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Jobisdone [24]
3 years ago
12

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)?
Chemistry
1 answer:
yulyashka [42]3 years ago
7 0

Answer:

0.0286 g O₂

Explanation:

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

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

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

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

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

                     1.97 moles of Na contains greater number of atoms than 6.8 × 10²² atoms of Li.

Explanation:

                    To solve this problem we will first calculate the number of atoms contained by 1.97 moles of Na and then will compare it with 6.8 × 10²² atoms of Li.

                    As we know that 1 mole of any substance contains exactly 6.022 × 10²³ particles which is also called as Avogadro's Number. So in order to calculate the number of atoms contained by 1.97 moles of Na, we will use following relation,

          Moles  =  Number of Atoms ÷ 6.022 × 10²³ atoms.mol⁻¹

Solving for Number of Atoms,

          Number of Atoms  =  Moles × 6.022 × 10²³ Atoms.mol⁻¹

Putting values,

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