Master the classification of chemical reactions including synthesis, decomposition, single and double replacement, combustion, and precipitation reactions.
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Master the classification of chemical reactions including synthesis, decomposition, single and double replacement, combustion, and precipitation reactions.
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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→AB
Examples:
Metal + Nonmetal → Ionic compound:
2Na(s)+Cl2(g)→2NaCl(s)
2Mg(s)+O2(g)→2MgO(s)
Nonmetal + Nonmetal → Molecular compound:
C(s)+O2(g)→CO2(g)
N2(g)+3H2(g)→2NH3(g) (Haber process)
Metal oxide + Water → Metal hydroxide:
CaO(s)+H2O(l)→Ca(OH)2(aq)
Na2O(s)+H2O(l)→2NaOH(aq)
Nonmetal oxide + Water → Acid:
CO2(g)+H2O(l)→H2CO3(aq) (carbonic acid)
SO3(g)+H2O(l)→H2SO4(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→A+B
Examples:
Heat decomposition:
2HgO(s)Δ2Hg(l)+O2(g)
CaCO3(s)ΔCaO(s)+CO2(g)
Electrolysis (electrical energy):
2H2O(l)→[electricity] 2H2(g) + O2(g)
2NaCl(l)→[electricity] 2Na(l) + Cl2(g)
Catalytic decomposition:
2H2O2(aq)MnO22H2O(l)+O2(g)
Binary compound decomposition:
2KClO3(s)Δ2KCl(s)+3O2(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→AC+B
Nonmetal replaces nonmetal:
A+BC→BA+C
Examples:
More reactive metal displaces less reactive:
Zn(s)+CuSO4(aq)→ZnSO4(aq)+Cu(s)
Mg(s)+2HCl(aq)→MgCl2(aq)+H2(g)
2Al(s)+3CuCl2(aq)→2AlCl3(aq)+3Cu(s)
More reactive halogen displaces less reactive:
Cl2(g)+2NaBr(aq)→2NaCl(aq)+Br2(l)
Br2(l)+2KI(aq)→2KBr(aq)+I2(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(s)+2AgNO3(aq)→?
Check: Cu is above Ag in activity series → YES, reaction occurs
Cu(s)+2AgNO3(aq)→Cu(NO3)2(aq)+2Ag(s)
Ag(s)+CuSO4(aq)→?
Check: Ag is below Cu in activity series → NO, no reaction
Ag(s)+CuSO4(aq)→No Reaction
4. Double Replacement (Metathesis) Reactions
Definition: Two compounds exchange partners
General form:
AB+CD→AD+CB
Cations and anions "switch partners"
Examples:
Precipitation reactions:
AgNO3(aq)+NaCl(aq)→AgCl(s)+NaNO3(aq)
Pb(NO3)2(aq)+2KI(aq)→PbI2(s)+2KNO3(aq)
Acid-base neutralization:
HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)
H2SO4(aq)+2KOH(aq)→K2SO4(aq)+2H2O(l)
Gas formation:
2HCl(aq)+Na2CO3(aq)→2NaCl(aq)+H2O(l)+CO2(g)
2HCl(aq)+Na2S(aq)→2NaCl(aq)+H2S(g)
Predicting if reaction occurs:
Double replacement reactions occur if one of these forms:
Precipitate (insoluble solid)
Water (in neutralization)
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):
CxHy+O2→CO2+H2O+energy
Complete combustion examples:
CH4(g)+2O2(g)→CO2(g)+2H2O(g) (methane)
C3H8(g)+5O2(g)→3CO2(g)+4H2O(g) (propane)
2C8H18(l)+25O2(g)→16CO2(g)+18H2O(g) (octane)
Complete combustion products:
CO₂ (carbon dioxide)
H₂O (water)
Energy (heat and light)
Incomplete combustion (insufficient O₂):
2CH4(g)+3O2(g)→2CO(g)+4H2O(g) (produces CO)
CH4(g)+O2(g)→C(s)+2H2O(g) (produces soot)
Other combustion reactions:
S(s)+O2(g)→SO2(g)
4Fe(s)+3O2(g)→2Fe2O3(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
A student adds a piece of aluminum metal to a solution containing copper(II) sulfate. A reddish-brown solid forms on the aluminum, and the blue solution fades. (a) Write the balanced molecular equation for this reaction. (b) Identify the oxidation and reduction half-reactions. (c) Calculate the mass of copper that forms when 5.40 g of aluminum reacts with excess CuSO₄. (Al = 27.0 g/mol, Cu = 63.5 g/mol)
💡 Show Solution
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:
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:
2Al+3CuSO4→Al2(SO4)3+Cu
Check Cu: 3 left, 1 right → need 3 Cu on right
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):
2Al(s)+3CuSO4(aq)→Al2(SO4)3(aq)
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(s)→Al3+(aq)+3e−
Al is oxidized (oxidation number increases)
Loses 3 electrons
Becomes Al³⁺
Reduction half-reaction (Cu gains electrons):
Cu2+(aq)+2e−−>Cu(s)
Cu²⁺ is reduced (oxidation number decreases)
Gains 2 electrons
Becomes Cu metal
Answer (b):
Oxidation: Al(s)→Al3+(aq)+3e
Reduction: Cu2+(aq)+2e−−
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→2Al3++6e−
Reduction × 3:
Add half-reactions:
2Al+3Cu2++6e−−>2Al
Cancel electrons:
2Al+3Cu2+−>2Al3++3Cu
Add spectator ions (SO₄²⁻):
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
nAl=molar massmass
nAl=27.0 g/mol5.40 g
Step 2: Use stoichiometry to find moles of Cu
From balanced equation:
2Al+3CuSO4→Al2(SO4)3+3Cu
Mole ratio: 2 mol Al : 3 mol Cu
3 mol Cu2 mol Al
Calculate moles of Cu:
nCu=0.200 mol Al×2 mol Al
nCu=0.200×23
Step 3: Convert moles of Cu to mass
massCu=nCu×M
massCu=0.300 mol×63.5 g/mol
massCu=19.05 g
massCu=19.1 g (3 sig figs)
Answer (c):
19.1 g Cu
Summary of Solution:
Calculation flow:
5.40 g Al÷27.0
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:
nAlnCu ✓ (matches equation)
Additional Insights:
Why does this reaction occur?
Activity series: Al more reactive than Cu
Electron transfer: Al readily loses electrons (good reducing agent)
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
Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.