Lesson overview
Concentration, gas volumes and yield: How much product can you expect?
Drop bigger and bigger pieces of magnesium ribbon into the same 10.0 cm³ of acid and catch the hydrogen in a gas syringe. Plot your five readings, find the point where the acid runs out, and use the equation to explain the volume where the graph levels off.
Suggested sequence
- Lesson 1 · How much gas will you get? Reacting masses, gas volumes at r.t.p. and volumes of solutions. Read together and discuss (15–20 min), practise with self-checking (20–25 min).
- Lesson 2 · When does the acid run out? Limiting reactants and the reactant in excess. Read together (10–15 min), show or explore the experiment (10 min), practise with self-checking (15–20 min).
- Lesson 3 · How much did you really make? Percentage yield, composition and purity; empirical and molecular formulae. Read together and discuss (15–20 min), practise with self-checking (20–25 min).
- End of module · final check. 8 new questions, 20–30 min, without hints or notes; pupils check their answers once they have answered them all.
- One to two weeks later · return questions. 5 questions, 15 min, lesson closed.
Learning goals
Part 1: How much gas will you get?
- Use the molar gas volume, taken as 24 dm³ at room temperature and pressure, r.t.p., in calculations involving gases
- Calculate stoichiometric reacting masses, volumes of gases at r.t.p. and volumes of solutions, including conversion between cm³ and dm³ (3.3.5)
Part 2: When does the acid run out?
- Use the molar gas volume, taken as 24 dm³ at room temperature and pressure, r.t.p., in calculations involving gases
- Calculate stoichiometric reacting masses, volumes of gases at r.t.p. and volumes of solutions, including conversion between cm³ and dm³ (3.3.5)
- Calculate limiting reactants (3.3.5)
Part 3: How much did you really make?
- Use the molar gas volume, taken as 24 dm³ at room temperature and pressure, r.t.p., in calculations involving gases
- Calculate stoichiometric reacting masses, volumes of gases at r.t.p. and volumes of solutions, including conversion between cm³ and dm³ (3.3.5)
- Calculate percentage yield, percentage composition by mass and percentage purity, given appropriate data
- Calculate empirical formulae and molecular formulae, given appropriate data
Misconceptions to listen for
- “One mole of gas is 24 cm³.” — It is 24 dm³, which is 24 000 cm³. For a volume in cm³, use 24 000.
- “A heavier gas takes up more space.” — A mole of any gas takes up 24 dm³ at r.t.p. The particles are far apart, so their size and mass hardly matter.
- “Multiply by the number in front of the substance you want.” — Use the ratio of the two numbers: O2 : H2 is 1 : 2, so halve.
- “The product has the same mass as the reactant.” — Only the total mass is conserved. Go through moles: change to moles, use the equation, change back.
- “The 2 in front of CaO doubles the answer.” — Both sides have a 2: Ca : CaO = 2 : 2 = 1 : 1.
- “Multiply a gas volume by 24 to get moles.” — Each mole takes up 24 dm³, so the number of moles is the volume divided by 24.
- “Use the mole ratio on the volumes of solutions.” — A volume of solution depends on its concentration too. Change to moles, use the equation, then V = n ÷ c.
- “The water makes 80 cm³ of gas too.” — At r.t.p. water is a liquid, so it has no gas volume in the mole ratio.
- “Use the ratio upside down.” — Write the ratio with the substance you want on top: NaN3 : N2 = 2 : 3, so moles of NaN3 = moles of N2 × 2/3.
- “Heavier particles take up more room.” — In a gas the particles are far apart, so their own size and mass hardly change the volume.
Scope
Moles in equations (3.3 Supplement 4, 5, 7 and 8): the molar gas volume of 24 dm³ at r.t.p., reacting masses, volumes of gases and volumes of solutions from equations, limiting reactants, percentage yield, composition by mass and purity, and empirical and molecular formulae. The mole, n = m/M and concentration are in the previous module; titration is left to topic 12.