If 3-kg of copper (I) sulfide reacts with excess oxygen, 3.9 kg of copper metal may be produced.
Let's consider the following balanced equation for the obtaining of pure copper.
Cu₂S(s) + O₂(g) ⇒ 2 Cu(s) + SO₂(g)
The molar mass of Cu₂S is 95.61 g/mol. The moles corresponding to 3 kg (3000 g) of Cu₂S are:
[tex]3000 g \times \frac{1mol}{95.61g} = 31 mol[/tex]
The molar ratio of Cu₂S to Cu is 1:2. The moles of Cu obtained from 31 moles of Cu₂S are:
[tex]31 mol Cu_2S \times \frac{2molCu}{1molCu_2S} =62 mol Cu[/tex]
The molar mass of Cu is 63.55 g/mol. The mass corresponding to 62 moles of Cu is:
[tex]62 mol \times \frac{63.55g}{mol} = 3.9 \times 10^{3} g = 3.9 kg[/tex]
If 3-kg of copper (I) sulfide reacts with excess oxygen, 3.9 kg of copper metal may be produced.
You can learn more about stoichiometry here: https://brainly.com/question/22288091
