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A 0.652-g sample of a pure strontium halide reacts with excess sulfuric acid. the solid strontium sulfate formed is separated, dried, and found to weigh 0.755 g. what is the formula of the original halide?

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superman1987

Answer:

The formula of the original halide is SrCl₂.

Explanation:

  • The balanced equation of this reaction is:

SrX₂ + H₂SO₄ → SrSO₄ + 2 HX, where X is the halide.

  • From the equation stichiometry, 1.0 mole of strontium halide will result in 1.0 mole of SrSO₄.
  • The number of moles of SrSO₄ (n = mass/molar mass) = (0.755 g) / (183.68 g/mole) = 4.11 x 10⁻³ mole.
  • The number of moles of SrX are  4.11 x 10⁻³ moles from the stichiometry of the balanced equation.
  • n = mass / molar mass, n =  4.11 x 10⁻³ moles and mass = 0.652 g.
  • The molar mass of SrX₂ = mass / n = (0.652) / (4.11 x 10⁻³ moles) = 158.62 g/mole.
  • The molar mass of SrX₂ (158.62 g/mole) = Atomic mass of Sr (87.62 g/mole) + (2 x Atomic mass of halide X).
  • The atomic mass of halide X = (158.62 g/mole) - (87.62 g/mole) / 2 = 71 / 2  g/mole = 35.5 g/mole.
  • This is the atomic mass of Cl.
  • So, the formula of the original halide is SrCl₂.

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