
Iron is removed from chalcopyrite as
(A) FeO
(B) FeS
(C) $F{{e}_{2}}{{O}_{3}}$
(D) $FeSi{{O}_{3}}$
Answer
552.3k+ views
Hint: Roasting is a process of heating of sulphide ore to a high temperature in presence of air. It is a step of the processing of certain ores. More specifically, roasting is a metallurgical process involving gas–solid reactions at elevated temperatures with the goal of purifying the metal component(s).
Complete step by step answer:
Chalcopyrite is a brass-yellow mineral with a chemical composition of$CuFe{{S}_{2}}$. It occurs in most sulfide mineral deposits throughout the world and has been the most important ore of copper for thousands of years. The surface of chalcopyrite loses its metallic luster and brass-yellow color upon weathering.
Copper metal can be extracted from the open air roasting of a mixture of chalcopyrite and silica sand, as shown in the following reaction:
\[2CuFe{{S}_{2}}~\left( s \right)\text{ }+\text{ }5{{O}_{2}}~\left( g \right)\text{ }+\text{ }2Si{{O}^{{}}}_{2}~\left( s \right)\text{ }\rightleftharpoons \text{ }2Cu\text{ }\left( l \right)\text{ }+\text{ }4S{{O}_{2}}~\left( g \right)\text{ }+\text{ }2FeSi{{O}_{3}}~\left( l \right)\]
Often before roasting, the ore has already been partially purified, e.g. by froth flotation. The concentrate is mixed with other materials to facilitate the process. The technology is useful but is also a serious source of air pollution.
The $FeSi{{O}_{3}}$ (ferrosilite) end member of the pyroxene series is not known as a pure mineral in nature and cannot be synthesized at atmospheric pressure. It is readily synthesized, however, at pressures from 18 to 45 kb and temperatures from \[1150{}^\circ to\text{ }1400{}^\circ C\] .
Therefore, iron is removed as $FeSi{{O}_{3}}$,
So, the correct answer is “Option D”.
Note: Partial roasting of sulphide ore is carried in order to convert it to the oxide form. The oxide form then reduces the remaining sulphide ore to the metallic form. This is called auto reduction. FeO is formed by roasting, which reacts according to the given reaction to form slag, which can be removed easily.
Complete step by step answer:
Chalcopyrite is a brass-yellow mineral with a chemical composition of$CuFe{{S}_{2}}$. It occurs in most sulfide mineral deposits throughout the world and has been the most important ore of copper for thousands of years. The surface of chalcopyrite loses its metallic luster and brass-yellow color upon weathering.
Copper metal can be extracted from the open air roasting of a mixture of chalcopyrite and silica sand, as shown in the following reaction:
\[2CuFe{{S}_{2}}~\left( s \right)\text{ }+\text{ }5{{O}_{2}}~\left( g \right)\text{ }+\text{ }2Si{{O}^{{}}}_{2}~\left( s \right)\text{ }\rightleftharpoons \text{ }2Cu\text{ }\left( l \right)\text{ }+\text{ }4S{{O}_{2}}~\left( g \right)\text{ }+\text{ }2FeSi{{O}_{3}}~\left( l \right)\]
Often before roasting, the ore has already been partially purified, e.g. by froth flotation. The concentrate is mixed with other materials to facilitate the process. The technology is useful but is also a serious source of air pollution.
The $FeSi{{O}_{3}}$ (ferrosilite) end member of the pyroxene series is not known as a pure mineral in nature and cannot be synthesized at atmospheric pressure. It is readily synthesized, however, at pressures from 18 to 45 kb and temperatures from \[1150{}^\circ to\text{ }1400{}^\circ C\] .
Therefore, iron is removed as $FeSi{{O}_{3}}$,
So, the correct answer is “Option D”.
Note: Partial roasting of sulphide ore is carried in order to convert it to the oxide form. The oxide form then reduces the remaining sulphide ore to the metallic form. This is called auto reduction. FeO is formed by roasting, which reacts according to the given reaction to form slag, which can be removed easily.
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