Question

In the extraction of copper, metal is formed is the Bessemer converter due to reaction:
A. \[2C{u_2}S + 2C{u_2}O \to 6Cu + S{O_2}\]
B. $2C{u_2}O \to 4Cu + {O_2}$
C. $C{u_2}S \to 2Cu + S$
D. $Fe + C{u_2}O \to 2Cu + FeO$

Answer 1

Hint: In bessemer converter when $C{u_2}S$ is oxidized with ${O_2}$, then $C{u_2}O$ is formed which reacts with the remaining $C{u_2}S$ and gives copper.

Complete step-by-step answer:
According to this question we have to find the reaction .
Hence, in the extraction of copper, when matte (mostly $C{u_2}S$) is transferred to the Bessemer converter and a blast of air is blown through the molten mass, a part of $C{u_2}S$ is oxidized which combines with remaining $C{u_2}S$ to for metal.
Hence the reaction will be :
$2C{u_2}S + 3{O_2} \to 2C{u_2}O + 2S{O_2}$
$C{u_2}S + 2C{u_2}O \to 6Cu + S{O_2}$

It is an example of auto-reduction, a method in which sulphide ores of less electropositive metals such as Hg, Pb, Cu etc. are heated in air to turn part of the ore into oxide or sulphate and then reacts with the remaining sulphide ore in the absence of air to give the metal and sulphur dioxide.

Note: The Bessemer converter was a machine and surrounding process involving the removal and conversion of impurities from pig iron into steel – a material that had historically been expensive and time consuming to produce. If you solve these questions when writing chemical equations, you always adjust the equation by adding a multiplier to change the number of atoms on one side to match the other.