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?

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%