Ask Question
19 February, 10:18

A 3.00 - g sample of an alloy (containing only pb and sn) was dissolved in nitric acid (hno3). Sulfuric acid was added to this solution, which precipitated 1.69 g of pbso4. Assuming that all of the lead was precipitated, what is the percentage of sn in the sample?

+4
Answers (1)
  1. 19 February, 12:05
    0
    The reaction between Pb and H2SO4 will be

    Pb+2 + SO4^-2 = PbSO4 (s)

    So each mole of Pb will give one mole of PbSO4

    Moles of PbSO4 obtained = mass / molar mass = 1.69 / 303.26 = 0.0056 moles

    Moles of Pb present = 0.0056

    Mass of Pb present = moles X atomic mass = 0.0056 X 207 = 1.16 grams

    So mass of Sn in sample = 3 - 1.16 = 1.84

    % of Sn = Mass of Sn X 100 / total mass = 1.84 X 100 / 3 = 61. 33%
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A 3.00 - g sample of an alloy (containing only pb and sn) was dissolved in nitric acid (hno3). Sulfuric acid was added to this solution, ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers