Chemical reaction: Process where reactants are transformed into products

• Bonds broken in reactants
• New bonds formed in products
• Atoms conserved (Law of Conservation of Mass)
• Energy changes accompany reactions

General representation:

$Reactants \rightarrow Products$

Balanced equation: Same number of each type of atom on both sides

Five Main Types of Reactions

1. Synthesis (Combination) Reactions

Definition: Two or more substances combine to form a single product

General form:

$A + B \rightarrow AB$

Examples:

Metal + Nonmetal → Ionic compound:

$2Na\text{(s)} + Cl2\text{(g)} \rightarrow 2NaCl\text{(s)}$

$2Mg\text{(s)} + O2\text{(g)} \rightarrow 2MgO\text{(s)}$

Nonmetal + Nonmetal → Molecular compound:

$C\text{(s)} + O2\text{(g)} \rightarrow CO2\text{(g)}$

$N2\text{(g)} + 3H2\text{(g)} \rightarrow 2NH3\text{(g)}$ (Haber process)

Metal oxide + Water → Metal hydroxide:

$CaO\text{(s)} + H2O\text{(l)} \rightarrow Ca(OH)2\text{(aq)}$

$Na2O\text{(s)} + H2O\text{(l)} \rightarrow 2NaOH\text{(aq)}$

Nonmetal oxide + Water → Acid:

$CO2\text{(g)} + H2O\text{(l)} \rightarrow H2CO3\text{(aq)}$ (carbonic acid)

$SO3\text{(g)} + H2O\text{(l)} \rightarrow H2SO4\text{(aq)}$ (sulfuric acid)

Characteristics:

• Multiple reactants → One product
• Often exothermic (releases energy)
• Opposite of decomposition

2. Decomposition Reactions

Definition: Single compound breaks down into two or more simpler substances

General form:

$AB \rightarrow A + B$

Examples:

Heat decomposition:

$2HgO\text{(s)} \xrightarrow{\Delta} 2Hg\text{(l)} + O2\text{(g)}$

$CaCO3\text{(s)} \xrightarrow{\Delta} CaO\text{(s)} + CO2\text{(g)}$

Electrolysis (electrical energy):

$2H2O\text{(l)} \rightarrow[\text{electricity] 2H2(g) + O2(g)}$

$2NaCl\text{(l)} \rightarrow[\text{electricity] 2Na(l) + Cl2(g)}$

Catalytic decomposition:

$2H2O2\text{(aq)} \xrightarrow{MnO_2} 2H2O\text{(l)} + O2\text{(g)}$

Binary compound decomposition:

$2KClO3\text{(s)} \xrightarrow{\Delta} 2KCl\text{(s)} + 3O2\text{(g)}$

Characteristics:

• One reactant → Multiple products
• Often requires energy input (endothermic)
• Need heat, electricity, or light
• Opposite of synthesis

3. Single Replacement (Displacement) Reactions

Definition: One element replaces another in a compound

General forms:

Metal replaces metal:

$A + BC \rightarrow AC + B$

Nonmetal replaces nonmetal:

$A + BC \rightarrow BA + C$

Examples:

More reactive metal displaces less reactive:

$Zn\text{(s)} + CuSO4\text{(aq)} \rightarrow ZnSO4\text{(aq)} + Cu\text{(s)}$

$Mg\text{(s)} + 2HCl\text{(aq)} \rightarrow MgCl2\text{(aq)} + H2\text{(g)}$

$2Al\text{(s)} + 3CuCl2\text{(aq)} \rightarrow 2AlCl3\text{(aq)} + 3Cu\text{(s)}$

More reactive halogen displaces less reactive:

$Cl2\text{(g)} + 2NaBr\text{(aq)} \rightarrow 2NaCl\text{(aq)} + Br2\text{(l)}$

$Br2\text{(l)} + 2KI\text{(aq)} \rightarrow 2KBr\text{(aq)} + I2\text{(s)}$

Predicting if reaction occurs:

Use Activity Series (reactivity order)

Metal Activity Series (most → least reactive):

• Li > K > Ba > Ca > Na > Mg > Al > Zn > Fe > Ni > Sn > Pb > H > Cu > Ag > Au

Rule: Metal higher in series displaces metal lower in series

Halogen Activity Series:

• F₂ > Cl₂ > Br₂ > I₂

Rule: Halogen higher in series displaces halogen lower in series

Will this reaction occur?

