Solution:

Given: Four balanced chemical equations

Find: Classify each reaction type

(a) 2Mg(s) + O₂(g) → 2MgO(s)

Step 1: Count reactants and products

• Reactants: 2 substances (Mg and O₂)
• Products: 1 substance (MgO)

Step 2: Identify pattern

• Multiple reactants → Single product
• This is COMBINATION pattern

Step 3: Classify

$\boxed{\text{Synthesis (Combination) Reaction}}$

Explanation:

• Two elements (Mg and O₂) combine to form one compound (MgO)
• Fits pattern: A + B → AB
• Also could be classified as combustion (metal + oxygen)

Additional notes:

• This is formation of ionic compound (metal + nonmetal)
• Exothermic (releases energy as heat and light)
• This is how magnesium burns with bright white flame

(b) CaCO₃(s) → CaO(s) + CO₂(g)

Step 1: Count reactants and products

• Reactants: 1 substance (CaCO₃)
• Products: 2 substances (CaO and CO₂)

Step 2: Identify pattern

• Single reactant → Multiple products
• This is BREAKDOWN pattern

Step 3: Classify

$\boxed{\text{Decomposition Reaction}}$

Explanation:

• One compound (CaCO₃) breaks down into two simpler substances
• Fits pattern: AB → A + B
• Requires heat (Δ) - thermal decomposition

Additional notes:

• This is how limestone (CaCO₃) is converted to lime (CaO)
• Important industrial process
• Endothermic (requires energy input)
• CO₂ gas escapes (driving force)

Real-world application:

• Making cement
• Occurs in caves forming stalactites/stalagmites (reverse reaction)

(c) Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

Step 1: Count reactants and products

• Reactants: 2 substances
• Products: 2 substances

Pattern so far: Could be single or double replacement

Step 2: Look more closely at reactants

• Zn(s): Free element (metal)
• HCl(aq): Compound

One reactant is free element!

Step 3: Identify what's happening

Rewrite to see the replacement:

$\ce{Zn + H-Cl -> Zn-Cl2 + H2}$

• Zn replaces H in HCl
• H is displaced as H₂ gas

Step 4: Classify

$\boxed{\text{Single Replacement (Displacement) Reaction}}$

Explanation:

• Free element (Zn) replaces another element (H) in a compound
• Fits pattern: A + BC → AC + B
• Zn is more reactive than H (check activity series)

Verification using activity series:

• Activity series: ... Zn > Fe > Ni > Sn > Pb > H > Cu > Ag > Au
• Zn is above H → can displace H
• Reaction occurs ✓

Additional notes:

• This produces H₂ gas (bubbling)
• Common lab reaction
• Used to test for reactive metals

Observable evidence:

• Zn metal dissolves
• H₂ gas bubbles form
• Solution gets warm (exothermic)

(d) AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

Step 1: Count reactants and products

• Reactants: 2 compounds
• Products: 2 compounds

Step 2: Identify what's happening

Look at ions switching:

Reactants:

• AgNO₃: Ag⁺ and NO₃⁻
• NaCl: Na⁺ and Cl⁻

Products:

• AgCl: Ag⁺ and Cl⁻ (Ag switched from NO₃⁻ to Cl⁻)
• NaNO₃: Na⁺ and NO₃⁻ (Na switched from Cl⁻ to NO₃⁻)

Both cations switched anions!

Pattern:

• AB + CD → AD + CB
• Ag⁺NO₃⁻ + Na⁺Cl⁻ → Ag⁺Cl⁻ + Na⁺NO₃⁻

Step 3: Classify

$\boxed{\text{Double Replacement (Metathesis) Reaction}}$

Explanation:

• Two compounds exchange ions
• Cations and anions "switch partners"
• Forms precipitate AgCl(s) - driving force!

Step 4: Verify precipitation using solubility rules

Check AgCl:

• Chloride (Cl⁻) compound
• Most chlorides soluble...
• BUT AgCl is exception (Ag⁺ is insoluble with Cl⁻)
• AgCl is insoluble → precipitates ✓

Check NaNO₃:

• Nitrate (NO₃⁻) compound
• All nitrates soluble
• NaNO₃ stays dissolved ✓

Additional notes:

• AgCl is white precipitate
• This is classic precipitation reaction
• Used in qualitative analysis to test for Cl⁻ ions

Observable evidence:

• White cloudy precipitate forms immediately
• Solution becomes opaque

Summary Table:

| Equation | Type | Key Feature | |----------|------|-------------| | (a) 2Mg + O₂ → 2MgO | Synthesis | 2 reactants → 1 product | | (b) CaCO₃ → CaO + CO₂ | Decomposition | 1 reactant → 2 products | | (c) Zn + 2HCl → ZnCl₂ + H₂ | Single Replacement | Free element replaces H | | (d) AgNO₃ + NaCl → AgCl + NaNO₃ | Double Replacement | Ions switch, precipitate forms |

Key skills practiced:

1. Counting reactants/products
2. Recognizing free elements
3. Identifying ion exchanges
4. Using activity series
5. Applying solubility rules

