Question

copper matte contains:
E. $C{u_2}S,C{u_2}O$ and silica
F. $C{u_2}S,CuO$ and silica
G. $C{u_2}S,FeO$ and silica
H. $C{u_2}S,FeS$ and silica

Answer 1

Hint: In a reverberatory furnace when copper is extracted from its oxide, two minerals get separated through this procedure. Try to remember the process and find out which minerals obtained after the procedure. Also these elements are present in the period 4 of the periodic table.

Step by step answer:
We can guess from the name that Copper matte is a compound of copper. It is a mixture of two minerals which got after the extraction of copper from its oxide. Matte is the procedure of extraction of copper before the final reduction.

Primarily, it is composed of copper, iron and lead sulphides. See the extraction process of copper. In a reverberatory furnace, when copper is extracted from cuprous oxide, cuprous oxide mixed with silica is heated. This leads to the separation of two sulphide elements named copper sulphide ($C{u_2}S$) and iron sulphide ($FeS$).
$C{u_2}S + FeS$
To remove the remaining $FeO$ and $FeS$ of the matter, it is put in the converter (silica- lined converter), $FeO$ oxidizes to $FeS$. This $FeO$ combines with silica to produce slag ($FeSi{O_3}$).

\[\begin{gathered}
  2FeS + 3{O_2} \to 2FeO + 2S{O_2} \\
FeO + Si{O_2} \to FeSi{O_3} \\
\end{gathered} \]

Thus, the correct answer is option D. i.e. $C{u_2}S,FeS$ and silica.

Additional information:
The composition of this matter includes copper sulphide (80%-90%), 5%-17% mixture of the alloy of $Fe - Co$, metallic copper etc. Thus we can say copper matte is a mixture of a large amount of copper sulphide and a small amount of iron sulphide.

Note: Remember the elements obtained during extraction of copper from cuprous oxide in a reverberatory furnace. Quantity of copper sulphide is very large but other alloy elements present in small quantities.