$Cu\text{(s)} + 2AgNO3\text{(aq)} \rightarrow ?$

Check: Cu is above Ag in activity series → YES, reaction occurs

$Cu\text{(s)} + 2AgNO3\text{(aq)} \rightarrow Cu(NO3)2\text{(aq)} + 2Ag\text{(s)}$

$Ag\text{(s)} + CuSO4\text{(aq)} \rightarrow ?$

Check: Ag is below Cu in activity series → NO, no reaction

$Ag\text{(s)} + CuSO4\text{(aq)} \rightarrow \text{No Reaction}$

4. Double Replacement (Metathesis) Reactions

Definition: Two compounds exchange partners

General form:

$AB + CD \rightarrow AD + CB$

Cations and anions "switch partners"

Examples:

Precipitation reactions:

$AgNO3\text{(aq)} + NaCl\text{(aq)} \rightarrow AgCl\text{(s)} + NaNO3\text{(aq)}$

$Pb(NO3)2\text{(aq)} + 2KI\text{(aq)} \rightarrow PbI2\text{(s)} + 2KNO3\text{(aq)}$

Acid-base neutralization:

$HCl\text{(aq)} + NaOH\text{(aq)} \rightarrow NaCl\text{(aq)} + H2O\text{(l)}$

$H2SO4\text{(aq)} + 2KOH\text{(aq)} \rightarrow K2SO4\text{(aq)} + 2H2O\text{(l)}$

Gas formation:

$2HCl\text{(aq)} + Na2CO3\text{(aq)} \rightarrow 2NaCl\text{(aq)} + H2O\text{(l)} + CO2\text{(g)}$

$2HCl\text{(aq)} + Na2S\text{(aq)} \rightarrow 2NaCl\text{(aq)} + H2S\text{(g)}$

Predicting if reaction occurs:

Double replacement reactions occur if one of these forms:

1. Precipitate (insoluble solid)
2. Water (in neutralization)
3. Gas (escapes from solution)

Use solubility rules to predict precipitates!

5. Combustion Reactions

Definition: Substance reacts rapidly with oxygen, releasing energy as heat and light

General form (hydrocarbon combustion):

$C_xH_y + O2 \rightarrow CO2 + H2O + \text{energy}$

Complete combustion examples:

$CH4\text{(g)} + 2O2\text{(g)} \rightarrow CO2\text{(g)} + 2H2O\text{(g)}$ (methane)

$C3H8\text{(g)} + 5O2\text{(g)} \rightarrow 3CO2\text{(g)} + 4H2O\text{(g)}$ (propane)

$2C8H18\text{(l)} + 25O2\text{(g)} \rightarrow 16CO2\text{(g)} + 18H2O\text{(g)}$ (octane)

Complete combustion products:

• CO₂ (carbon dioxide)
• H₂O (water)
• Energy (heat and light)

Incomplete combustion (insufficient O₂):

$2CH4\text{(g)} + 3O2\text{(g)} \rightarrow 2CO\text{(g)} + 4H2O\text{(g)}$ (produces CO)

$CH4\text{(g)} + O2\text{(g)} \rightarrow C\text{(s)} + 2H2O\text{(g)}$ (produces soot)

Other combustion reactions:

$S\text{(s)} + O2\text{(g)} \rightarrow SO2\text{(g)}$

$4Fe\text{(s)} + 3O2\text{(g)} \rightarrow 2Fe2O3\text{(s)}$ (rusting)

Characteristics:

• Always involves O₂ as reactant
• Exothermic (releases energy)
• Often produces flame
• Important for energy production

Solubility Rules

Essential for predicting precipitation in double replacement reactions

Soluble Compounds (form aqueous solutions)

