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How many moles of oxygen atoms are there in 0.500 mol of hydrated iron(II) ammonium sulfate, \((NH_4)_2Fe(SO_4)_2 \cdot 6H_2O(s)\)?

A. 4.00
B. 7.00
C. 8.00
D. 14.00

Answer :

To determine how many moles of oxygen atoms are in 0.500 mol of hydrated iron(II) ammonium sulfate, (NH₄)₂Fe(SO₄)₂·6H₂O, we need to break down the components of this compound.

  1. Identify the Composition of the Compound:
    The formula (NH₄)₂Fe(SO₄)₂·6H₂O indicates that the compound consists of:

    • 2 ammonium ions, (NH₄)₂
    • 1 iron ion, Fe
    • 2 sulfate ions, SO₄
    • 6 water molecules, H₂O

  2. Count the Oxygen Atoms in Each Component:

    • From 2 sulfate ions (SO₄): Each sulfate ion contains 4 oxygen atoms. Therefore,
      • 2 SO₄ gives us: 2 × 4 = 8 oxygen atoms.
    • From 6 water molecules (H₂O): Each water molecule contains 1 oxygen atom. Therefore,
      • 6 H₂O gives us: 6 × 1 = 6 oxygen atoms.

  3. Total Moles of Oxygen in One Mole of Compound:

    • Total oxygen from both sources = 8 (from sulfate) + 6 (from water) = 14 oxygen atoms.

  4. Calculate Oxygen Atoms in 0.500 mol:

    • If there are 14 oxygen atoms in 1 mole of the hydrated compound, then in 0.500 mol, the number of moles of oxygen is:
      [tex]ext{Moles of Oxygen} = 0.500 ext{ mol} \times 14 = 7.00 ext{ mol}[/tex]

Final Answer: So, there are 7.00 moles of oxygen atoms in 0.500 mol of hydrated iron(II) ammonium sulfate.

Thus, the correct option is B. 7.00.

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