Removal of Iron from Chalcopyrite Ore

To determine how iron is removed during the extraction of copper from chalcopyrite ($\text{CuFeS}_2$), let us analyze the chemical reactions involved in the metallurgical process:

1. Partial Roasting:
   When chalcopyrite is heated in a reverberatory furnace in the presence of air, it undergoes partial oxidation:
   $2\text{CuFeS}_2 + \text{O}_2 \xrightarrow{\Delta} \text{Cu}_2\text{S} + 2\text{FeS} + \text{SO}_2 \uparrow$

   The ferrous sulfide ($\text{FeS}$) formed is further partially oxidized to iron(II) oxide ($\text{FeO}$):
   $2\text{FeS} + 3\text{O}_2 \xrightarrow{\Delta} 2\text{FeO} + 2\text{SO}_2 \uparrow$

2. Removal of Iron (Slag Formation):
   To remove iron as an impurity, silica ($\text{SiO}_2$) is added as an acidic flux during the smelting process. The basic iron oxide ($\text{FeO}$) reacts with the flux ($\text{SiO}_2$) to form iron(II) silicate ($\text{FeSiO}_3$), which acts as a slag:
   $\text{FeO (s)} + \text{SiO}_2 \text{ (s)} \xrightarrow{\Delta} \underset{\text{Slag}}{\text{FeSiO}_3 \text{ (l)}}$

   Since the slag ($\text{FeSiO}_3$) is lighter than the molten matte ($\text{Cu}_2\text{S} + \text{FeS}$), it floats on the top surface and is easily skimmed off and removed.

Therefore, iron is removed from chalcopyrite as slag, which is $\text{FeSiO}_3$.

Correct Option: D