Question

Pure copper may be produced by the reaction of copper(I) sulfide with oxygen gas as follows: `Cu_2S(s) + O_2(g) -> 2Cu(s) + SO_2(g)`. What mass of copper(I) sulfide is required in order to prepare 0.650 kg of copper metal?

Answer 1

We require `~=732*g` ore...................

Explanation:

You have given the stoichiometric equation that represents reduction of cuprous sulfide:

`Cu_2S(s) + O_2(g) rarr 2Cu(s) + SO_2(g)`

Sulfur is oxidized; copper and oxygen are reduced.

`"Moles of copper"=(650*g)/(63.55*g*mol^-1)=10.3*mol`.

Given the equation, there is HALF AN EQUIVALENT of copper ore, i.e. `5.11*mol` of `Cu_2S` are required.

And this represents a mass of `5.1*molxx143.1*g*mol^-1`

`~=732*g`

I think the ore is the mineral `"chalcosite"`, which is approx. 80% metal by mass. Because of the high metal content, this is one of the most valuable copper ores.