Inorganic Chemistry

7 marks · Easy

You will use combustion and displacement to answer the following questions.

(a)

A student combusts magnesium in pure oxygen and collects 4.8 g of magnesium oxide as product. Write a balanced equation for this reaction.

[1 mark]

(b)

Calculate the minimum mass of magnesium that could have reacted to produce 4.8 g of magnesium oxide (MgO). [Mg = 24, O = 16]

[2 marks]

(c)

If the products from part (a) are mixed with copper(II) oxide and heated, a reaction occurs. Write an equation for the displacement reaction.

[1 mark]

(d)

State which metal is reduced in this reaction, and explain your answer.

[1 mark]

(e)

Calculate the maximum mass of copper that can be formed if 4.8 g of MgO (from part (a)) originally came from enough Mg, and all the Mg is used to displace copper from CuO. [Cu = 64]

[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.

Show answer

(a) 2Mg(s) + O2(g) -> 2MgO(s)

(b) 2.88 g

(d) Copper is reduced because it gains electrons and forms copper metal from copper(II) oxide.

(e) 4.8 g MgO contains 2.88 g Mg. That produces 2.88 g Cu (from stoichiometry, 1:1 molar). Maximum mass = 2.88 g. g

Common mistakes: forgetting the 1:1 molar ratio for MgO and Mg; misidentifying reduced species; missing units

Show mark scheme and worked solution

Marking points

A1 (a) · 1 mark · Writes correct balanced equation for magnesium combustion

B1 (b) · 1 mark · Uses molar mass of MgO to find moles

B2 (b) · 1 mark · Calculates and states the correct mass of magnesium (follow-through)

C1 (c) · 1 mark · Gives correct equation for displacement of CuO by Mg

D1 (d) · 1 mark · Identifies copper as reduced and explains electron gain

E1 (e) · 1 mark · Correctly calculates moles or ratio for copper produced (follow-through)

E2 (e) · 1 mark · Correctly gives final mass (2.88 g) (follow-through)

Worked solution

a) 2Mg(s) + O2(g) → 2MgO(s)

b) Mr(MgO) = 24 + 16 = 40, moles MgO = 4.8 / 40 = 0.12 mol. Moles Mg needed = 0.12 mol. Mass = 0.12 × 24 = 2.88 g

c) Mg(s) + CuO(s) → MgO(s) + Cu(s)

d) Copper is reduced because it is converted from CuO (oxidation state +2) to Cu(0) by gaining electrons.

e) From (b) there is 0.12 mol Mg. 1 mol Mg produces 1 mol Cu, so 0.12 mol Cu is formed. Mass Cu = 0.12 × 64 = 7.68 g