Đáp án:
FexOy+2yHCl−−−>xFeCl2yx+yH2O
Ta có: mddHCl=52,14.1,05=54,747(g)
=>mHCl=(C%HCl.mddHCl) : 100=(10x54,747) : 100=5,4747(g)
=>nHCl=5,4747 : 36,5=0,15(mol)
Theo PTHH: nFexOy=0,15/2y=0,075/y(mol)
Ta có: 4=0,075/y.(56x+16y)
=>4y=4,2x+1,2y
<=>4,2x=2,8y
=>x/y=2,8/4,2=2/3
=> x=2 và y=3
=>CT:Fe2O3