Question

Give the steps involved in the extraction of metals of low and medium reactivity from their respective sulphide ores.

Answer 1

Steps involved in the extraction of metals of low and medium reactivity from their respective sulfide ores.

• For extraction of metals of low and medium reactivity from their respective sulfide ores the following two mechanisms are involved:

1. Roasting
2. Reduction

• Roasting is the mechanism in which the sulfide ore of the respective metal is strongly heated in the presence of high amount of oxygen to convert it into oxides.
• The obtained oxides are reduced to respective metals of interest using a suitable reducing agent.

Chemical equation involved in the extraction of medium reactivity metals such as zinc from its sulfide ore is as follows:

Roasting:

$2 Z n S+3 O_{2} \stackrel{\Delta}{\rightarrow} 2 Z n O+2 S O_{2}$

Reduction:

In the case of zinc oxide, carbon is used as reducing agent.

$Z n O+C \rightarrow Z n+C O$

Chemical equation involved in the extraction of low reactivity metals such as copper from its sulfide ore is as follows:

Roasting:

$2 \mathrm{Cu}_{2} \mathrm{S}+3 \mathrm{O}_{2} \stackrel{\Delta}{\rightarrow} 2 \mathrm{Cu}_{2} \mathrm{O}+2 \mathrm{SO}_{2}$

Reduction:

In the case of copper oxide, coke is used as reducing agent.

$Cu_2O+C\rightarrow2Cu+CO$

Learn more about such concept

Describe three steps involved in extraction of metals from their ores

https://brainly.in/question/11881890

How does the method used for extracting a metal from its ore depend on the metal’s position in the reactivity series ? Explain with examples.

https://brainly.in/question/11756621

Answer 2

Answer:

your answer is

During extraction of metals of low and medium reactivity, a respective sulphide ore is first heated in air. This helps in obtaining oxide of metal. It is easier to extract a metal from its oxide rather than from its sulphide. Hence, this step is used. Mercury is a metal of low reactivity.

Mercury sulphide (cinnabar) is heated in air. Mercury sulphide gets oxidized to produce mercury oxide.

2HgS+3O

2

→2HgO+2SO

2

After that, mercury oxide is reduced to obtain mercury.

2HgO→2Hg+O

2

Zinc is a metal of medium reactivity. It is found as zinc blende (ZnS).

Zinc blende is roasted to be converted into zinc oxide. Zinc spar is put under calcination to be converted into zinc oxide.

2ZnS+3O

2

→2ZnO+2SO

2

ZnCO

3

→ZnO+CO

2

Zinc oxide so obtained is reduced to zinc metal by heating with carbon (a reducing agent).

ZnO+C→Zn+CO

this you right answer