PTHH:
4FeCO3 + O2 ---> 2Fe2O3 + 4CO2(1)
4FexOy + (3x-2y) O2-----> 2xFe2O3(2)
CO2 + Ba(OH)2 ---> BaCO3 + H2O(3)
CO2 + BaCO3 + H2O ----> Ba(HCO3)2(4)
n kết tủa = 0.02 mol
TH1 nCO2 = n BaCO3 = 0.02 mol
=> m FeCO3 = 0.02*116=2.32 g
=> m FexOy= 12.64-2.32=10.32 g
=> nFexOy= 10.32/56x+16y(*)
=> mFe2O3(2)=11.2-0.01*160=9.6g
=> n Fe2O3 = 9.6/160=0.06mol
=> n FexOy = 0.12/x (**)
Từ (*)

**)
=> x/y=8/15 (loai)
TH2
nBa(OH)2= 0.15*0.2=0.03 mol
CO2 + Ba(OH)2 ---> BaCO3 + H2O(3)
0.03____0.03_______0.03_____
CO2 + BaCO3 + H2O ----> Ba(HCO3)2(4)
0.01____0.01____
=> n CO2=0.04 mol
làm tượng tự
=>x/y = 2/3
=> Fe2O3