Question

When copper ore is mixed with silica, in a reverberatory furnace copper matte is produced. The copper matte contains:
A. Sulphides of copper (II) and iron (II)
B. Sulphides of copper (II) and iron (III)
C. Sulphides of copper (I) and iron (II)
D. Sulphides of copper (I) and iron (III)

Answer 1

Hint: Copper is extracted from its ores by various methods. We know the various copper ores include chalcopyrite(\[CuFe{S_2}\]), chalcocite ($C{u_2}S$), etc. Various gangue particles (unwanted materials in the form of impurities) are also present in the ore. It is removed by concentration and then it is heated in a reverberatory furnace.

Complete step by step answer:
As we know the concentration is the removal of the gangue from the ore particles. The various concentration methods include froth floatation, washing, etc.
After the concentration process is done, the concentrated ore is heated with silicon dioxide or silica in the presence of oxygen in a reverberatory furnace.
$2CuFe{S_2} + 2Si{O_2} + 4{O_2} \to C{u_2}S + 2FeSi{O_3} + 3S{O_2}$
Here chalcopyrite when treated with oxygen is called roasting. The copper ion in chalcopyrite is in $ + 2$ an oxidation state. It is reduced to $ + 1$ an oxidation state in $C{u_2}S$.
The iron in \[CuFe{S_2}\], chalcopyrite is in $ + 3$ an oxidation state and is reduced to $ + 2$ in $FeSi{O_3}$.
The copper matte contains sulphides of copper and iron with oxidation state $ + 1$ and $ + 2$ respectively.

So, the correct answer is Option C.

Note: We know that oxidation is the increase in the oxidation number while reduction is the decrease in oxidation number. Here, it is reduced. Roasting is heating in the presence of oxygen whereas heating in the absence of oxygen is calcination. When the copper sulphide formed by the roasting is again heated with the presence of a blast of air, copper is formed. It is known as blister copper which contains porous copper. It can be further purified by electrolytic refining and other refining processes.