Chemical Reactions and Stoichiometry

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your WAEC exam.

Mole Concept: 1 mole = $6.02 \times 10^{23}$ particles (Avogadro’s number, $N_A$)

Molar Mass: Mass of 1 mole of substance (numerically equal to relative molecular/formula mass in grams)

Key Equations: $$n = \frac{m}{M} \quad \text{(moles = mass/molar mass)}$$

$$n = \frac{V}{V_m} \quad \text{(moles = volume/gas molar volume)}$$

$$n = \frac{C \times V}{1000} \quad \text{(for solutions, V in cm}^3\text{)}$$

Gas molar volume at RTP (room temperature and pressure) = $24,\text{dm}^3/\text{mol}$

⚡ WAEC Tip: For gases, 1 mole occupies 24 dm³ at room temperature (or 22.4 dm³ at STP). Always check which condition is specified in the question.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding.

Balancing Chemical Equations:

Example: $\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3$

Balance Fe: 2Fe + O₂ → Fe₂O₃ Balance O: 2Fe + 3/2 O₂ → Fe₂O₃ Multiply by 2: 4Fe + 3O₂ → 2Fe₂O₃

Types of Chemical Reactions:

1. Combination (Synthesis): A + B → AB Example: 2Na + Cl₂ → 2NaCl

2. Decomposition: AB → A + B Example: 2H₂O → 2H₂ + O₂

3. Displacement (Single replacement): A + BC → AC + B Example: Zn + 2HCl → ZnCl₂ + H₂

4. Double displacement (Metathesis): AB + CD → AD + CB Example: AgNO₃ + NaCl → AgCl + NaNO₃

5. Combustion: Organic + O₂ → CO₂ + H₂O Example: CH₄ + 2O₂ → CO₂ + 2H₂O

6. Redox (Oxidation-Reduction): Transfer of electrons Example: Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag

Oxidation Number Rules:

• Free elements = 0
• Monatomic ion = charge
• Oxygen = -2 (except in peroxides = -1)
• Hydrogen = +1 (except in metal hydrides = -1)
• Sum in compound = 0; sum in polyatomic ion = charge

Stoichiometric Calculations:

Worked Example 1: Find the mass of oxygen required to completely burn 12g of carbon.

Solution: C + O₂ → CO₂ Moles of C = 12/12 = 1 mol From equation: 1 mol C reacts with 1 mol O₂ Moles of O₂ = 1 mol Mass of O₂ = 1 × 32 = 32 g

Worked Example 2: 25 cm³ of 0.2M HCl reacts with Na₂CO₃. Find volume of CO₂ produced at RTP.

Solution: 2HCl + Na₂CO₃ → 2NaCl + H₂O + CO₂ Moles of HCl = (0.2 × 25)/1000 = 0.005 mol From equation: 2 mol HCl gives 1 mol CO₂ Moles of CO₂ = 0.005/2 = 0.0025 mol Volume of CO₂ = 0.0025 × 24 = 0.06 dm³ = 60 cm³

⚡ Common Mistake: Forgetting to use the stoichiometric ratio from the balanced equation. Always check how many moles of product are formed per mole of reactant.

Limiting Reagent: The reactant that is completely consumed first, limiting the amount of product.

Worked Example 3: 10g of H₂ reacts with 80g of O₂ to form H₂O. Which is limiting?

Solution: 2H₂ + O₂ → 2H₂O Moles of H₂ = 10/2 = 5 mol Moles of O₂ = 80/32 = 2.5 mol Required ratio: 2 mol H₂ : 1 mol O₂ For 2.5 mol O₂, need 5 mol H₂ — exactly available So both are completely consumed (stoichiometric amounts)

Empirical and Molecular Formula:

Empirical formula = simplest whole number ratio Molecular formula = actual number of atoms

Example: A compound contains 40% C, 6.7% H, 53.3% O. Find empirical formula.