1. Group 1 (alkali metals) compounds:

• All Li⁺, Na⁺, K⁺, Rb⁺, Cs⁺ salts are soluble
• Example: NaCl, K₂SO₄, LiNO₃

2. Ammonium (NH₄⁺) compounds:

• All NH₄⁺ salts are soluble
• Example: (NH₄)₂SO₄, NH₄Cl

3. Nitrates (NO₃⁻), acetates (C₂H₃O₂⁻), perchlorates (ClO₄⁻):

• All soluble
• Example: AgNO₃, Ca(NO₃)₂, Pb(C₂H₃O₂)₂

4. Most chlorides (Cl⁻), bromides (Br⁻), iodides (I⁻):

• Soluble EXCEPT with Ag⁺, Pb²⁺, Hg₂²⁺
• AgCl, PbBr₂, Hg₂I₂ are insoluble

5. Most sulfates (SO₄²⁻):

• Soluble EXCEPT with Sr²⁺, Ba²⁺, Pb²⁺
• SrSO₄, BaSO₄, PbSO₄ are insoluble

Insoluble Compounds (form precipitates)

1. Carbonates (CO₃²⁻), phosphates (PO₄³⁻), chromates (CrO₄²⁻):

• Insoluble EXCEPT with Group 1 and NH₄⁺
• CaCO₃, Ag₃PO₄, PbCrO₄ are insoluble
• But: Na₂CO₃, K₃PO₄, (NH₄)₂CrO₄ are soluble

2. Hydroxides (OH⁻):

• Insoluble EXCEPT with Group 1, Ba²⁺, Sr²⁺, Ca²⁺ (slightly)
• Fe(OH)₃, Al(OH)₃, Cu(OH)₂ are insoluble
• But: NaOH, KOH, Ba(OH)₂ are soluble

3. Sulfides (S²⁻):

• Insoluble EXCEPT with Group 1, Group 2, NH₄⁺
• CuS, FeS, PbS are insoluble
• But: Na₂S, (NH₄)₂S are soluble

Using Solubility Rules

Example: Will precipitation occur?

$AgNO3\text{(aq)} + NaCl\text{(aq)} \rightarrow ?$

Step 1: Identify potential products (switch partners)

• Ag⁺ with Cl⁻ → AgCl
• Na⁺ with NO₃⁻ → NaNO₃

Step 2: Check solubility

• AgCl: Cl⁻ compound with Ag⁺ → Insoluble (exception to Cl⁻ rule)
• NaNO₃: Group 1 compound → Soluble

Step 3: Write equation

• AgCl precipitates → (s)
• NaNO₃ stays dissolved → (aq)

$AgNO3\text{(aq)} + NaCl\text{(aq)} \rightarrow AgCl\text{(s)} + NaNO3\text{(aq)}$

Yes, precipitation occurs! (AgCl is white precipitate)

Identifying Reaction Types

Strategy: Look for patterns

Synthesis: Multiple → One

• 2 or more reactants, 1 product
• Example: $2H2 + O2 \rightarrow 2H2O$

Decomposition: One → Multiple

• 1 reactant, 2 or more products
• Often needs energy input
• Example: $2H2O \rightarrow 2H2 + O2$

Single Replacement: Element + Compound → Element + Compound

• Free element replaces element in compound
• Check activity series
• Example: $Zn + CuSO4 \rightarrow ZnSO4 + Cu$

Double Replacement: Compound + Compound → Compound + Compound

• Ions switch partners
• Forms precipitate, water, or gas
• Example: $AgNO3 + NaCl \rightarrow AgCl + NaNO3$

Combustion: Substance + O₂ → CO₂ + H₂O

• Always has O₂ as reactant
• Produces CO₂ and H₂O (if hydrocarbon)
• Exothermic, flame
• Example: $CH4 + 2O2 \rightarrow CO2 + 2H2O$

Note: Some reactions fit multiple categories!

$2H2\text{(g)} + O2\text{(g)} \rightarrow 2H2O\text{(g)}$

This is both synthesis AND combustion!

Driving Forces for Reactions

Why do reactions occur? What makes them "go"?

