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Answer :
Final answer:
Approximately 62.70 kg of lead must be added to 73 kg of tin to yield a Pb-Sn alloy with a composition of 67 at % Sn-33 at % Pb.
Explanation:
To find out how many kilograms of lead must be added to 73 kg of tin to yield an alloy with a composition of 67 at% Sn and 33 at% Pb, we first need to determine the ratio of atoms of tin and lead that this alloy would have. Since we are dealing with atomic percentages, we use the atomic masses of tin and lead, which are approximately 118.7 u and 207 u, respectively.
Moles of Sn = Mass of Sn / Molar mass of Sn = 73 kg / 118.71 g/mol (molar mass of Sn) ≈ 0.615 moles. Since the desired ratio is 67% Sn and 33% Pb:
Moles of Sn = (Moles of Sn) * (33/67) ≈ 0.615 moles * (33/67) ≈ 0.303 moles. Mass of Pb = Moles of Pb * Molar mass of Pb ≈ 0.303 moles * 207.2 g/mol (molar mass of Pb) ≈ 62.70 kg.
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