C: 40/12 = 3.33; H: 6.7/1 = 6.7; O: 53.3/16 = 3.33 Divide by smallest (3.33): C:H:O = 1:2:1 Empirical formula = CH₂O

If molecular mass = 180 g/mol: n = 180/30 = 6 Molecular formula = (CH₂O)₆ = C₆H₁₂O₆

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🔴 Extended — Deep Study (3mo+)

Comprehensive theory for serious exam preparation.

Yield of Reactions:

Theoretical Yield: Maximum amount of product possible (from stoichiometry)

Actual Yield: Amount actually obtained in experiment

Percentage Yield: $$\text{% yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$$

Atom Economy (WAEC Higher Level): $$\text{Atom economy} = \frac{\text{Molar mass of desired product}}{\text{Sum of molar masses of all reactants}} \times 100$$

Higher atom economy = greener process (less waste)

Concentration Calculations:

Molarity (M): Moles per dm³ (mol/L) $$M = \frac{n}{V} = \frac{m}{M \times V}$$

Standard Solution: A solution of known concentration

Dilution Formula: $$C_1V_1 = C_2V_2$$

Example: How to make 250 cm³ of 0.1M HCl from 2M HCl? $C_1V_1 = C_2V_2$ $2 \times V_1 = 0.1 \times 250$ $V_1 = 12.5,\text{cm}^3$ Dilute 12.5 cm³ of 2M HCl to 250 cm³

Titrations:

$$C_A \times V_A \times n_B = C_B \times V_B \times n_A$$

Where $n_A, n_B$ = number of H⁺ or OH⁻ per formula unit

Back Titration: Used when one reactant is:

• Insoluble (e.g., CaCO₃ in marble)
• Volatile (e.g., NH₃)
• Weak acid/base

Electrochemistry:

Redox Reactions:

• Oxidation = loss of electrons (LEO: Lose Electrons = Oxidation)
• Reduction = gain of electrons (GER: Gain Electrons = Reduction)
• Oxidising agent = gets reduced (accepts electrons)
• Reducing agent = gets oxidised (donates electrons)

Standard Electrode Potentials ($E°$):

Half-reaction	$E°$ (V)
Li⁺ + e⁻ → Li	-3.04
Mg²⁺ + 2e⁻ → Mg	-2.37
2H⁺ + 2e⁻ → H₂	0.00
Cu²⁺ + 2e⁻ → Cu	+0.34
Br₂ + 2e⁻ → 2Br⁻	+1.07
F₂ + 2e⁻ → 2F⁻	+2.87

More positive $E°$ = stronger oxidising agent More negative $E°$ = stronger reducing agent

Galvanic Cell (Voltaic):

• Spontaneous reaction (positive $E°_{cell}$)
• Anode = oxidation = negative electrode
• Cathode = reduction = positive electrode
• $E°{cell} = E°{cathode} - E°_{anode}$

Electrolytic Cell:

• Non-spontaneous, requires external electrical energy
• Used for electrolysis
• Electroplating: metal to be plated is the cathode

⚡ WAEC Previous Year Pattern:

Year	Question	Concept
2023	Limiting reagent calculation	Stoichiometry
2022	Titration calculation	Acid-base
2021	Electroplating mass calculation	Faraday’s laws

Faraday’s Laws of Electrolysis:

First Law: Mass deposited $m = \frac{Q \times M}{n \times F}$

Where $Q$ = charge (coulombs), $M$ = molar mass, $n$ = electrons per ion, $F$ = 96,500 C/mol

Since $Q = It$ (current × time): $$m = \frac{It \times M}{n \times F}$$

Second Law: Same charge deposits different masses proportional to equivalent masses.

⚡ Exam Strategy: Always write the balanced equation first. Convert all volumes to dm³ (divide cm³ by 1000). For gas calculations, check if RTP (24 dm³/mol) or STP (22.4 dm³/mol) is specified.

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