8H+ + 3Fe2+ + CrO4 2- = 3Fe3+ + Cr3+ + 4H2O
no of moles of K2CrO4 = molarity* volume in L
                                       Â
= 0.04322*0.04568Â Â = 0.00197 moles
K2CrO4 ---------> 2K+ + CrO42-
0.00197moles            Â
0.00197moles
from balanced equation
1 mole of CrO42- react with 3 moles
Fe2+
0.00197 moles CrO42- react with = 3*0.00197/1 =
0.00591moles of Fe+2
mass of Fe  = no of moles* gram atomic mass
                  Â
= 0.00591*56 = 0.33096g of Fe
mass percentage of Fe = 0.33096*100/0.9087Â Â =
36.42%