Identify `(B)` and `( C)` :

Options

A. `FeS + FeO`
B. `FeO + Cu_(2) S`
C. `FeS + Cu_(2) O`
D. `Cu_(2) O + Cu_(2) S`

Answer

B

Explanation

`underset((A)) (2CuFeS_(2) + O_(2)) overset ("Roasting") rarr Cu_(2)S + 2FeS + SO_(2)`
`Cu_(2) S` and `FeS` are further partially oxidised.
`2Cu_(2) S + 3O_(2) rarr 2Cu_(2) O + 2SO_(2)`
`2FeS + 3O_(2) rarr 2FeO + 2SO_(2)`
At iron is more reactive than copper, `FeS` is preferentially oxidised to `FeO` than `Cu_(2)S` to `Cu_(2)O`. If at all any `Cu_(2)O` is formed, it reacts with `FeS` to give `Cu_(2) S` again.
`Cu_(2)O + FeS rarr Cu_(2) S + FeO`
Hence, `(B)` and `( C)` are `FeO` and `Cu_(2)S`.
On smelting `(B) + (C)`, i.e., heating with `SiO_(2)` (flux) and soem powdered coke, `FeO` combines with silica to form fusible ferrous silicate slag.
`FeO +SiO_(2) rarr underset("Ferrous silicate (slag)")(FeSiO_(3))`
Slag being lighter, floats on the lower layer which comprises of copper matte. Copper matte mainly consists of `Cu_(2) S` and some unchanged `FeS`.
Copper matte is separated from slag, and on electrolysis gives pure `Cu` metal. Hence,
`(B)` and `(C)` are `FeO` and `Cu_(2) S` respectively.