1. Formation of Precipitate

Ions in solution combine to form insoluble solid

$Pb^{2+(aq) + 2I^-(aq) -> PbI2(s)}$

Driving force: Decrease in solubility (ions prefer solid state)

2. Formation of Water

Neutralization reactions

$H^+\text{(aq)} + OH^-\text{(aq)} \rightarrow H2O\text{(l)}$

Driving force: Formation of very stable H₂O molecule

3. Formation of Gas

Gas escapes from solution

2H^+\text{(aq)} + CO3^{2-(aq) -> H2O(l) + CO2(g) ^}

Driving force: Entropy increase (gas more disordered than aqueous)

Common gases formed:

• CO₂ from carbonates + acid
• H₂S from sulfides + acid
• NH₃ from ammonium salts + base
• SO₂ from sulfites + acid

4. Transfer of Electrons (Redox)

Oxidation-reduction reactions

$Zn\text{(s)} + Cu^{2+(aq) -> Zn^{2+}(aq) + Cu(s)}$

Driving force: Electron transfer to more stable configuration

5. Energy Release

Exothermic reactions (ΔH < 0)

$CH4\text{(g)} + 2O2\text{(g)} \rightarrow CO2\text{(g)} + 2H2O\text{(g)} + \text{energy}$

Driving force: System goes to lower energy state

Note: Most spontaneous reactions have at least one driving force!

Summary Table

Tips for Success

1. Learn to recognize patterns

• Count reactants and products
• Look for free elements
• Identify ions switching

2. Use activity series

• Predicts single replacement
• Higher replaces lower

3. Memorize solubility rules

• Essential for double replacement
• Predicts precipitates

4. Check for driving forces

• Precipitate, water, gas, redox
• No driving force → likely no reaction

5. Balance equations

• Atoms must be conserved
• Check your work!

• F₂ > Cl₂ > Br₂ > I₂

Cl₂ is above Br₂ → Cl₂ is more reactive

• Cl₂ can displace Br⁻ from compounds
• Reaction occurs!

Step 3: Predict products

Cl₂ replaces Br in NaBr:

• Na⁺ pairs with Cl⁻ → NaCl
• Br is displaced as Br₂

Unbalanced:

$Cl2\text{(g)} + NaBr\text{(aq)} \rightarrow NaCl\text{(aq)} + Br2\text{(l)}$

Step 4: Balance equation

Check atoms:

• Left: 2 Cl, 1 Na, 1 Br
• Right: 1 Cl, 1 Na, 2 Br

Br not balanced!

Balance Br: Need 2 NaBr on left

$Cl2\text{(g)} + 2NaBr\text{(aq)} \rightarrow NaCl\text{(aq)} + Br2\text{(l)}$

Check atoms:

• Left: 2 Cl, 2 Na, 2 Br
• Right: 1 Cl, 1 Na, 2 Br

Cl and Na not balanced!

Balance Cl and Na: Need 2 NaCl on right

$Cl2\text{(g)} + 2NaBr\text{(aq)} \rightarrow 2NaCl\text{(aq)} + Br2\text{(l)}$

Final check:

• Left: 2 Cl, 2 Na, 2 Br ✓
• Right: 2 Cl, 2 Na, 2 Br ✓

Answer (a):

$\boxed{Cl2\text{(g)} + 2NaBr\text{(aq)} \rightarrow 2NaCl\text{(aq)} + Br2\text{(l)}}$

Observation: Solution turns brown/orange (color of Br₂)

(b) Cu(s) + HCl(aq) →

Step 1: Identify reaction type

• Cu: Free element (metal)
• HCl: Compound (acid)
• Pattern: Element + Compound

This would be SINGLE REPLACEMENT if it occurs

Step 2: Check if reaction occurs

Use metal activity series:

• ... Zn > Fe > Ni > Sn > Pb > H > Cu > Ag > Au

Cu is BELOW H in activity series

• Cu is less reactive than H
• Cu cannot displace H from acids
• NO REACTION

Answer (b):

$\boxed{Cu\text{(s)} + HCl\text{(aq)} \rightarrow NR \text{ (No Reaction)}}$

Explanation:

• Copper is too unreactive (noble)
• Cannot displace hydrogen from HCl
• This is why copper pipes are safe for water systems

Contrast:

• If we used Zn instead: $Zn + 2HCl \rightarrow ZnCl2 + H2$ ✓ (Zn above H)
• If we used Au: $Au + HCl \rightarrow NR$ (Au below H)

To dissolve copper, need:

