Question

Extraction of copper from copper pyrite $\left( {CuFe{S_2}} \right)$ involves :
A. Crushing followed by concentration of the ore by froth – flotation
B. Removal of iron as slag
C. Self-reduction step to produce ‘blister copper’ following evolution of $S{O_2}$
D. Refining of ‘blister copper’ by carbon reduction

Answer 1

. In this particular question write the complete process for the extraction of copper from copper pyrite with reactions involved in each step because from that we will know which of the given steps are involved as the question is multi choice.

Complete step by step answer:
Process for extraction of copper from copper pyrite
At the starting copper pyrite is roasted in a reverberatory furnace because on roasting we will get copper oxide from the copper pyrite from which the copper can be extracted by doing reduction.
$2CuFe{S_2} + {O_2} \to C{u_2}S + 2FeS + S{O_2}$

Now as we know that the iron is more reactive than the copper.
So, ferrous sulphide and cuprous sulphide must react with oxygen to form ferrous oxide, copper oxide and sulphur dioxide.
$2FeS + 3{O_2} \to 2FeO + 2S{O_2}$
$C{u_2}S + {O_2} \to C{u_2}O + Fe{O_2}$
$C{u_2}O + FeS \to C{u_2}S + FeO$

Now ferrous sulphide is not allowing cuprous sulphide to oxidise as iron is more reactive.
So, compounds left in the furnace will be,
$C{u_2}S$, $FeO$ and some unreacted $FeS$.
So, now we are smelting the roasted ore in the furnace.
So, we put roasted ore in a blast furnace with coke and flux (silica flux).
So, the reaction in the blast furnace will be,
$FeO + Si{O_2} \to FeSi{O_3}\left( {{\text{slag}}} \right)$

As the temperature of the blast furnace is very high. So, everything changes to molten form
So, there will be two molten layers
First layer will be of slag because it is light, and the second layer will be of Copper matte ($C{u_2}S$ and some unchanged
Now copper matte is put in a Bessemer converter to recover copper from copper matte and remove all impurities and this process is known as Bessemerisation process.
So, impurities like sulphur, arsenic and antimony are oxidised to their respective volatile oxides.

$$FeS + {O_2} \to FeO + S{O_2} \\\ FeO + Si{O_2} \to FeSi{O_3}\left( {{\text{slag}}} \right) \\\$$

As the slag is floating on the top. So, here slag is removed. So, iron has been removed as slag.
Now oxidation of copper sulphide starts to form cuprous oxide and sulphur dioxide.
And then cuprous oxide reacts with copper sulphide to form copper metal and this process is called the auto reduction process.

$$2C{u_2}S + 3{O_2} \to 2C{u_2}O + 2S{O_2} \\\ 2C{u_2}O + C{u_2}S \to 6Cu + S{O_2} \\\$$

Now copper metal is below. So, the Bessemer converter is tilted so that it falls in sand mounds and then is left to solidify and cool.
Now sulphur dioxide starts removing but some are trapped inside.
So, the appearance of the solidified copper will be like blisters.
Therefore the copper we get from here is known as blister copper.
So, we do electrolytic refining of that copper.
So, in anode there will be blister copper and pure copper will be in cathode.
So, after refining, pure copper starts collecting at cathode.
So, at the end of the process we will get the pure copper.
Now as we can see from the above process that the,
Correct option will be A, B and C.
But D is not correct because refining of blister copper is done by electrolytic refining not by carbon reduction.
So, the correct answer is “Option A, B and C”.

Note: Whenever we come up with this type of problem then first, we had to write the complete process and then check which steps are involved in the process and we should remember that in copper extraction process at last when we have blister copper than we change that to the pure copper by electrolytic refining not by copper reduction and they both are two different process and here copper reduction cannot occur. So, refining of copper by copper reduction will be an incorrect option.