$A_3: RCl_n$

$n_{AgCl}=\dfrac{2,87}{143,5}=0,02 mol$

$RCl_n+nAgNO_3\to nAgCl+R(NO_3)_2$

$\Rightarrow n_{RCl_n}=\dfrac{0,02}{n}$

$M_{RCl_n}=\dfrac{1,27n}{0,02}=63,5n=R+35,5n$

$\Leftrightarrow R=28n$

$n=2\Rightarrow R=56(Fe)$

Vậy $A_3: FeCl_2$

$A_1: Fe$

$A_2: HCl$

$A_4: H_2$

$A_5: NaOH$

$A_6: Fe(OH)_2$

$A_7: NaCl$

$A_8: H_2O$

$A_9: O_2$

$A_{10}: Fe(OH)_3$

$A_{11}: Fe_2O_3$

PTHH:

$Fe+2HCl\to FeCl_2+H_2$

$FeCl_2+2NaOH\to Fe(OH)_2+2NaCl$

$4Fe(OH)_2+2H_2O+O_2\to 4Fe(OH)_3$

$2Fe(OH)_3\buildrel{{t^o}}\over\to Fe_2O_3+3H_2O$

$Fe_2O_3+3H_2\buildrel{{t^o}}\over\to 2Fe+3H_2O$

Nếu lấy 1,27 gam A3 tác dụng với dung dịch AgNO3 dư thì thu được 2,87 gam kết tủa.
⇒ A3 là FeCl2

A1:Fe

A2:HCl

A3:FeCl2
A4:H2
A5:NaOH
A6:Fe(OH)2
A7:NaCl
A8:H2O
A9:O2
A10:Fe(OH)3
A11:Fe2O3