CHEM10007 · Fundamentals Of Chemistry
Stoichiometry, Solutions & Gases
This is the most calculation-heavy and most-tested chapter. You master the mass-mass stoichiometry roadmap, find limiting reagents and percentage yields, and report answers to the correct significant figures. You then handle solutions (molarity c = n/V, dilution and solution stoichiometry) and gases (the gas laws, partial pressures, and the ideal-gas equation pV = nRT with molar volume). These archetypes dominate Section B.
What this chapter covers
- 01Mass-mass roadmap: mass A → n(A) ÷ M → n(B) × mole ratio → mass B × M
- 02Limiting reagent: divide each reactant's moles by its coefficient; the smallest is limiting
- 03Percentage yield = (actual yield / theoretical yield) × 100
- 04Significant figures: × ÷ take the fewest sig figs; + − take the fewest decimal places
- 05Solubility (g per 100 g water); saturated, unsaturated and supersaturated solutions
- 06Molarity c = n/V (mol L−1); dilution c1V1 = c2V2; solution/titration stoichiometry
- 07Gas laws: Boyle (p ∝ 1/V), Charles (V ∝ T), combined p1V1/T1 = p2V2/T2
- 08Ideal gas pV = nRT (T in kelvin), Dalton's partial pressures, molar volume (24.8 L mol−1 at SLC), gas stoichiometry
Limiting reagent and mass of precipitate
- 1 mark — balanced equation with statesBalanced equation: BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq).
- 1 mark — moles of each reactantn(BaCl2) = 18.0 ÷ 208.2 = 0.0864 mol; n(Na2SO4) = 22.0 ÷ 142.0 = 0.155 mol.
- 1 mark — correct limiting reagentDivide by coefficients (both 1): 0.0864 vs 0.155, so BaCl2 is the limiting reagent.
- 1 mark — moles of productMole ratio BaCl2 : BaSO4 = 1 : 1, so n(BaSO4) = 0.0864 mol.
- 1 mark — mass with correct sig figsm(BaSO4) = 0.0864 × 233.4 = 20.2 g (3 sig figs).
Key terms
- Limiting reagent
- The reactant that is fully consumed first and therefore determines the maximum amount of product; found by dividing each reactant's moles by its coefficient and taking the smallest.
- Percentage yield
- The actual yield as a percentage of the theoretical (stoichiometric) yield: %yield = (actual / theoretical) × 100. Always ≤ 100 % in practice.
- Molarity (c)
- Concentration in moles of solute per litre of solution, c = n/V (mol L−1). The basis of n = cV used in titration arithmetic.
- Ideal gas equation
- pV = nRT, relating pressure, volume, moles and temperature, with R = 8.314 L kPa K−1 mol−1. Temperature must be in kelvin.
- Molar volume
- The volume occupied by one mole of an ideal gas: 24.8 L mol−1 at SLC (25 °C, 100 kPa) and 22.7 L mol−1 at STP (0 °C, 100 kPa).
Stoichiometry, Solutions & Gases FAQ
How do I decide which reactant is limiting?
Convert each reactant's mass to moles, then divide each by its balancing coefficient. The reactant giving the smallest value is limiting; use its moles for all product calculations. The other reactant is in excess.
Which value of R should I use in pV = nRT?
Match R to your units. Use R = 8.314 L kPa K−1 mol−1 with pressure in kPa, or R = 0.08206 L atm K−1 mol−1 with pressure in atm. Volume is in litres and temperature must always be in kelvin (K = °C + 273.15).
How do significant figures work in multi-step problems?
Carry extra digits through intermediate steps and round only at the end. For × and ÷ the final answer takes the fewest sig figs among the inputs; for + and − it takes the fewest decimal places. This is explicitly marked.
Exam move
This chapter rewards repetition: work through limiting-reagent, percentage-yield, molarity and gas-law problems until the roadmap is reflexive. Make two habits automatic — temperature in kelvin and volume in litres — because most lost marks here are unit errors, not chemistry errors. Practise quoting significant figures out loud as you go, since Section B awards a dedicated mark for them, and use the supplied formula sheet to choose the right relationship rather than guessing.
