a)PTHH Fe + 2HCl -> FeCl2 + H2↑(1)

  (mol) 0,1    0,2         0,1     <-0,1

         Fe2O3 + 6HCl -> 2FeCl3 + 3H2O (2)

  (mol)0,015->  0,09        0,03        0,045

nH2=2,24/22,4=0,1(mol)

=>Theo PTHH (1) ta có nFe=0,1(mol)

=>mFe=0,1x56=5,6(g)

=>%mFe=5,6x100%/8=70%

     %mFe2O3=100-70=30%

b)mFe2O3=8-5,6=2,4(g)

=>nFe2O3=2,4/160=0,015(mol)

=>Theo PTHH =>∑nHCl=nHCl(1)+nHCl(2)=0,2+0,09=0,29(mol)

=>mHCl=0,29x36,5=10,585(g)

=>mddHCl=10,585x100/10=105,85(g)

=>VddHCl=Vddsp=105,85/1,1≈96(ml)=0,96(lit)

-Theo PTHH (1) ta có nFeCl2=0,1(mol)=>CmFeCl2=0,1/0,96≈0,104(M)

-Theo PTHH (2) ta có nFeCl3=0,03(mol)=>CmFeCl2=0,03/0,96≈0,03125(M)