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?

Md Ariful Islam

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Md Ariful Islam

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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.

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