w326 Types of Reactions Worksheet | PDF - Free Printable

Let’s solve each problem one by one. We’ll balance the chemical equations and identify the reaction type for each.

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Problem 2:
__ FePO₄ + __ BaCl₂ → __ Ba₃(PO₄)₂(s) + __ FeCl₃

Step 1: Look at phosphate (PO₄). On the right, Ba₃(PO₄)₂ has 2 PO₄ groups. So we need 2 FePO₄ on the left.

→ 2 FePO₄ + __ BaCl₂ → __ Ba₃(PO₄)₂ + __ FeCl₃

Step 2: Now iron (Fe): 2 Fe on left → need 2 FeCl₃ on right.

→ 2 FePO₄ + __ BaCl₂ → __ Ba₃(PO₄)₂ + 2 FeCl₃

Step 3: Barium (Ba): Right side has 3 Ba in Ba₃(PO₄)₂ → need 3 BaCl₂ on left.

→ 2 FePO₄ + 3 BaCl₂ → 1 Ba₃(PO₄)₂ + 2 FeCl₃

Step 4: Check chlorine (Cl): Left = 3 × 2 = 6 Cl; Right = 2 × 3 = 6 Cl → balanced!

Balanced equation:
2 FePO₄ + 3 BaCl₂ → Ba₃(PO₄)₂ + 2 FeCl₃

Reaction type: Two compounds swap partners → Double Replacement

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Problem 3:
__ Al(OH)₃ + __ H₂SO₄ → __ Al₂(SO₄)₃(aq) + __ H₂O

Step 1: Aluminum (Al): Right has 2 Al → need 2 Al(OH)₃ on left.

→ 2 Al(OH)₃ + __ H₂SO₄ → __ Al₂(SO₄)₃ + __ H₂O

Step 2: Sulfate (SO₄): Right has 3 SO₄ → need 3 H₂SO₄ on left.

→ 2 Al(OH)₃ + 3 H₂SO₄ → 1 Al₂(SO₄)₃ + __ H₂O

Step 3: Hydrogen and oxygen — count H atoms:

Left: From 2 Al(OH)₃ → 2×3 = 6 H; from 3 H₂SO₄ → 3×2 = 6 H → total 12 H

Right: Water (H₂O) has 2 H per molecule → need 6 H₂O to get 12 H.

Check oxygen too? Let’s verify:

Left O: 2×3 (from OH) + 3×4 (from SO₄) = 6 + 12 = 18 O
Right O: 3×4 (in sulfate) + 6×1 (in water) = 12 + 6 = 18 O → good!

Balanced equation:
2 Al(OH)₃ + 3 H₂SO₄ → Al₂(SO₄)₃ + 6 H₂O

Reaction type: Acid + Base → Salt + Water → Neutralization (a type of Double Replacement)

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Problem 4:
__ HCl + __ FeCO₃ → __ CO₂ + __ H₂O + __ FeCl₂

Step 1: Iron (Fe): 1 on each side → okay.

Step 2: Carbonate (CO₃): becomes CO₂ → so 1 FeCO₃ gives 1 CO₂.

Step 3: Chlorine: FeCl₂ has 2 Cl → need 2 HCl on left.

→ 2 HCl + 1 FeCO₃ → 1 CO₂ + __ H₂O + 1 FeCl₂

Step 4: Hydrogen: 2 H from HCl → need 1 H₂O (which has 2 H)

Check oxygen: Left: 3 O from FeCO₃; Right: 2 from CO₂ + 1 from H₂O = 3 → good!

Balanced equation:
2 HCl + FeCO₃ → CO₂ + H₂O + FeCl₂

Reaction type: Acid + Carbonate → Gas + Water + Salt → Gas Evolution (also a type of Double Replacement)

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Problem 5:
__ AgNO₃ + __ Na₂CO₃ → __ Ag₂CO₃ + __ NaNO₃

Step 1: Silver (Ag): Right has 2 Ag → need 2 AgNO₃ on left.

→ 2 AgNO₃ + __ Na₂CO₃ → __ Ag₂CO₃ + __ NaNO₃

Step 2: Sodium (Na): Left has 2 Na → need 2 NaNO₃ on right.

→ 2 AgNO₃ + 1 Na₂CO₃ → 1 Ag₂CO₃ + 2 NaNO₃

Step 3: Check nitrate (NO₃): Left = 2, Right = 2 → good.

Carbonate: 1 on each side → good.

Balanced equation:
2 AgNO₃ + Na₂CO₃ → Ag₂CO₃ + 2 NaNO₃

Reaction type: Two compounds swap ions → Double Replacement

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Problem 6:
__ Rb₂O + heat → __ Rb + __ O₂

This is decomposition with heat.

Step 1: Oxygen: Left has 1 O, right has 2 O → need 2 Rb₂O on left.

→ 2 Rb₂O → __ Rb + __ O₂

Step 2: Rubidium: Left has 4 Rb → need 4 Rb on right.

→ 2 Rb₂O → 4 Rb + 1 O₂

Check: O: 2 on left, 2 on right → good.

Balanced equation:
2 Rb₂O → 4 Rb + O₂

Reaction type: One compound breaks into elements → Decomposition

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Problem 7:
__ Rb + __ HCl → __ H₂ + __ RbCl

Step 1: Hydrogen: H₂ has 2 H → need 2 HCl on left.

→ __ Rb + 2 HCl → 1 H₂ + __ RbCl

Step 2: Chlorine: 2 Cl on left → need 2 RbCl on right.

→ __ Rb + 2 HCl → 1 H₂ + 2 RbCl

Step 3: Rubidium: 2 Rb needed on left.

→ 2 Rb + 2 HCl → H₂ + 2 RbCl

Balanced equation:
2 Rb + 2 HCl → H₂ + 2 RbCl

Reaction type: Element replaces another in compound → Single Replacement

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Problem 8:
__ Cu(OH)₂ → __ H₂O + __ CuO

Step 1: Copper: 1 on each side → okay.

Step 2: Oxygen: Left: 2 O from OH + ? Wait — Cu(OH)₂ has 2 O and 2 H from hydroxide, plus Cu.

Actually: Cu(OH)₂ → CuO + H₂O

Count atoms:

Left: Cu=1, O=2, H=2
Right: Cu=1, O=1 (CuO) + 1 (H₂O) = 2, H=2 → already balanced!

Balanced equation:
Cu(OH)₂ → H₂O + CuO

Reaction type: One compound breaks down → Decomposition

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Final Answer:

2) 2 FePO₄ + 3 BaCl₂ → Ba₃(PO₄)₂ + 2 FeCl₃ — Double Replacement
3) 2 Al(OH)₃ + 3 H₂SO₄ → Al₂(SO₄)₃ + 6 H₂O — Neutralization (Double Replacement)
4) 2 HCl + FeCO₃ → CO₂ + H₂O + FeCl₂ — Gas Evolution (Double Replacement)
5) 2 AgNO₃ + Na₂CO₃ → Ag₂CO₃ + 2 NaNO₃ — Double Replacement
6) 2 Rb₂O → 4 Rb + O₂ — Decomposition
7) 2 Rb + 2 HCl → H₂ + 2 RbCl — Single Replacement
8) Cu(OH)₂ → H₂O + CuO — Decomposition