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