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Date	May 2018	Marks available	1	Reference code	18M.2.sl.TZ2.1
Level	SL	Paper	2	Time zone	TZ2
Command term	Formulate	Question number	1	Adapted from	N/A

Question

A student determined the percentage of the active ingredient magnesium hydroxide, Mg(OH)₂, in a 1.24 g antacid tablet.

The antacid tablet was added to 50.00 cm³ of 0.100 mol dm⁻³ sulfuric acid, which was in excess.

Calculate the amount, in mol, of H₂SO₄.

[1]

Formulate the equation for the reaction of H₂SO₄ with Mg(OH)₂.

[1]

The excess sulfuric acid required 20.80 cm³ of 0.1133 mol dm⁻³ NaOH for neutralization.

Calculate the amount of excess acid present.

[1]

Calculate the amount of H₂SO₄ that reacted with Mg(OH)₂.

[1]

Determine the mass of Mg(OH)₂ in the antacid tablet.

[2]

Calculate the percentage by mass of magnesium hydroxide in the 1.24 g antacid tablet to three significant figures.

[1]

Markscheme

n(H₂SO₄) «= 0.0500 dm³ × 0.100 mol dm^–3» = 0.00500/5.00 × 10^–3«mol»

[1 mark]

H₂SO₄(aq) + Mg(OH)₂(s) → MgSO₄(aq) + 2H₂O(l)

Accept an ionic equation.

[1 mark]

«n(H₂SO₄) = $\frac{1}{2}$ × n(NaOH) = $\frac{1}{2}$ (0.02080 dm³ × 0.1133 mol dm^–3)»

0.001178/1.178 × 10^–3 «mol»

[1 mark]

n(H₂SO₄) reacted «= 0.00500 – 0.001178» = 0.00382/3.82 × 10^–3 «mol»

[1 mark]

n(Mg(OH)₂) «= n(H₂SO₄) =» = 0.00382/3.82 × 10^–3 «mol»

m(Mg(OH)₂) «= 0.00382 mol × 58.33 g mol^–1» = 0.223 «g»

Award [2] for correct final answer.

[2 marks]

% Mg(OH)₂ «= $\frac{{0.223{\text{ g}}}}{{1.24{\text{ g}}}}$ × 100» = 18.0 «%»

Answer must show three significant figures.

[1 mark]

Examiners report

[N/A]

Syllabus sections

Core » Topic 1: Stoichiometric relationships » 1.1 Introduction to the particulate nature of matter and chemical change

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