Chemistry • Year 11 • Module 2 • Lesson 16
Stoichiometry in Solution
Lock in the key vocabulary, the full solution-stoichiometry pathway, and the critical unit-conversion rule before tackling harder calculations.
1. Term–definition match
The definitions below are shuffled. In the right-hand column write the matching term from this list: n = cV, solution stoichiometry, mole ratio, titration stoichiometry, dilution calculation, units consistency, limiting reagent, excess reagent, molar mass, precipitate. 10 marks (1 each)
| # | Definition | Matching term |
|---|---|---|
| 1.1 | The fundamental equation linking concentration (mol L−1) and volume (L) to give moles of solute. | |
| 1.2 | Calculations that use moles derived from the concentration and volume of reacting solutions, combined with stoichiometric mole ratios. | |
| 1.3 | The ratio of stoichiometric coefficients from the balanced equation; used to relate moles of one species to moles of another. | |
| 1.4 | A calculation type where n(acid) × coefficient ratio = n(base) at the equivalence point; used to find unknown concentration. | |
| 1.5 | The relationship c1V1 = c2V2; moles of solute remain constant when a solution is diluted. | |
| 1.6 | The requirement that volume must be in litres and concentration in mol L−1 before substituting into n = cV. | |
| 1.7 | The reactant that is completely consumed first and determines the theoretical yield of products. | |
| 1.8 | The reactant that remains after the limiting reagent is fully consumed. | |
| 1.9 | The mass per mole of a substance (g mol−1); used to convert between moles and mass via m = n × MM. | |
| 1.10 | An insoluble solid that forms and settles out when two solutions containing reacting ions are mixed. |
2. True or false — with correction
Circle T or F for each statement. If the statement is false, write the corrected version on the line below it. 12 marks (1 T/F + 1 correction each)
2.1 When using n = cV, the volume must be converted to litres before calculation. T / F
2.2 In solution stoichiometry, you can use the total combined volume of two mixed solutions to calculate the number of moles of each individual reactant. T / F
2.3 To find the concentration of a product formed when two solutions are mixed, you divide the moles of product by the total combined volume of the two solutions. T / F
2.4 The limiting reagent is always the reactant with the smallest mass in grams. T / F
2.5 25.0 mL converted to litres equals 0.0250 L. T / F
2.6 The full solution-stoichiometry pathway is: moles from n = cV → apply mole ratio → convert to mass or concentration. T / F
3. Fill-in-the-blank paragraph
Use the word bank to complete the passage. Each word is used once. 8 marks (1 per blank)
Word bank:
concentration · litres · limiting · molar mass · mole ratio · moles · total · volume
Solution stoichiometry combines two skills: using n = c × V to find ___________ of solute, and applying the ___________ from the balanced equation to relate moles of reactant to moles of product. Before substituting into n = cV, the ___________ must always be expressed in ___________. When finding the mass of a product, multiply moles by the ___________. When finding the ___________ of a product formed in a mixture, divide moles by the ___________ combined volume of the two solutions. If two reactants are present, first identify the ___________ reagent by comparing n ÷ coefficient for each species; the smaller value indicates the reagent that runs out first.
4. Function recall
Answer each question in 1–2 sentences using precise chemical terms. 8 marks (2 each)
4.1 What is the purpose of writing “V = ___ mL = ___ L” as the very first step in a solution-stoichiometry calculation?
4.2 When two solutions are mixed and a product forms, why must you use the total combined volume when calculating the concentration of that product?
4.3 What is the final step of solution stoichiometry when the question asks “what mass of precipitate forms?”
4.4 Describe in one sentence the method used to identify which of two solution reactants is the limiting reagent.
5. Build a concept map
Draw labelled arrows between the six terms below to show how they connect. Each arrow must carry a linking phrase (e.g. “gives”, “divided by”, “multiplied by”). Aim for at least 6 labelled arrows. 6 marks (1 per valid labelled arrow)
Supplied terms: concentration · volume (L) · n(reactant) · mole ratio · n(product) · mass (g).
Q1 — Term–definition match
1.1 n = cV • 1.2 solution stoichiometry • 1.3 mole ratio • 1.4 titration stoichiometry • 1.5 dilution calculation • 1.6 units consistency • 1.7 limiting reagent • 1.8 excess reagent • 1.9 molar mass • 1.10 precipitate.
Q2 — True / false with correction
2.1 True.
2.2 False. You must calculate moles for each reactant using its own volume and concentration separately: n1 = c1 × V1 and n2 = c2 × V2. Only use total volume when calculating the concentration of a product in the final mixed solution.
2.3 True.
2.4 False. The limiting reagent is the reactant that runs out first based on the mole ratio comparison (n ÷ coefficient), not the smallest mass. A smaller mass can belong to the excess reagent if its molar mass is low or its coefficient is small.
2.5 True.
2.6 True.
Q3 — Cloze paragraph
In order: moles / mole ratio / volume / litres / molar mass / concentration / total / limiting.
Q4.1 — Purpose of writing the mL → L conversion first
n = cV only produces the correct number of moles when V is in litres. Writing the conversion explicitly as the first step prevents the most common error in solution stoichiometry: substituting mL directly, which gives an answer 1000× too large.
Q4.2 — Why total combined volume is used for product concentration
When two solutions are mixed, the product dissolves in the entire resulting solution. The volume of that solution is V1 + V2. Using only one of the original volumes would give a concentration that is too high, because the actual solution is more dilute than either original solution alone.
Q4.3 — Final step when question asks for mass of precipitate
Multiply the moles of the precipitate (n) by its molar mass (MM): m = n × MM. The molar mass is calculated from the periodic table by summing the atomic masses of all atoms in the formula of the precipitate.
Q4.4 — Identifying the limiting reagent
Calculate n = cV for each reactant independently, then divide each by the stoichiometric coefficient from the balanced equation. The reactant with the smaller (n ÷ coefficient) value is the limiting reagent — it runs out first.
Q5 — Sample concept map
Correct maps should include arrows such as:
- concentration — × → n(reactant) (combined with volume)
- volume (L) — × concentration gives → n(reactant)
- n(reactant) — × mole ratio gives → n(product)
- mole ratio — applied to n(reactant) to get → n(product)
- n(product) — × molar mass gives → mass (g)
- n(product) — ÷ total volume gives → concentration
Award 1 mark per valid labelled arrow (minimum 6, maximum 6 marked).