• Oxidizing acid (HNO₃ or hot concentrated H₂SO₄)
• These work by different mechanism (not simple single replacement)

(c) Ba(NO₃)₂(aq) + K₂SO₄(aq) →

Step 1: Identify reaction type

• Both are ionic compounds in solution
• Pattern: Compound + Compound

This is DOUBLE REPLACEMENT

Step 2: Predict products (switch partners)

Identify ions:

• Ba(NO₃)₂: Ba²⁺ and NO₃⁻
• K₂SO₄: K⁺ and SO₄²⁻

Switch partners:

• Ba²⁺ with SO₄²⁻ → BaSO₄
• K⁺ with NO₃⁻ → KNO₃

Unbalanced:

$Ba(NO3)2\text{(aq)} + K2SO4\text{(aq)} \rightarrow BaSO4 + KNO3$

Step 3: Check if reaction occurs (solubility)

Check BaSO₄:

• Sulfate (SO₄²⁻) compound
• Most sulfates soluble...
• BUT BaSO₄ is exception (Ba²⁺ makes insoluble sulfate)
• BaSO₄ is INSOLUBLE → precipitates! ✓

Check KNO₃:

• Nitrate (NO₃⁻) compound
• All nitrates soluble
• KNO₃ stays dissolved

Reaction occurs because precipitate forms!

Step 4: Assign states

$Ba(NO3)2\text{(aq)} + K2SO4\text{(aq)} \rightarrow BaSO4\text{(s)} + KNO3\text{(aq)}$

Step 5: Balance equation

Check atoms:

• Ba: 1 left, 1 right ✓
• N: 2 left, 1 right ✗
• O: 6 + 4 = 10 left, 4 + 3 = 7 right ✗
• K: 2 left, 1 right ✗
• S: 1 left, 1 right ✓

Balance K and N: Need 2 KNO₃

$Ba(NO3)2\text{(aq)} + K2SO4\text{(aq)} \rightarrow BaSO4\text{(s)} + 2KNO3\text{(aq)}$

Final check:

• Ba: 1, 1 ✓
• N: 2, 2 ✓
• O: 10, 10 ✓
• K: 2, 2 ✓
• S: 1, 1 ✓

Answer (c):

$\boxed{Ba(NO3)2\text{(aq)} + K2SO4\text{(aq)} \rightarrow BaSO4\text{(s)} + 2KNO3\text{(aq)}}$

Observation: White precipitate (BaSO₄) forms immediately

Note: BaSO₄ is used in medical imaging (barium meals)

(d) C₃H₈(g) + O₂(g) →

Step 1: Identify reaction type

• C₃H₈: Hydrocarbon (propane)
• O₂: Oxygen gas
• Pattern: Hydrocarbon + O₂

This is COMBUSTION

Step 2: Predict products

Complete combustion of hydrocarbon:

• C → CO₂
• H → H₂O
• Energy released

$C3H8\text{(g)} + O2\text{(g)} \rightarrow CO2\text{(g)} + H2O\text{(g)}$

Step 3: Balance equation

Strategy: Balance in order C, H, then O

Balance C: 3 carbons left → need 3 CO₂

$C3H8 + O2 \rightarrow 3CO2 + H2O$

Balance H: 8 hydrogens left → need 4 H₂O (each has 2 H)

$C3H8 + O2 \rightarrow 3CO2 + 4H2O$

Balance O:

• Right side: 3 CO₂ = 6 O, plus 4 H₂O = 4 O
• Total O on right = 6 + 4 = 10 O atoms
• Need 10 O atoms on left
• Each O₂ has 2 O → need 10/2 = 5 O₂

$C3H8 + 5O2 \rightarrow 3CO2 + 4H2O$

Final check:

• C: 3, 3 ✓
• H: 8, 8 ✓
• O: 10, 10 ✓

Answer (d):

$\boxed{C3H8\text{(g)} + 5O2\text{(g)} \rightarrow 3CO2\text{(g)} + 4H2O\text{(g)} + \text{energy}}$

Additional information:

• This is complete combustion (sufficient O₂)
• Highly exothermic (releases heat and light)
• Blue flame when burning
• Used in gas grills, camping stoves, torches

If insufficient O₂ (incomplete combustion):

