Why partial roasting of sulphide ore is done in the metallurgy of copper ?

Options

A. 1.Auto-reduction of `Cu_(2)O` formed is carried out by remaining `Cu_(2)S` in the reaction.
B. 2.Cu is separated out by partial reduction due to sedimentation
C. 3.Due to difference in gravity `Cu_(2)O` and `Cu_(2)S` are separated.
D. 4.Complete roasting cannot be done in one step hence partial roasting is done.

Explanation

### Step-by-Step Solution:

1. **Understanding Roasting**:
- Roasting is a metallurgical process that involves heating an ore in the presence of a strong supply of oxygen. This process is crucial for the extraction of metals from their ores.

2. **Identifying the Sulphide Ore**:
- In the metallurgy of copper, one of the primary sulphide ores is cuprous sulphide (Cu2S).

3. **Partial Roasting Process**:
- During partial roasting, cuprous sulphide (Cu2S) is heated with oxygen (O2). The reaction can be represented as:
\[ 2 \text{Cu}_2\text{S} + \text{O}_2 \rightarrow 2 \text{Cu}_2\text{O} + 2 \text{SO}_2 \]
- This reaction produces cuprous oxide (Cu2O) and sulfur dioxide (SO2) as a byproduct.

4. **Role of Cuprous Oxide**:
- The cuprous oxide (Cu2O) formed in the reaction can undergo further reduction. This is where the remaining cuprous sulphide (Cu2S) plays a critical role.

5. **Reduction to Obtain Copper**:
- The cuprous oxide (Cu2O) can be reduced back to copper (Cu) by reacting with the remaining cuprous sulphide (Cu2S):
\[ \text{Cu}_2\text{O} + \text{Cu}_2\text{S} \rightarrow 6 \text{Cu} + \text{SO}_2 \]
- This step is essential as it allows for the recovery of pure copper from the ore.

6. **Conclusion**:
- Partial roasting is performed to convert the sulphide ore into a form (Cu2O) that can be easily reduced to obtain pure copper. The process also helps in managing the sulfur content by producing sulfur dioxide as a byproduct.