Chemical Reactions, Stoichiometry and the Mole
Equations and the mole
A balanced symbol equation has the same number of each type of atom on both sides. The mole is the amount containing 6.02 × 10^23 particles; moles = mass ÷ molar mass (Mr). Concentration = moles ÷ volume in dm3. The limiting reactant is used up first and controls the amount of product. Percentage yield = actual yield ÷ theoretical yield × 100, and atom economy compares the mass of useful product with the total mass of products.
Reaction types and rates
Common reactions include combustion, neutralisation, displacement, precipitation, thermal decomposition and redox. Rate increases with higher concentration, higher temperature, greater surface area and a catalyst. Catalysts lower the activation energy without being used up.
Worked example
Hydrogen burns: 2H2 + O2 → 2H2O. Calculate the mass of water formed from 4 g of hydrogen. Mr of H2 is 2, so moles of H2 = 4 ÷ 2 = 2 mol. The equation shows 2 mol H2 gives 2 mol H2O. Mr of water is 18, so mass = 2 × 18 = 36 g.
Energy changes
Exothermic reactions release heat and raise the temperature; endothermic reactions absorb heat and lower it. Bond breaking absorbs energy and bond making releases energy.
Traps that cost marks
Always balance equations before using mole ratios. Use the Mr of the formula as written, including diatomic gases such as O2 and Cl2. Concentration in mol/dm3 is not the same as mass concentration in g/dm3. Do not call the limiting reactant the one present in the smallest mass; compare moles.