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 2.37 g of PbSO₄.Assuming that all of the lead was precipitated,what is the percentage of Sn in the sample? (molar mass of PbSO₄ = 303.3 g/mol)

Question

Quizplus · Accepted Answer

To find the percentage of Sn in the sample, first calculate the amount of Pb in the sample using the mass of PbSO4 precipitated. The molar mass of PbSO4 is 303.3 g/mol, so 2.37 g of PbSO4 corresponds to 2.37 g / 303.3 g/mol = 0.00782 mol of PbSO4. Since each mole of PbSO4 contains one mole of Pb, there are 0.00782 mol of Pb in the sample. The molar mass of Pb is approximately 207.2 g/mol, so the mass of Pb in the sample is 0.00782 mol * 207.2 g/mol = 1.62 g. The sample initially weighed 3.00 g, so the mass of Sn in the sample is 3.00 g - 1.62 g = 1.38 g. The percentage of Sn in the sample is then (1.38 g / 3.00 g) * 100% = 46.0%.

A 300-G Sample of an Alloy (Containing Only Pb and Sn)was