Gọi `X` là `FeCl_n`
`FeCl_n+nAgNO_3->Fe(NO_3)_n+nAgCl`
`n_(AgCl)=\frac{4,305}{143,5}=0,03(mol)`
Theo phương trình
`n_(FeCl_n)=1/n .0,03`
`=>M_(FeCl_n)=\frac{1,905}{\frac{0,03}{n}}=63,5n`
`=>56+35,5n=63,5n`
`=>n=2`
`=>X` là `FeCl_2`
`Fe+2HCl->FeCl_2+H_2`
`FeCl_2+2NaOH->Fe(OH)_2+2NaCl`
`4Fe(OH)_2+O_2+2H_2O->4Fe(OH)_3`
$2Fe(OH)_3\xrightarrow{t^o}Fe_2O_3+3H_2O$
$Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O$
`A:Fe`
`B:HCl`
`X:FeCl_2`
`D:H_2`
`E:NaOH`
`F:Fe(OH)_2`
`G:NaCl`
`H:H_2O`
`I:O_2`
`K:Fe(OH)_3`
`L:Fe_2O_3`
