Question

The chief ore of zinc is the sulphide, ZnS. The ore is concentrated by floatation process and then heated in air, which converts the ZnS to ZnO.
a.)\[2ZnS\,+\,3{{O}_{2}}\,\xrightarrow{90.6\%}\,2ZnO\,+\,2S{{O}_{2}}\]
The ZnO is then treated with dilute $H_2SO_4$.
b.)\[2ZnO\,+\,2{{H}_{2}}S{{O}_{4}}\,\xrightarrow{100\%}\,2ZnS{{O}_{4}}\,+\,2{{H}_{2}}O\]
Zn metal is produced on electrolysis.
c.)\[2ZnS{{O}_{4}}\,+\,2{{H}_{2}}O\,\xrightarrow{98.2\%}\,2Zn\,+\,2{{H}_{2}}S{{O}_{4}}\,+{{O}_{2}}\]
Efficiencies of the processes have been indicated above the arrow mark. Mass of Zn will be obtained from an ore containing 225 kg of ZnS in kg is (nearest integer value):

Answer 1

Hint: Zinc is extracted from its ore called Zinc Blende (chemically ZnS). It involves multiple steps to obtain the pure zinc metal. Following are the processes involved:
Concentration
Roasting
Reduction
Purification

Complete step by step solution:

In the flotation process, \[ZnS\] is converted to \[ZnO\]. The reaction is
\[2ZnS\,+\,3{{O}_{2}}\,\xrightarrow{90.6\%}\,2ZnO\,+\,2S{{O}_{2}}\]
The efficiency of the flotation process is given to be 90.6 %. Again, when the obtained \[ZnO\]is reacted with dilute \[{{H}_{2}}S{{O}_{4}}\] we get the following reaction with efficiency of 100%:
\[2ZnO\,+\,2{{H}_{2}}S{{O}_{4}}\,\xrightarrow{100\%}\,2ZnS{{O}_{4}}\,+\,2{{H}_{2}}O\]
Finally, an electrolytic process is performed on the \[ZnS{{O}_{4}}\]and Zinc is obtained. The efficiency of the electrolytic step is given to be 98.2 %. The equation for the reaction is:
\[2ZnS{{O}_{4}}\,+\,2{{H}_{2}}O\,\xrightarrow{98.2\%}\,2Zn\,+\,2{{H}_{2}}S{{O}_{4}}\,+{{O}_{2}}\]
First, we will calculate how many moles of \[ZnS\]are there in 225kg (225000g) of \[ZnS\]. Molar mass of \[ZnS\] is 97.38g/mol.
\[\begin{align}
  & Moles\,of\,ZnS\,=\,\dfrac{Given\,mass\,of\,ZnS\,in\,grams}{Molar\,mass\,of\,ZnS}=\,\dfrac{22500}{97.38} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,=\,231\,0.53\,moles\,of\,ZnS \\
\end{align}\]
Now, we will use the first equation to calculate the number of moles of \[ZnO\,\]produced. We can clearly see that there is a 1:1 ratio in the equation, \[ZnS\]reacted: \[ZnO\,\]produced. Hence, theoretical 2310.53 moles of \[ZnO\,\]should be produced but this reaction has the efficiency of only 90.6%. So, the number of moles of \[ZnO\,\]produced will actually be:
\[\begin{align}
  & Moles\,of\,ZnO\,=\,Moles\,of\,ZnS\,produces\,\times \,efficiency\,of\,reaction \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\,2310.53\,\times \,\dfrac{90.6}{100} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\,2162.66\,moles\,of\,ZnO \\
\end{align}\]
Again, in the second equation these moles of \[ZnO\,\] will react with \[{{H}_{2}}S{{O}_{4}}\]and produce \[ZnS{{O}_{4}}\]. Here also we can see that the ratio of \[ZnO\,\] reacted: \[ZnS{{O}_{4}}\] produced is 1:1 and the efficiency of the reaction is 100%. Therefore, the number of moles of \[ZnS{{O}_{4}}\]produced will be equal to the number of moles of \[ZnO\,\] i.e. 2162.66.
Finally, in the third equation, there will be 2162.66 moles of \[ZnS{{O}_{4}}\] present to react with water. Again, in the third equation also the ratio of \[ZnS{{O}_{4}}\] reacted and \[Zn\] produced is 1:1. Therefore, theoretical 2162.66 mole of \[Zn\] should be produced but this reaction has an efficiency of only 92.2% . So, the number of moles of \[Zn\] produced would actually be:
\[\begin{align}
  & Moles\,of\,Zn\,=\,Moles\,of\,ZnS{{O}_{4}}\,produces\,\times \,efficiency\,of\,reaction \\
 & \,\,\,\,\,\,\,\,\,\,\,=\,2162.66\,\times \,\dfrac{98.2}{100} \\
 & \,\,\,\,\,\,\,\,\,\,\,=\,2123.7\,moles\,of\,Zn \\
\end{align}\]
Now, we will calculate the total mass of \[Zn\] produced in grams where the number of moles of \[Zn\] is 2123.7. And the molar mass of \[Zn\] is 65.38g.
\[\begin{align}
  & Mass\,of\,Zn\,=\,Moles\,of\,Zn\,\times \,Molar\,mass\,of\,Zn \\
 & \,\,\,\,\,\,\,\,\,=\,2123.7\,\times \,65.38 \\
 & \,\,\,\,\,\,\,\,\,=\,138847.5\,g \\
 & \,\,\,\,\,\,\,\,\,=139\,kg \\
\end{align}\]
Since, 1 kg = 1000 g.
Therefore, the amount of Zinc produced in the reaction is 139 kg.

Note: Zinc is quite reactive and hence does not occur in the free state. It reacts with sulphur present in the atmosphere to form Zinc sulphide (ZnS), commonly known as zinc blende. We have to process this ore to finally get the pure Zinc.