Question

Pure copper may be produced by the reaction of copper(I) sulfide with oxygen gas as follows: Cu2S(s) + O2(g)  2Cu(s) + SO2(g) If 0.680 kg of copper(I) sulfide reacts with excess oxygen, what mass of copper metal may be produced ? A) 0.680 kg B) 0.136 kg C) 0.271 kg D) 0.543 kg E) 1.36 kg

Answer 1

Answer:

The correct answer is option D.

Explanation:

$Cu_2S(s) + O_2(g)\rightarrow 2Cu(s) + SO_2(g)$

Mass of copper(I) sulfide = 0.680 kg = 680 g

1 kg = 1000 g

Moles of copper (I) sulfide = $\frac{680 g}{159 g/mol}=4.277 mol$

According to reaction,1 mole of copper(I) sulfide gives 2 moles of copper,then 4.277 moles of copper(I) sulfide will give :

$\frac{2}{1}\times 4.277 mol=8.554 mol$ copper

Mass of 8.554 moles of copper:

8.554 mol × 63.5 g/mol = 543 g  = 0.543 kg

0.543 kg is mass of copper metal may be produced.