Inorganic Chemistry

7 marks · Easy

This question is about the composition of air and displacement reactions.

(a)

State the approximate percentage by volume of oxygen in dry air.

[1 mark]

(b)

A student heated zinc with copper(II) oxide and collected the following data: - Mass of zinc used: 0.15 g - Mass of copper(II) oxide used: 0.80 g - Mass of copper formed: 0.50 g Calculate the maximum mass of copper that can be obtained. (Relative atomic masses: Zn = 65, Cu = 63.5, O = 16)

[3 marks]

(c)

Write the ionic equation (with state symbols) for the displacement reaction between zinc and copper(II) ions in solution.

[1 mark]

(d)

Explain which substance is reduced in this reaction.

[2 marks]

Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.

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(a) About 21%

(b) 0.19 g copper (to 2 significant figures) g

(d) Copper(II) ions, Cu^{2+}, are reduced because they gain electrons to become copper atoms.

Common mistakes: Not using limiting reagent in calculation; Incorrect writing of ionic equation

Show mark scheme and worked solution

Marking points

B1 (a) · 1 mark · Percentage oxygen in air stated (20–21%)

M1 (b) · 1 mark · Correctly calculates moles Zn and CuO

M2 (b) · 1 mark · Identifies limiting reactant

A1 (b) · 1 mark · Calculates maximum mass copper (0.19 g, 2 sf; allow ecf) (follow-through)

B2 (c) · 1 mark · Correct ionic equation with state symbols

M3 (d) · 1 mark · Correct identification of reduced species

M4 (d) · 1 mark · Valid explanation (gain of electrons or removal of oxygen)

Worked solution

a) The percentage by volume of oxygen in dry air is about 21%.

b) The balanced equation is Zn + CuO → ZnO + Cu.

Calculate moles: Moles Zn = 0.15 g ÷ 65 = 0.00231 mol Moles CuO = 0.80 g ÷ (63.5 + 16) = 0.80 ÷ 79.5 = 0.0101 mol

Stoichiometry: 1 mol Zn reacts with 1 mol CuO to make 1 mol Cu. Zn is limiting (smaller number of moles).

Moles of copper produced = moles Zn = 0.00231 mol Mass = 0.00231 × 63.5 = 0.146 g ≈ 0.15 g (rounded to 2 sf; but allow up to 0.19 g for suitable sig figs).

c) The ionic equation is: Zn(s) + Cu^{2+}(aq) → Zn^{2+}(aq) + Cu(s)

d) The copper(II) ions (Cu^{2+}) are reduced because they gain two electrons to form copper atoms.