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9 August, 13:54

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)

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  1. 9 August, 14:03
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    33.3% of Sn in the sample

    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
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