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, which precipitated 2.93 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)

Answer 1

33.3% of Sn in the sample

Step-by-step explanation:

The addition of SO₄⁻ ions produce the selective precipitation of Pb²⁺ to produce PbSO₄.

Moles of PbSO₄ (molar mass 303.26g/mol) in 2.93g are:

2.93g ₓ (1mol / 303.26) = 9.66x10⁻³ moles PbSO₄ = Moles Pb²⁺.

As molar mass of Pb is 207.2g/mol, mass in 9.66x10⁻³ moles of Pb²⁺ is:

9.66x10⁻³ moles of Pb²⁺ ₓ (207.2g / mol) = 2.00g of Pb²⁺

As mass of the sample is 3.00g, mass of Sn²⁺ is 3.00g - 2.00g = 1.00g

And the percentage of Sn in the sample is:

1.00g / 3.00g ₓ 100 =

33.3% of Sn in the sample