称取含FE（M=55.85)试样0.4000克,溶解后将其中的FE2+还原为fe2+,

6Fe²+ + Cr₂O₇²+ + 14H+ = 6Fe³+ + 2Cr³+ + 7H₂O
6 1
x 0.02000*0.02500 = 0.0005000
x = 6*0.0005000/1 = 0.003000 mol
1 mol Fe2+相当于 0.5molFe2O3,0.003000 molFe2+相当于0.001500molFe2O3
Fe2O3质量:0.001500*159.7 = 0.2396 g
质量分数:0.2396/0.4000 = 0.5989 = 59.89%