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)#
