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?

Question

Destinee Velez

Chemistry

13 January, 04:23

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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Answers (1)

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Felicity Whitehead · Answer 1

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.