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Using stoichiometry, find the molarity of [tex]SO_4^{2-}[/tex] in a solution prepared by mixing 2.15 g of alum, KAl(SO₄)₂•12H₂O, with 200 mL of water. (Hint: the stoichiometry is NOT 1:1.) The molecular weight of alum is 474.39 g/mol. Report your answer to three significant figures.

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Answer:

0.0453 M

Explanation:

First, we calculate the number of moles of alum:

  • 2.15 g alum ÷ 474.39 g/mol = 4.53x10⁻³ mol alum

Looking at the chemical formula, we see that there are two SO₄²⁻ species for each alum molecule.

  • 4.53x10⁻³ mol alum * [tex]\frac{2molSO_{4}^{-2}}{1molAlum}[/tex] = 9.06x10⁻³ mol SO₄⁻²

Now we calculate the molarity of SO₄⁻², dividing the moles by the volume:

  • 200 mL ⇒ 200 / 1000 = 0.200 L
  • 9.06x10⁻³ mol SO₄⁻² / 0.200 L = 0.0453 M

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