AxOy+yH2->xA+yH2O
0,12/y...0,12
nH2=2,688/22,4=0,12mol=>mH2=0,12.2=0,24g
nH2=nH2O=0,12mol=>mH2O=0,12.18=2,16g
mA=6,4+0,24-2,16=4,48g(theo ĐLBTKL)
A+2yHCl->xACl2y/x+yH2
0,08/y............................0,08
Có: 0,08/y.A=4,48
=>A=56n
Biện luận n=1, A=56(Fe)
Có: 0,12/y.(56x+16y)=6,4
6,72x/y=4,48
x/y=4,48/6,72=2/3
=>x=2;y=3
CT: Fe2O3