Introduction
In the last lesson you learned how to use a balanced equation to find the mass of one substance from the mass of another. This lesson builds on that skill, adding limiting reactants, percentage yield, and calculations involving concentrations and volumes of solutions. These are the core skills examiners expect for "how much?" questions in Stoichiometry (Talbot, p.418).
Recap: Why equations work for calculations
A balanced equation shows the ratio of moles of reactants and products, not just atoms. This works because mass is always conserved — atoms are rearranged, never created or destroyed (Talbot, p.82). So if we know the moles of one substance, the equation's ratio tells us the moles of everything else.
Step-by-step method
- Write a balanced equation.
- Convert given data into moles (using n = m / M for solids, or n = c × V for solutions).
- Use the mole ratio from the equation to find moles of the unknown.
- Convert back to the units asked for (mass, volume, concentration).
Worked Example 1 – Reacting masses
Calculate the mass of calcium oxide produced when 50.0 g of calcium carbonate is fully decomposed.
CaCO₃(s) → CaO(s) + CO₂(g)
- M(CaCO₃) = 100.09 g mol⁻¹ → n = 50.0 / 100.09 = 0.500 mol
- Mole ratio CaCO₃ : CaO = 1 : 1 → n(CaO) = 0.500 mol
- M(CaO) = 56.08 g mol⁻¹ → m = 0.500 × 56.08 = 28.0 g
Worked Example 2 – Limiting reactant
3.00 g of magnesium reacts with 10.0 cm³ of 2.00 mol dm⁻³ hydrochloric acid.
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
- n(Mg) = 3.00 / 24.31 = 0.123 mol
- n(HCl) = c × V = 2.00 × 0.0100 = 0.0200 mol
- Required ratio Mg : HCl is 1 : 2, so 0.123 mol Mg would need 0.247 mol HCl — but only 0.0200 mol is available.
- HCl is the limiting reactant.
- n(H₂) formed = ½ × n(HCl) = 0.0100 mol → mass = 0.0100 × 2.02 = 0.0202 g
Worked Example 3 – Percentage yield
Using the limiting reactant above, a student actually collects 0.016 g of H₂ gas.
percentage yield = (actual yield / theoretical yield) × 100%
percentage yield = (0.016 / 0.0202) × 100% = 79.2%
Percentage yield is always less than 100% in real experiments because of losses, side reactions, or incomplete reactions.
Concentrations and volumes
For reactions in solution, moles are found using n = c × V (V in dm³). This lets you calculate an unknown concentration or volume using exactly the same mole-ratio method — just substitute n = c × V wherever mass would normally appear.
Key takeaways
- Always balance the equation first — the mole ratio comes from the coefficients.
- Convert all given quantities to moles before using the equation's ratio.
- The limiting reactant runs out first and determines the maximum (theoretical) amount of product.
- Percentage yield compares actual yield to theoretical yield and is almost always below 100%.
- The same method works for masses, volumes of gases, and concentrations/volumes of solutions — only the conversion step changes.