Question

Give the steps involved in the extraction of metals of low and medium reactivity

from their respective sulphide ores. Write the necessary equations.​

Answer 1

Answer:

low in the reactivity series like mercury (Hg), the following steps are involved <br> (i) Roasting (heating the ore strongly in the presence of excess of air) Metal sulphide is converted into metal oxide. <br>

<br>

<br> (ii) Reduction Metal oxide is then reduced to metal by heating. <br>

<br>

<br> (iii) Refining of metal. For metals, medium in reactivity series like zinc (Zn), the following steps are involved <br> (i) Roasting (for sulphide ores) The sulphide ore is converted into metal oxide by heating the sulphide are strongly in the presence of air. <br>

<br> (ii) Reduction Metal oxides are reduced to metals by using a suitable reducing agent. This can be done by either of two. <br> (a) Reduction by heating with carbon (smelting) <br>

<br> (b) Reduction by heating with aluminium <br>

Answer 2

Explanation:

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:

Roasting

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}2ZnS+3O

2

→

Δ

2ZnO+2SO

2

Reduction:

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

Z n O+C \rightarrow Z n+C OZnO+C→Zn+CO

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}2Cu

2

S+3O

2

→

Δ

2Cu

2

O+2SO

2

Reduction:

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

Cu_2O+C\rightarrow2Cu+COCu

2

O+C→2Cu+CO

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