Calculate the amount, in mol, of H₂SO₄.

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

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

Calculate the amount of excess acid present.

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

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

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

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]