Stoichiometry
Stoichiometry
This domain tests quantitative relationships in reactions: the mole, molar mass, limiting reagents, percentage yield, empirical formulas and solution calculations. You must show a clear unit-bearing calculation.
Core rules
- n = m/M, where M is molar mass in g mol^-1.
- One mole contains 6.0 × 10^23 particles. At STP (273 K, 1 × 10^5 Pa), one mole of an ideal gas occupies 22.4 L.
- For solutions, n = cV with c in mol L^-1 and V in litres.
- Use the balanced equation to convert moles of one substance to moles of another.
- The limiting reagent is the reactant that produces the least product, not necessarily the one with the smallest mass.
- Percentage yield = actual yield/theoretical yield × 100.
Worked example
5.00 g of magnesium reacts with 100 mL of 1.00 M HCl: Mg + 2HCl → MgCl2 + H2. n(Mg) = 5.00/24.3 = 0.206 mol. n(HCl) = 1.00 × 0.100 = 0.100 mol. HCl needs 0.200 mol Mg for complete reaction, so HCl is limiting. From 2HCl : 1H2, n(H2) = 0.0500 mol. Volume at STP = 0.0500 × 22.4 = 1.12 L.
Traps
Do not compare masses directly to find the limiting reagent. Convert to moles and use coefficients. Keep volume in litres in cV; 100 mL is 0.100 L. If a question gives a gas volume not at STP, use PV = nRT rather than 22.4 L.