Chemical Reactions and Stoichiometry
This domain tests the mole as the currency of chemistry: balancing equations, identifying the limiting reagent, calculating yields and atom economy, and handling concentrations, dilutions and gas volumes confidently.
Core relationships
- n = m/M; n = cV with V in dm3; n = V/24.0 for a gas at room temperature and pressure, using the value given in the data provided.
- One mole contains 6.02 × 1023 particles.
- The balanced equation supplies the molar ratio; never use the ratio from an unbalanced equation.
- For solutions, c in mol dm−3 = mol/dm3; convert cm3 to dm3 by dividing by 1000.
Limiting reagents
- Convert every reactant to moles, divide each by its coefficient, and the smallest quotient is limiting.
- Product amount is fixed by the limiting reagent; the excess reagent remains unreacted and can be calculated by subtraction.
Yield and atom economy
- Percentage yield = actual mass or moles ÷ theoretical × 100; it measures how much of the expected product was obtained.
- Atom economy = Mr of desired product ÷ total Mr of all reactants × 100; it measures how much reactant mass ends up in the desired product, independent of how well the reaction runs.
Dilution and titration
- Dilution: c1V1 = c2V2 because moles of solute are unchanged.
- Titration: moles from cV, then use the equation ratio to reach the unknown.
Worked example
4.60 g of sodium (0.200 mol) reacts with 8.00 g of chlorine (0.113 mol) in 2Na + Cl2 → 2NaCl. The sodium needs 0.100 mol of chlorine, which is available, so sodium is limiting. n(NaCl) = 0.200 mol, so m = 0.200 × 58.5 = 11.7 g. The chlorine is in excess by 0.013 mol.
Traps that cost marks
- Forgetting to balance before applying a mole ratio.
- Quoting product mass from the excess reagent, which always overestimates the yield.
- Adding masses or volumes of solutions instead of moles when deciding which is limiting.
- Omitting water of crystallisation from a hydrated molar mass.
- Treating atom economy and percentage yield as the same quantity.