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Smelters and Galena Conversion

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A small smelter converts galena (lead sulfide) into pure lead by heating the ore with an excess of oxygen. Sulfur dioxide is produced as an air pollutant, as a result of the reaction. To remove sulfur dioxide, the flue gas is passed through a slurry of lime (calcium oxide) to produce calcium sulfate. The smelter processes 120 kg of galena ore per hour. Show a flow diagram of the process, write all chemical equations involved and calculate the following:

A. How much lead is produced in a 24 hour day?
B. How much air is required in one hour to completely react all of the galena? Air is 21% oxygen.
C. Assuming the exit temperature of the flue gas is 650 ºC and the barometric pressure is 0.9 atm, what is the volume of the sulfur dioxide produced?
D. How much dry lime is required to completely remove all of the sulfur dioxide from the flue gas?

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Solution Summary

The solution answers several inquiries on the lead production, air requirement, exit temperature, and dry lime requirement with respect to a small smelter converting galena, which is lead sulphide. Answering all these inquiries brings forth an outlook into the reaction of the galena conversion itself.

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Every 120 kg of lead ore or PbS are processed in 1 hour.
The molar mass of PbS is (207+32)=239
This means that the number of moles of PbS produced per hour is (mass/molarmass=number of moles)
So n=120000/239=502.1 moles.
From the first equation above, we can see that the stoichiometric coefficients of lead and PbS are 1:1, which means every one mole of PbS gives one mole of Pb. Hence, we produce 502.1 moles of Pb in one hour
Again mass= molarmass x number of ...

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