Strategy for identifying reaction types:

1. Count substances (reactants and products)
2. Look for free elements
3. Check if ions are switching
4. Identify driving forces (precipitate, gas, water)

Solution:

Task: Predict products, write balanced equations, or indicate no reaction

(a) Cl₂(g) + NaBr(aq) →

Step 1: Identify reaction type

• Cl₂: Free element (halogen)
• NaBr: Compound
• Pattern: Element + Compound

This is SINGLE REPLACEMENT

Step 2: Determine if reaction occurs

Check halogen activity series:

• 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:

$\ce{Cl2(g) + NaBr(aq) -> NaCl(aq) + Br2(l)}$

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

$\ce{Cl2(g) + 2NaBr(aq) -> NaCl(aq) + Br2(l)}$

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

$\ce{Cl2(g) + 2NaBr(aq) -> 2NaCl(aq) + Br2(l)}$

Final check:

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

Answer (a):

$\boxed{\ce{Cl2(g) + 2NaBr(aq) -> 2NaCl(aq) + Br2(l)}}$

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):

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

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: $\ce{Zn + 2HCl -> ZnCl2 + H2}$ ✓ (Zn above H)
• If we used Au: $\ce{Au + HCl -> NR}$ (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:

$\ce{Ba(NO3)2(aq) + K2SO4(aq) -> BaSO4 + KNO3}$

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

$\ce{Ba(NO3)2(aq) + K2SO4(aq) -> BaSO4(s) + KNO3(aq)}$

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

$\ce{Ba(NO3)2(aq) + K2SO4(aq) -> BaSO4(s) + 2KNO3(aq)}$

Final check:

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

Answer (c):

$\boxed{\ce{Ba(NO3)2(aq) + K2SO4(aq) -> BaSO4(s) + 2KNO3(aq)}}$

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

$\ce{C3H8(g) + O2(g) -> CO2(g) + H2O(g)}$

Step 3: Balance equation

Strategy: Balance in order C, H, then O

Balance C: 3 carbons left → need 3 CO₂

$\ce{C3H8 + O2 -> 3CO2 + H2O}$

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

$\ce{C3H8 + O2 -> 3CO2 + 4H2O}$

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

$\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$

Final check:

• C: 3, 3 ✓
• H: 8, 8 ✓
• O: 10, 10 ✓

Answer (d):

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

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):

$\ce{2C3H8 + 7O2 -> 6CO + 8H2O}$ (produces toxic CO)

Or:

$\ce{C3H8 + 2O2 -> 3C + 4H2O}$ (produces soot)

Yellow/orange flame indicates incomplete combustion

Summary of Answers:

| Reaction | Balanced Equation | Type | |----------|-------------------|------| | (a) | $\ce{Cl2 + 2NaBr -> 2NaCl + Br2}$ | Single replacement ✓ | | (b) | $\ce{Cu + HCl -> NR}$ | Would be SR, but no reaction | | (c) | $\ce{Ba(NO3)2 + K2SO4 -> BaSO4(s) + 2KNO3}$ | Double replacement ✓ | | (d) | $\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$ | Combustion ✓ |

Key concepts applied:

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

Solution:

Given:

• Aluminum metal (Al) added to copper(II) sulfate solution (CuSO₄)
• Observations: Reddish-brown solid forms, blue solution fades
• Mass of Al = 5.40 g
• Molar masses: Al = 27.0 g/mol, Cu = 63.5 g/mol

Find: (a) Balanced equation, (b) Half-reactions, (c) Mass of Cu formed

Part (a): Write balanced molecular equation

Step 1: Identify reaction type

• 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:

$\ce{Al(s) + CuSO4(aq) -> Al2(SO4)3(aq) + Cu(s)}$

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:

$\ce{2Al + 3CuSO4 -> Al2(SO4)3 + Cu}$

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

$\ce{2Al(s) + 3CuSO4(aq) -> Al2(SO4)3(aq) + 3Cu(s)}$

Final atom check:

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

Answer (a):

$\boxed{\ce{2Al(s) + 3CuSO4(aq) -> Al2(SO4)3(aq) + 3Cu(s)}}$

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):

$\ce{Al(s) -> Al^{3+}(aq) + 3e^-}$

Al is oxidized (oxidation number increases)

• Loses 3 electrons
• Becomes Al³⁺

Reduction half-reaction (Cu gains electrons):

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

Cu²⁺ is reduced (oxidation number decreases)

• Gains 2 electrons
• Becomes Cu metal

Answer (b):

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

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

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

Add half-reactions:

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

Cancel electrons:

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

Add spectator ions (SO₄²⁻):

$\ce{2Al + 3CuSO4 -> Al2(SO4)3 + 3Cu}$ ✓

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:

$\ce{2Al + 3CuSO4 -> Al2(SO4)3 + 3Cu}$

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:

| Substance | Molar Mass | Moles | Mass | |-----------|------------|-------|------| | Al (reactant) | 27.0 g/mol | 0.200 mol | 5.40 g (given) | | Cu (product) | 63.5 g/mol | 0.300 mol | 19.1 g (answer) |

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)