$2C3H8 + 7O2 \rightarrow 6CO + 8H2O$ (produces toxic CO)

Or:

$C3H8 + 2O2 \rightarrow 3C + 4H2O$ (produces soot)

Yellow/orange flame indicates incomplete combustion

Summary of Answers:

Key concepts applied:

1. Activity series (halogens and metals)
2. Solubility rules
3. Combustion product prediction
4. Balancing equations
5. State symbols

• Al: Free element (metal)
• CuSO₄: Compound
• Pattern: Metal + Compound

This is SINGLE REPLACEMENT

Step 2: Check activity series

Metal activity series:

• ... Mg > Al > Zn > Fe > ... > Cu > Ag > Au

Al is ABOVE Cu → Al is more reactive

• Al can displace Cu from compounds
• Reaction occurs! ✓

Step 3: Predict products

Al replaces Cu in CuSO₄:

• Al pairs with SO₄²⁻ → Al₂(SO₄)₃
• Cu is displaced as Cu metal

Why Al₂(SO₄)₃?

• Al forms Al³⁺ ions
• SO₄²⁻ is 2- charge
• Need 2 Al³⁺ and 3 SO₄²⁻ to balance charges
• Formula: Al₂(SO₄)₃

Unbalanced equation:

$Al\text{(s)} + CuSO4\text{(aq)} \rightarrow Al2(SO4)3\text{(aq)} + Cu\text{(s)}$

Step 4: Balance the equation

Method: Balance atoms systematically

Current atoms:

• Al: 1 left, 2 right → need 2 Al on left
• Cu: 1 left, 1 right → balanced for now
• S: 1 left, 3 right → need 3 CuSO₄ on left
• O: 4 left, 12 right → will balance with sulfates

Try balancing:

$2Al + 3CuSO4 \rightarrow Al2(SO4)3 + Cu$

Check Cu: 3 left, 1 right → need 3 Cu on right

$2Al\text{(s)} + 3CuSO4\text{(aq)} \rightarrow Al2(SO4)3\text{(aq)} + 3Cu\text{(s)}$

Final atom check:

• Al: 2, 2 ✓
• Cu: 3, 3 ✓
• S: 3, 3 ✓
• O: 12, 12 ✓

Answer (a):

$\boxed{2Al\text{(s)} + 3CuSO4\text{(aq)} \rightarrow Al2(SO4)3\text{(aq)} + 3Cu\text{(s)}}$

Verification of observations:

• Reddish-brown solid: Cu metal forming on Al ✓
• Blue solution fades: Cu²⁺ ions (blue) being reduced to Cu(s) ✓

Part (b): Identify oxidation and reduction half-reactions

Step 1: Determine oxidation states

Reactants:

• Al(s): 0 (free element)
• CuSO₄: Cu is +2, S is +6, O is -2

Products:

• Al₂(SO₄)₃: Al is +3, S is +6, O is -2
• Cu(s): 0 (free element)

Step 2: Identify changes

Aluminum:

• Al: 0 → +3
• Loses 3 electrons
• Oxidation (LEO: Lose Electrons = Oxidation)

Copper:

• Cu: +2 → 0
• Gains 2 electrons
• Reduction (GER: Gain Electrons = Reduction)

Step 3: Write half-reactions

Oxidation half-reaction (Al loses electrons):

$Al\text{(s)} \rightarrow Al^{3+(aq) + 3e^-}$

Al is oxidized (oxidation number increases)

• Loses 3 electrons
• Becomes Al³⁺

Reduction half-reaction (Cu gains electrons):

$Cu^{2+(aq) + 2e^- -> Cu(s)}$

Cu²⁺ is reduced (oxidation number decreases)

• Gains 2 electrons
• Becomes Cu metal

Answer (b):

$\boxed{\text{Oxidation: } Al\text{(s)} \rightarrow Al^{3+(aq) + 3e^-}}$

$\boxed{\text{Reduction: } Cu^{2+(aq) + 2e^- -> Cu(s)}}$

Step 4: Verify overall equation

To get overall reaction, balance electrons:

• Oxidation produces 3 e⁻ per Al
• Reduction uses 2 e⁻ per Cu
• LCM of 3 and 2 = 6

