Lead is found in earth’s crust in several different lead ores. suppose a certain rock is composed of 38.0% pbs (galena), 25.0% pbco3 (cerussite), and 17.4% pbso4 (anglesite). the remainder of the rock is composed of substances containing no lead. how much of this rock (in kg) must be processed to obtain 5.2 metric tons of lead? (a metric ton is 1000 kg.)

To solve this problem, we can the molar masses to calculate for the total percent of lead in the rock.

For PbS:

percent Pb in PbS = [(207 g/mol Pb) / (239.3 g/mol PbS)] * 100%

percent Pb in PbS = 86.6% 
So, amount of Pb in the rock is:

percent Pb in rock = 86.6 % of 38.0%

percent Pb in rock = 32.91 %

For PbSO4:
percent Pb in PbSO4 = [(207 g/mol Pb) / (303.3 g/mol PbSO4)] * 100%

percent Pb in PbSO4 = 68.3% 
So, amount of Pb in the rock is:

percent Pb in rock = 68.3% of 17.4%

percent Pb in rock = 11.88 %

For PbCO3:
percent Pb in PbCO3= [(207 g/mol Pb) / (267.2 g/mol PbCO3)] * 100%

percent Pb in PbCO3 = 77.54% 
So, amount of Pb in the rock is:

percent Pb in rock = 77.54 % of 25.0%

percent Pb in rock = 19.38 %

The sum is:

Total percent Pb in rock = 19.38 + 11.88 + 32.91
Total percent Pb in rock = 64.2 %

Therefore the amount of rock needed is: 

Amount rock needed = 5.2 MT / 0.642

Amount rock needed = 8.1 MT

or

Amount rock needed = 8100 kg

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