. 1, 2, 1, 2

Final Answer:
35. 2, 3, 2, 3
36. 1, 1, 1, 1
37. 1, 1, 1, 2
38. 1, 3, 2, 3
39. 3, 1, 1, 3
40. 3, 2, 3, 2
41. 1, 1, 1, 2
42. 2, 1, 2, 1
43. 1, 2, 1, 2
44. 1, 2, 1, 1
45. 1, 2, 1, 2
46. 3, 2, 1, 6
47. 1, 2, 1, 2