3. Stoichiometry (Part 5) (5/6) (Cambridge IGCSE Chemistry 0620 for 2023, 2024 & 2025)

Limiting Reactants

• Definition:

  • The limiting reactant is the substance in a chemical reaction that is consumed first, thereby limiting the amount of product formed.

  • Analogous to a recipe where the ingredient that runs out first determines how much food can be made.

• Key Concept:

  • The limiting reactant is insufficient for producing additional products, while other reactants may remain in excess.

Example: Potassium and Oxygen Reaction

• Chemical Reaction:

  • Reaction: Potassium + Oxygen → Potassium Oxide

  • Balanced equation shows 4 moles of potassium react with 1 mole of oxygen.

• Given Masses:

  • Potassium: 12 g

  • Oxygen: 16 g

• Calculating Moles:

  • Use the mole formula to find actual moles:

    • Moles of Potassium = 12 g / (molar mass of K) ≈ 0.31 mol

    • Moles of Oxygen = 16 g / (molar mass of O2) ≈ 0.5 mol

• Determining Limiting Reactant:

  • 0.31 mol K needs only 0.08 mol O to react (according to the ratio).

  • There is 0.5 mol O available, making oxygen the excess reactant.

  • Conclusion: Potassium is the limiting reactant.

Concentration

• Definitions:

  • Solute: Substance that is dissolved (e.g., salt in water).

  • Solvent: Substance that dissolves the solute (e.g., water).

  • Solution: Resulting mixture of solute and solvent.

• Concentration:

  • Measures how much solute is present in a given amount of solution.

  • Increased concentration = more solute in a specific volume.

  • Calculated as: Concentration (g/dm³ or mol/dm³) = Mass of solute (g) / Volume of solution (dm³).

• Conversion Units:

  • To convert from dm³ to cm³: (Volume (cm³) = Volume (dm³) \times 1000)

  • To convert from cm³ to dm³: (Volume (dm³) = Volume (cm³) / 1000)

Example: Concentration Calculation

• Given: 12 g of salt dissolved in 1500 cm³ water.

• Convert Volume:

  • 1500 cm³ = 1.5 dm³

• Calculate Concentration:

  • Concentration = 12 g / 1.5 dm³ = 8 g/dm³

Molar Concentration Calculation

• Molar Mass Calculation:

  • To find moles: Moles = Mass (g) / Molar Mass (g/mol)

• Example: Hydrochloric Acid

  • 146 g in 250 cm³ water.

  • Molar Mass of HCl = 1 + 35.5 = 36.5 g/mol.

  • Moles of HCl = 146 g / 36.5 g/mol = 4 moles.

• Convert Volume:

  • 250 cm³ = 0.25 dm³

• Calculate Concentration:

  • Concentration = 4 moles / 0.25 dm³ = 16 mol/dm³.

Titration

• Definition:

  • A technique to find the concentration of a solution by reacting it with a solution of known concentration.

• Applications:

  • Acid-base titrations to determine the neutralization volume or concentration of the opposite solution.

Example: Titration Calculation

• Given: 30 cm³ of 0.025 mol/dm³ sulfuric acid reacts with 60 cm³ sodium hydroxide.

• Convert Volumes:

  • 30 cm³ = 0.03 dm³ and 60 cm³ = 0.06 dm³.

• Calculating Moles of Sulfuric Acid:

  • Moles = Concentration × Volume = 0.025 mol/dm³ × 0.03 dm³ = 0.00075 moles.

• Stoichiometry of the Reaction:

  • 1 mole H2SO4 reacts with 2 moles NaOH.

  • Moles of NaOH = 0.00075 moles H2SO4 × 2 = 0.0015 moles NaOH.

• Calculating Concentration of Sodium Hydroxide:

  • Concentration = Moles / Volume = 0.0015 moles / 0.06 dm³ = 0.025 mol/dm³.

Conclusion

• This concludes part five of topic three, focused on Stoichiometry.

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