Extraction of copper by smelting uses silica as an additive to remove.

During the extraction of copper from its sulphide ores, the concentrated ore is first roasted to convert most of the $$FeS$$ and a part of $$Cu_2S$$ into their oxides. This produces a molten mass containing mainly $$Cu_2S$$, $$Cu_2O$$ and $$FeO$$, along with some unchanged $$FeS$$.

At this stage the mass is transferred to a smelting furnace. Here we deliberately add powdered silica, which is chemically represented as $$SiO_2$$. First, let us recall a fundamental fact from metallurgy: an acidic oxide such as $$SiO_2$$ will combine readily with a basic oxide to form a fusible salt called slag. The overall goal is to remove unwanted basic oxides in the form of this slag, thereby preventing them from contaminating the impure copper (called matte).

Among the oxides present, $$FeO$$ is strongly basic, whereas $$Cu_2O$$ is only weakly basic and, in addition, is actually useful later because it reacts with $$Cu_2S$$ to yield metallic copper. Therefore our primary target for removal is $$FeO$$.

We thus rely on the following chemical reaction, which expresses exactly how silica removes iron(II) oxide:

$$FeO + SiO_2 \;\rightarrow\; FeSiO_3$$

Here $$FeSiO_3$$ is iron(II) silicate, a fusible, lighter slag that floats over the molten mixture and can be separated off easily. Notice that neither $$FeS$$, $$Cu_2S$$ nor $$Cu_2O$$ reacts with $$SiO_2$$ under the furnace conditions because they are not basic oxides; hence they remain in the melt or, in the case of $$Cu_2O$$, participate in a later self-reduction step.

Therefore, the specific species that silica removes is $$FeO$$.

Hence, the correct answer is Option C.