Final answer:
To determine the number of Fe(II) ions in 10.0 g of FeSO₄, we first convert the mass to moles using the molar mass of FeSO₄, then use Avogadro's number to find that there are approximately 3.96 x 10²² Fe(II) ions.
Step-by-step explanation:
Calculating the Number of Fe(II) Ions in FeSO₄
To find out how many Fe(II) ions are in 10.0 g of FeSO₄, we need to use the molar mass of FeSO₄ and Avogadro's number. The molar mass of FeSO₄ (Iron(II) Sulfate) is roughly 151.91 g/mol. Therefore, the first step is to convert the mass of FeSO₄ to moles:
10.0 g FeSO₄ * (1 mol FeSO₄ / 151.91 g FeSO₄) = 0.0658 mol FeSO₄
Since each mole of FeSO₄ contains one mole of Fe(II) ions, there are 0.0658 moles of Fe(II) ions. Using Avogadro's number (6.022 x 10²³ particles/mol), we can then calculate the number of Fe(II) ions:
0.0658 mol Fe(II) * (6.022 x 10²³ ions/mol) ≈ 3.96 x 10²² Fe(II) ions
Thus, there are approximately 3.96 x 10²² Fe(II) ions in 10.0 g of FeSO₄.