REACTION TYPES
| Type | Pattern |
|---|
| Synthesis | A + B → AB (two → one) |
| Decomposition | AB → A + B (one → two) |
| Single disp. | A + BC → AC + B (activity series) |
| Double disp. | AB + CD → AD + CB (ion swap) |
| Combustion | CₓHᵧ + O₂ → CO₂ + H₂O |
BALANCING RULES
Law of Conservation of Mass: atoms cannot be created or destroyed — atom count must match on both sides.
- Change COEFFICIENTS only — never subscripts
- Balance: metals → nonmetals → H → O
- For combustion: balance C, then H, then O last
- Always check every atom after balancing
Always balance FIRST before any calculation.
ACTIVITY SERIES (partial)
Most → least reactive: Li, K, Na, Mg, Al, Zn, Fe, Ni, Pb, H, Cu, Ag, Au
Reaction only occurs if displacing element is higher than element it replaces.
THE MOLE
Avogadro
Nₐ = 6.022 × 10²³ /mol
Molar mass M
sum of atomic masses (g/mol)
N = n × Nₐ
number of particles
MOLAR MASS EXAMPLES
M(H₂O)
2(1.008) + 16.00 = 18.02 g/mol
M(NaCl)
22.99 + 35.45 = 58.44 g/mol
M(CO₂)
12.01 + 2(16.00) = 44.01 g/mol
M(Fe₂O₃)
2(55.85) + 3(16.00) = 159.7 g/mol
MOLE RATIOS
Coefficients in the balanced equation = mole ratios.
2H₂ + O₂ → 2H₂O: 2 mol H₂ : 1 mol O₂ : 2 mol H₂O
Conversion
mol B = mol A × (coeff B / coeff A)
STOICHIOMETRY ROADMAP
g A → mol A → mol B → g B
- Step 1: Balance the equation
- Step 2: g A ÷ M(A) = mol A
- Step 3: mol A × [coeff B / coeff A] = mol B
- Step 4: mol B × M(B) = g B
LIMITING REAGENT
The reactant that runs out first and determines how much product forms.
- Convert all reactants to mol
- Use ratios: find mol product from each reactant
- Smallest mol product → limiting reagent
- Use limiting reagent value for final answer
Warning: limiting reagent ≠ reactant with smaller mass — always use mol and ratios.
PERCENT YIELD
% yield
(actual / theoretical) × 100%
Theoretical
calculated from stoichiometry
% yield ≤ 100% always. Above 100% = measurement error.
COMMON MISTAKES
- Forgetting to balance FIRST
- Using molar mass of element, not compound
- Taking mole ratio from masses, not coefficients
- Assuming smaller mass = limiting reagent
- % yield > 100% — always an error
- Skipping mol step: never use g in ratio directly
BALANCING WORKED EXAMPLE
C₃H₈ + O₂ → CO₂ + H₂O
C: 3 → put 3 in front of CO₂
H: 8 → put 4 in front of H₂O
O: 6+4=10 on right → 5 O₂
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O ✓
QUICK CONVERSION REMINDER
1 mol =
6.022 × 10²³ particles
1 mol =
M grams (molar mass)