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)

Answers

Answer 1

Answer:

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

Related Questions

which process is used to produce gases from solutions of salts dissolved in water or another liquid?

A.Electrolysis
B.Metallic bonding
C.Ionic bonding
D. Polar covalent bonding

Answers

Answer:

A.Electrolysis

Explanation:

A.Electrolysis

For example,  electrolysis of solution of NaCl in water gives H2 and O2.

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