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Date May 2018 Marks available 2 Reference code 18M.2.sl.TZ2.1
Level SL Paper 2 Time zone TZ2
Command term Determine 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]
a.

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

[1]
b.

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]
c.

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

[1]
d.

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

[2]
e.

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

[1]
f.

Markscheme

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

[1 mark]

a.

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

 

Accept an ionic equation.

[1 mark]

b.

«n(H2SO4) = 1 2 × n(NaOH) = 1 2  (0.02080 dm3 × 0.1133 mol dm–3)»

0.001178/1.178 × 10–3 «mol»

[1 mark]

c.

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

[1 mark]

d.

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]

e.

% Mg(OH)2 «= 0.223  g 1.24  g × 100» = 18.0 «%»

 

Answer must show three significant figures.

[1 mark]

f.

Examiners report

[N/A]
a.
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b.
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c.
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d.
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e.
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f.

Syllabus sections

Core » Topic 1: Stoichiometric relationships » 1.2 The mole concept
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