Molar Mass of Lead

Smelters and Galena Conversion

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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