Multiply to balance electrons:

• Oxidation × 2: $2Al \rightarrow 2Al^{3+ + 6e^-}$
• Reduction × 3: $3Cu^{2+ + 6e^- -> 3Cu}$

Add half-reactions:

$2Al + 3Cu^{2+ + 6e^- -> 2Al^{3+} + 3Cu + 6e^-}$

Cancel electrons:

$2Al + 3Cu^{2+ -> 2Al^{3+} + 3Cu}$

Add spectator ions (SO₄²⁻):

$2Al + 3CuSO4 \rightarrow Al2(SO4)3 + 3Cu$ ✓

Matches our answer from part (a)!

Terminology:

• Al is the reducing agent (causes reduction of Cu²⁺, gets oxidized itself)
• Cu²⁺ is the oxidizing agent (causes oxidation of Al, gets reduced itself)

Part (c): Calculate mass of Cu formed

Given:

• Mass of Al = 5.40 g
• Excess CuSO₄ (Al is limiting reactant)

Step 1: Convert mass of Al to moles

$n_{Al} = \frac{\text{mass}}{\text{molar mass}}$

$n_{Al} = \frac{5.40 \text{ g}}{27.0 \text{ g/mol}} = 0.200 \text{ mol Al}$

Step 2: Use stoichiometry to find moles of Cu

From balanced equation:

$2Al + 3CuSO4 \rightarrow Al2(SO4)3 + 3Cu$

Mole ratio: 2 mol Al : 3 mol Cu

$\frac{2 \text{ mol Al}}{3 \text{ mol Cu}}$

Calculate moles of Cu:

$n_{Cu} = 0.200 \text{ mol Al} \times \frac{3 \text{ mol Cu}}{2 \text{ mol Al}}$

$n_{Cu} = 0.200 \times \frac{3}{2} = 0.300 \text{ mol Cu}$

Step 3: Convert moles of Cu to mass

$\text{mass}_{Cu} = n_{Cu} \times M_{Cu}$

$\text{mass}_{Cu} = 0.300 \text{ mol} \times 63.5 \text{ g/mol}$

$\text{mass}_{Cu} = 19.05 \text{ g}$

$\text{mass}_{Cu} = 19.1 \text{ g}$ (3 sig figs)

Answer (c):

$\boxed{19.1 \text{ g Cu}}$

Summary of Solution:

Calculation flow:

$5.40 \text{ g Al} \xrightarrow{÷27.0} 0.200 \text{ mol Al} \xrightarrow{×\frac{3}{2}} 0.300 \text{ mol Cu} \xrightarrow{×63.5} 19.1 \text{ g Cu}$

Stoichiometry summary:

Mole ratio check:

• $\frac{n_{Cu}}{n_{Al}} = \frac{0.300}{0.200} = \frac{3}{2}$ ✓ (matches equation)

Additional Insights:

Why does this reaction occur?

1. Activity series: Al more reactive than Cu
2. Electron transfer: Al readily loses electrons (good reducing agent)
3. Energy favorable: Overall reaction exothermic

Practical applications:

• Demonstrates reactivity of metals
• Shows electron transfer visually
• Thermite reaction (Fe₂O₃ + Al) uses same principle

Observable changes:

• Al surface becomes pitted/dissolved
• Reddish-brown Cu deposits on Al
• Blue Cu²⁺ solution becomes colorless
• Solution may get warm (exothermic)

Safety notes:

• Reaction is exothermic
• If large amounts used, significant heat generated
• Al₂(SO₄)₃ remains dissolved (not visible)

Percent yield consideration:

• Calculated: 19.1 g (theoretical yield)
• Actual yield in lab might be less
• Some Cu might not adhere to Al
• Some might remain suspended in solution

Mass comparison:

• Started with 5.40 g Al
• Produced 19.1 g Cu
• Mass of Cu > mass of Al used
• This makes sense: Cu has larger molar mass (63.5 vs 27.0)
• Even though we make fewer moles of Cu (0.300 vs 0.200 used), the mass is greater

Complete answer check: ✓ Equation balanced ✓ Half-reactions identified correctly ✓ Stoichiometry calculation accurate ✓ Significant figures appropriate (3 sig figs)