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Fe(II) can be precipitated from a slightly basic aqueous solution by bubbling oxygen through the solution, which converts Fe(II) to insoluble Fe(III):

4Fe(OH)+(aq) + 4OH−(aq) + O2​(g) + 2H2​O(l) ------> 4Fe(OH)​3(s)

How many grams of O2 are consumed to precipitate all of the iron in 65.0 mL of 0.0550 M Fe(II)?

Respuesta :

maacastrobr

Answer:

0.0286 g O₂

Explanation:

  • 4Fe(OH)⁺(aq) + 4OH⁻(aq) + O₂(g) + 2H₂O(l) → 4Fe(OH)​₃(s)

First we calculate the moles of Fe(II) present in the given volume of the solution:

  • 65.0 mL ⇒ 65.0/1000 = 0.065 L
  • 0.0550 M * 0.065 L = 3.575x10⁻³ mol Fe(II)

Now we convert moles of Fe(II) into moles of O₂ and finally into grams using its molar mass (32 g/mol):

  • 3.575x10⁻³ mol Fe(II) * 1molO₂/4molFe(II) * 32g/mol = 0.0286 g O₂