DP Chemistry (last assessment 2024)

Question 20N.2.hl.TZ0.1b(i)

b(i).

[Maximum mark: 1]

20N.2.hl.TZ0.1b(i)

Chlorine undergoes many reactions.

$2.67 g$ of manganese(IV) oxide was added to $200.0 {cm}^{3}$ of $2.00 mol {dm}^{- 3} HCl$ .

${MnO}_{2} (s) + 4 HCl (aq) \to {Cl}_{2} (g) + 2 H_{2} O (l) + {MnCl}_{2} (aq)$

(b(i))

Calculate the amount, in $mol$ , of manganese(IV) oxide added.

[1]

Markscheme

$« \frac{2.67 g}{86.94 g {mol}^{- 1}} = » 0.0307 « mol »$ ✔

Examiners report

Stoichiometric calculations were generally well done and over 90% could calculate mol from a given mass.

Syllabus sections