Stoichiometry, the Mole Concept and Chemical Equations
Stoichiometry, the mole concept and chemical equations
The GCE O-Level exam tests balancing equations, mole calculations, limiting reagents, concentration, gas volumes and percentage yield. You must show units and use the equation ratios correctly.
Core rules
- One mole contains 6.02 x 10^23 particles. Moles = mass / molar mass.
- A balanced equation gives the mole ratio. Never compare masses directly unless the molar masses are equal.
- Concentration in mol/dm3 = moles / volume in dm3. Convert cm3 to dm3 by dividing by 1000.
- For gases, use the molar volume stated in the question. At room temperature and pressure it is commonly 24 dm3/mol; at standard temperature and pressure it is 22.4 dm3/mol.
- The limiting reagent is used up first and controls the maximum product. Excess reagent remains.
Worked example
4.8 g of magnesium reacts completely with excess dilute hydrochloric acid. Calculate the volume of hydrogen produced at room temperature and pressure, taking the molar volume as 24 dm3/mol. Ar(Mg) = 24.
Reasoning: Mg + 2HCl -> MgCl2 + H2. Moles of Mg = 4.8/24 = 0.20 mol. The equation ratio Mg:H2 is 1:1, so moles of H2 = 0.20 mol. Volume = 0.20 x 24 = 4.8 dm3. The acid is in excess, so magnesium is limiting and determines the hydrogen amount.
Traps that cost marks
An unbalanced equation gives wrong ratios. Do not use mass ratios such as 24 g Mg to 2 g H2 without converting to moles. Check whether the question gives r.t.p. or s.t.p. before choosing 24 or 22.4 dm3/mol. In titration calculations, convert cm3 to dm3. Percentage yield = actual yield/theoretical yield x 100; percentage purity compares pure mass with sample mass.
Key points
- Moles = mass/molar mass; use Avogadro's number for particles.
- Balanced equations give mole ratios, not mass ratios.
- Concentration = moles/volume in dm3.
- Limiting reagent controls product amount.
- Use the molar volume stated in the question.