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2.4 Reacting masses and volumes (of solutions and gases)

Syllabus
9701–2028–2029
Topic
2.4
Level
AS

Learning objectives

2.4.1Perform calculations• Perform calculations including use of the mole concept, involving:- (a) reacting masses (from formulas and equations) including percentage yield calculations- (b) volumes of gases (e.g. in the burning of hydrocarbons)- (c) volumes and concentrations of solutions- (d) limiting reagent and excess reagent (When performing calculations, candidates’ answers should reflect the number of significant figures- given or asked for in the question. When rounding up or down, candidates should ensure that significant figures are neither lost unnecessarily nor used beyond what is justified (see also Mathematical requirements section).)- (e) deduce stoichiometric relationships from calculations such as those in 2.4.1(a)–(d)

Use moles as the bridge in every stoichiometric calculation

Start with a balanced equation. Convert each supplied quantity to moles, use the equation coefficients as a mole ratio, then convert the required moles to the requested mass, gas volume or solution quantity. Coefficients relate amounts in moles, not masses directly.

Quantity known Convert to moles Convert back
mass, m n = m ÷ M m = nM
solution concentration, c, and volume, V n = cV, with V in dm³ c = n ÷ V
gas volume, V, at stated conditions n = V ÷ Vₘ V = nVₘ

Use the molar volume stated or justified by the conditions: Vₘ = 24.0 dm³ mol⁻¹ at room conditions, or 22.4 dm³ mol⁻¹ at s.t.p. (101 kPa and 273 K). Convert 1000 cm³ = 1 dm³ before substituting where necessary.

CaCO3+2HCl⟶CaCl2+CO2+H2O\mathrm{CaCO_3 + 2HCl \longrightarrow CaCl_2 + CO_2 + H_2O}

If 25.0 cm³ of 0.200 mol dm⁻³ HCl reacts with excess CaCO₃, n(HCl) = 0.200 × 0.0250 = 0.00500 mol. The 2:1 ratio gives 0.00250 mol CO₂, so at room conditions V(CO₂) = 0.00250 × 24.0 = 0.0600 dm³ = 60.0 cm³.

percentage yield=actual yieldtheoretical yield×100%\mathrm{percentage\ yield=\frac{actual\ yield}{theoretical\ yield}\times100\%}

When two reactant amounts are supplied, compare n ÷ coefficient for each reactant. The smaller value identifies the limiting reagent and fixes the maximum product amount; the other reagent is in excess. Calculate unused excess only after subtracting the amount that reacts.

To deduce an unknown stoichiometric relationship, convert measured quantities to moles and reduce the mole amounts to the simplest justified ratio. Keep unrounded values through the working, then report the final answer to the significant figures given or requested. Check units, conditions, ratio direction and that percentage yield is actual ÷ theoretical.