The student dissolved 0.30 moles of iron (III) sulfate.
Because iron (III) sulfate is soluble in aqueous solution, it will dissociate into Fe^(3+) and SO_4^(-) ions according to the balanced chemical equation
Fe_2(SO_4)_(3(aq)) -> color(red)(2)Fe_text((aq])^(3+) + 3SO_(4(aq))^(2-)
Notice that have a 1:color(red)(2) mole ratio between Fe_2(SO_4)_3 and Fe^(3+), which means that 1 mole of iron (III) sulfate produces 2 moles of iron (III) ions in solution.
You can use the molarity of the ions to determine how many moles were produced
C = n/V => n = C * V
n_(Fe^(3+)) = "0.60 M" * "1.000 L" = "0.60 moles" Fe^(3+)
Therefore, the number of moles of iron (III) sulfate needed to produce this many moles of iron (III) ions will be
0.60cancel("moles"Fe^(3+)) * ("1 mole"Fe_2(SO_4)_3)/(color(red)(2)cancel("moles"Fe^(3+))) = color(green)("0.30 moles" color(green)(Fe_2(SO_4)_3)
