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?

Mar 14, 2017

We require $\cong 732 \cdot g$ ore...................

Explanation:

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

$C {u}_{2} S \left(s\right) + {O}_{2} \left(g\right) \rightarrow 2 C u \left(s\right) + S {O}_{2} \left(g\right)$

Sulfur is oxidized; copper and oxygen are reduced.

$\text{Moles of copper} = \frac{650 \cdot g}{63.55 \cdot g \cdot m o {l}^{-} 1} = 10.3 \cdot m o l$.

Given the equation, there is HALF AN EQUIVALENT of copper ore, i.e. $5.11 \cdot m o l$ of $C {u}_{2} S$ are required.

And this represents a mass of $5.1 \cdot m o l \times 143.1 \cdot g \cdot m o {l}^{-} 1$

$\cong 732 \cdot g$

I think the ore is the mineral $\text{chalcosite}$, which is approx. 80% metal by mass. Because of the high metal content, this is one of the most valuable copper ores.