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Step 1:I'll solve this problem step by step, focusing on determining the maximum mass of lead that can be extracted from galena (PbS).
Step 2:: Identify the given information
- Mass of galena = $$2.5 \times 10^{3} \mathrm{~g}$$ (2.5 kg converted to grams)
- Ore: Galena (PbS)
Step 3:: Calculate the molar masses
- Molar mass of PbS = $$207.2 + 32.1 = 239.3 \mathrm{~g/mol}
- Molar mass of S = 32.1 \mathrm{~g/mol}
Step 4:: Determine the molar ratio of Pb in PbS
- Molar mass of Pb in PbS = $$207.2 \mathrm{~g/mol}
- PbS contains 1 mol of Pb per 1 mol of compound
Step 5:: Calculate the maximum mass of lead
\frac{207.2 \mathrm{~g~Pb}}{239.3 \mathrm{~g~PbS}} \times 2.5 \times 10^{3} \mathrm{~g~PbS} = 2.2 \times 10^{3} \mathrm{~g~Pb}
Final Answer
2.2 \times 10^{3} \mathrm{~g} or 2.2 \mathrm{~kg} of pure lead
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