nH2 = 0,01 mol
pt :
Fe +2 HCl ---> FeCl2 + H2
0,01...0,02.........0,01.........0,01
FexOy + 2yHCl ---> xFeCl 2y/x +y H2O
=> mFexOy = 0,72 (g)
mFe = 0.01*56 = 0.56g
=>%mFe = 0.56*100/1.28=43.75%
%m oxit Fe= 56.25%
FexOy----->xFe
0.05/x---------0.05
1.28g------>0.56gFe...