Stoichiometry and Chemical Reactions

The Mole Concept

A mole = 6.022 × 10²³ particles (Avogadro's number).

Molar mass = mass in grams of one mole of a substance.

Conversions:

• moles = mass (g) / molar mass (g/mol)
• moles = particles / 6.022×10²³
• moles = volume (L) / 22.4 L/mol (at STP for gas)

Balancing Chemical Equations

Atoms must be conserved (law of conservation of mass).

Example: Combustion of methane: CH₄ + 2O₂ → CO₂ + 2H₂O

Check: C:1=1 ✓, H:4=4 ✓, O:4=4 ✓

Stoichiometric Calculations

Mole ratios from balanced equation:

Example: How many grams of CO₂ from burning 32 g of CH₄?

Moles CH₄ = 32/16 = 2 mol
Ratio: 1 mol CH₄ → 1 mol CO₂
Moles CO₂ = 2 mol
Grams CO₂ = 2 × 44 = 88 g

Limiting Reagent

The reactant that runs out first — determines the amount of product.

1. Convert each reactant to moles
2. Divide by stoichiometric coefficient
3. Smallest ratio = limiting reagent

Types of Chemical Reactions

| Type | General Form | Example | |------|-------------|--------| | Synthesis | A + B → AB | 2H₂ + O₂ → 2H₂O | | Decomposition | AB → A + B | 2H₂O → 2H₂ + O₂ | | Single displacement | A + BC → AC + B | Zn + 2HCl → ZnCl₂ + H₂ | | Double displacement | AB + CD → AD + CB | NaCl + AgNO₃ → AgCl + NaNO₃ | | Combustion | Fuel + O₂ → CO₂ + H₂O | — |

Engineering Relevance

• Combustion: energy from fuels, CO₂ emissions
• Corrosion: electrochemical reactions on steel
• Water chemistry: pH, precipitation reactions
• Concrete chemistry: hydration of cement

Summary

Stoichiometry is the foundation of chemical calculations. Balance the equation, use mole ratios, and track units carefully.