Topic 3 Practice Paper: Stoichiometry — mark scheme (print)

QuizQuipPractice Series

QQ/IGCSE-CHEM/T3

IGCSE · Cambridge IGCSE Chemistry (Extended (Core + Supplement))

Mark scheme

Stoichiometry

Topic 3 Practice Paper · Maximum mark: 30

Notes for markers

Award one mark for each point marked [1], up to the marks shown for the part.
Accept answers worded differently that show the same understanding.
Units are required where the answer line shows one. ecf: allow an error carried forward.

Section A: Multiple-choice questions[10 marks]

Question	Answer
1	ARelative formula mass = 23 + 35.5 = 58.5.	1
2	BAluminium ions carry a 3+ charge and oxide ions a 2− charge, so the ratio is two to three.	1
3	BAmount = mass ÷ molar mass = 20 ÷ 100 = 0.20 mol.	1
4	DVolume = amount × molar gas volume = 3.0 × 24 = 72 dm³.	1
5	CConcentration = amount ÷ volume = 0.50 ÷ 2.0 = 0.25 mol/dm³.	1
6	DMᵣ of NH₃ = 17 and nitrogen contributes 14, so the percentage is (14 ÷ 17) × 100 = 82%.	1
7	ADividing each subscript by six gives the simplest ratio, CH₂O.	1
8	CMass = amount × molar mass = 0.50 × 18 = 9.0 g.	1
9	AHalving the nitrogen halves the ammonia, so the mass is 34 ÷ 2 = 17 g.	1
10	DThe hydrogen ion from the acid and the hydroxide ion from the alkali combine to form water, shown as H⁺ + OH⁻ → H₂O.	1

Section B: Structured questions[20 marks]

Question	Answer and marking points	Marks
11(a)	hydrogen [1]	1
11(b)	It has the same number of atoms of each element on both sides of the arrow. same number of atoms of each element on both sides [1]	1
11(c)	Mᵣ = 1 + 35.5 = 36.5. 1 + 35.5 [1] = 36.5 [1]	2
	Total for Question 11	4
12(a)	Mass = 0.050 × 80 = 4.0 g. mass = 0.050 × 80 [1] = 4.0 g [1]	2
12(b)	0.050 mol, because the equation shows a one-to-one ratio. 0.050 mol (1:1 ratio) [1]	1
12(c)	Mass = 0.050 × 160 = 8.0 g. mass = 0.050 × 160 [1] = 8.0 g [1]	2
12(d)	So that all of the copper(II) oxide reacts; the oxide is the limiting reagent. so that all the copper(II) oxide reacts [1]	1
	Total for Question 12	6
13(a)	Amount = 0.100 × (20.0 ÷ 1000) = 0.00200 mol. amount = 0.100 × 20.0/1000 [1] = 0.00200 mol [1]	2
13(b)	The ratio is one to one, so 0.00200 mol of HCl is needed: volume = 0.00200 ÷ 0.100 = 0.0200 dm³ = 20.0 cm³. volume = 0.00200 ÷ 0.100 = 0.0200 dm³ [1] = 20.0 cm³ [1]	2
13(c)	pH 7 [1]	1
	Total for Question 13	5
14(a)	Volume = 2.0 × 24 = 48 dm³. volume = 2.0 × 24 [1] = 48 dm³ [1]	2
14(b)	Percentage yield = (51 ÷ 68) × 100 = 75%. (51 ÷ 68) × 100 [1] = 75% [1]	2
14(c)	The reaction is reversible, so not all the reactants are converted, or some ammonia is lost during separation. reaction is reversible / some product is lost during separation [1]	1
	Total for Question 14	5

— End of mark scheme —