Answer :

In 5.00 g of FeSO4 (iron(II) sulfate), there are approximately 0.0257 moles of Fe2+ ions. This is calculated by converting the mass of FeSO4 to moles using its molar mass and then considering the stoichiometry of the compound.

1. To determine the number of iron(II) ions (Fe2+) in 5.00 g of FeSO4, we need to follow a series of steps. Firstly, we calculate the molar mass of FeSO4, which consists of one iron atom (Fe), one sulfur atom (S), and four oxygen atoms (O). The atomic masses are 55.845 g/mol for Fe, 32.06 g/mol for S, and 16.00 g/mol for O. Adding them up, we get a molar mass of 151.91 g/mol for FeSO4.

2. Next, we convert the mass of FeSO4 (5.00 g) to moles by dividing it by the molar mass. Thus, 5.00 g / 151.91 g/mol gives us approximately 0.0329 moles of FeSO4.

3. Considering the balanced chemical equation for the formation of FeSO4, we can see that each FeSO4 molecule contains one iron(II) ion (Fe2+). Therefore, the number of Fe2+ ions is equal to the number of FeSO4 molecules.

4. Consequently, we have approximately 0.0329 moles of Fe2+ ions in 5.00 g of FeSO4. To find the number of Fe2+ ions, we multiply the number of moles by Avogadro's number (6.022 x 10^23 ions per mole). Thus, 0.0329 moles x 6.022 x 10^23 ions/mole gives us around 1.98 x 10^22 Fe2+ ions in 5.00 g of FeSO4.

5. In summary, there are approximately 1.98 x 10^22 iron(II) ions (Fe2+) in 5.00 g of FeSO4. This calculation is based on converting the mass of FeSO4 to moles, considering the stoichiometry of the compound, and using Avogadro's number to determine the number of ions.

Learn more about Avogadro's number here: brainly.com/question/28812626

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