Answer
3.075 g Fe²⁺
Procedure
The formula of iron (II) sulfite is FeS, to get the grams of Fe²⁺ present, we need to determine the ratio of the elements that form the molecule.
Fe = 55.845 g/mol
S = 32.065 g/mol
FeS = 55.845 g/mol + 32.065 g/mol = 87.91 g/mol
Ratio Fe = 55.845/87.91= 0.6353
Ratio S= 32.065/89.91 = 0.3647
Grams of Fe²⁺ = 4.84 (0.6353) = 3.075 g Fe²⁺