S5 Chemistry: Ionic Precipitation (Paper 3: Challenge)

2026-11-12

Ten challenging items: redox or not, limiting reagent, remaining ion concentration, reagent volume and percentage yield, with answers and notes.

Direct answers　Answers in order: not redox, the reactant used up first, treat as a precipitate, 0.00250 mol, 0.359 g, about 0.0167 M, 25 cm3, 2.00 g, 90.0 per cent, chloride, a slight excess is enough.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10
Suggested time	20
Difficulty mix	Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1
Chapter	Chapter: S5 Chemistry: Ionic Precipitation
Level	S5

02　Questions (10 items)

FoundationIs a precipitation reaction a redox reaction? Why?
FoundationWhat is meant by the limiting reagent?
FoundationCalcium sulphate is slightly soluble. How should it be treated in ionic equations?
IntermediateWhen 25.0 cm3 of 0.10 M silver nitrate is mixed with 35.0 cm3 of 0.10 M sodium chloride, how many moles of silver chloride form?
IntermediateFrom the previous question, what mass of silver chloride forms, using Mr = 143.5?
IntermediateFrom that mixture, with a total volume of 60.0 cm3, what is the concentration of the remaining chloride ions?
ChallengeWhat volume of 0.20 M sodium sulphate solution is needed to precipitate 0.0050 mol of barium ions completely?
Challenge2.22 g of calcium chloride, Mr = 111, is dissolved and treated with excess sodium carbonate. What mass of calcium carbonate, Mr = 100, is expected?
ChallengeFrom the previous question, if only 1.80 g of calcium carbonate is actually collected, what is the percentage yield?
Error redoA student argues that adding more precipitating agent always gives a more complete reaction, so more is better. How should this be corrected?

03　Answers and one-line notes

#	Answer	One-line note
1	No. It is only an exchange of ions, so the oxidation number of every element is unchanged.	Redox requires electron transfer and a change in oxidation number, while precipitation only recombines ions.
2	The reactant that is used up first and therefore fixes the maximum amount of product.	The other reactant is left in excess and the surplus takes no further part in the reaction.
3	Treat it as a precipitate: a white calcium sulphate precipitate forms.	Slightly soluble substances still precipitate from reasonably concentrated solutions, so in test-tube work they are treated as insoluble.
4	0.00250 mol	There are 0.00250 mol of silver ions and 0.00350 mol of chloride ions, so silver is limiting and 0.00250 mol forms.
5	0.359 g	0.00250 times 143.5 equals 0.35875 g, which rounds to 0.359 g.
6	About 0.0167 M	0.00100 mol of chloride remains and dividing by 0.0600 dm3 gives 0.0167 mol dm-3.
7	25 cm3	Barium and sulphate react one to one, so 0.0050 mol of sulphate is needed; dividing by 0.20 M gives 0.025 dm3 or 25 cm3.
8	2.00 g	There is 0.0200 mol of calcium chloride and calcium gives calcium carbonate one to one, so 0.0200 mol or 2.00 g forms.
9	90.0 per cent	Percentage yield is actual mass divided by theoretical mass, and 1.80 over 2.00 gives 0.900.
10	Only a slight excess is needed; a large excess wastes reagent, raises cost and contaminates the product.	Completeness depends on solubility and the limiting reagent, so blind excess is neither economical nor helpful.

04　What to do after marking

After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
Repeat the same item two days later; move to the next paper only when both attempts are correct.
Parents only need the answer and note columns; there is no need to explain every item.

Use one set of data to work out the limiting reagent, the mass of product and the remaining concentration, then check the three answers agree.Tutor Registration →

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