Giải thích các bước giải:

a. PTHH:

$Fe+2HCl\to FeCl_2+H_2$

$Al_2O_3 + 6HCl\to 2AlCl_3  +3H_2O$

$NaOH+HCl\to NaCl+H_2O$

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

$AlCl_3 + 3NaOH\to 3NaCl + Al(OH)_3$

$Al(OH)_3+NaOH\to NaAlO_2+2H_2O$

$4Fe(OH)_2+O_2\xrightarrow{t^\circ} 2Fe_2O_3 + 4H_2O$

b. Kết tủa thu được là $Fe(OH)_2$

$n_{Fe_2O_3}=\frac 4{160}=0,025\ mol$

$\Rightarrow n_{Fe}=2n_{Fe_2O_3}=0,05\ \text{mol}$

$\Rightarrow n_{Al_2O_3}=\frac{6,88-0,05.56}{102}=0,04\ \text{mol}$

Ta có: $n_{H_2}=n_{Fe}=0,05\ \text{mol}$

$\Rightarrow V=0,05.22,4=1,12\ \text{l}$

$n_{NaOH}=0,24.2=0,48\ \text{mol}$
$n_{HCl}=2n_{Fe}+6n_{Al_2O_3}=2.0,05+6.0,04=0,34\ \text{mol}$

Ta có: $n_{FeCl_2}=n_{Fe}=0,05\ mol; n_{AlCl_3}=2n_{Al_2O_3}=0,08\ mol$

$n_{NaOH\ (FeCl_2+AlCl_3)}=2.n_{FeCl_2} + 4n_{AlCl_3}=2.0,05+4.0,08=0,42\ \text{mol}$

$\Rightarrow n_{HCl\ du}=n_{NaOH (+HCl)}=0,48-0,42=0,06\ \text{mol}$

$\Rightarrow\sum n_{HCl}=0,34+0,06=0,4\ \text{mol}$

$\Rightarrow \text C\%_{dd\ HCl}=\frac{0,4.36,5}{250}\cdot 100\%=5,84\%$

c. Chất tan trong A gồm HCl dư: $0,06$ mol, $FeCl_2:0,05\ mol, AlCl_3:\ 0,08\ mol$

BTKL: $m_{cr}+m_{dd\ HCL}=m_{dd\ A}+m_{H_2}$

$\Rightarrow 6,88+250=m_{dd\ A}+0,05.2$

$\Rightarrow m_{dd\ A}=256,78\ \text{gam}$

$\Rightarrow \text C\%_{HCl\ du}=\frac{0,06.36,5}{256,78}\cdot 100\%=0,85\%$

$\text C\%_{FeCl_2}=\frac{0,05.127}{256,78}\cdot 100\%=2,47\%$

$\text C\%_{AlCl_3}=\frac{0,08.133,5}{256,78}\cdot 100\%=4,16\%$

Fe+2HCl→FeCl2+H2

Al2O3+6HCl→2AlCl3+3H2O

FeCl2+2NaOH→Fe(OH)2+2NaCl

AlCl3+3NaOH→Al(OH)3+3NaCl

Al(OH)3+NaOH→NaAlO2+2H2O

4Fe(OH)2+O2→2Fe2O3+4H2O

nFe2O3=4/160=0,025mol

→nFe=0,025.2=0,05mol

nAl2O3=(6,88-0,05.56)/102=0,04mol

VH2=0,05.22,4=1,12l

nNaOH=0,24.2=0,48l

nHCl=0,05.2+0,04.6=0,34mol

nNaOH td với 2 muối=0,05.2+4.0,08=0,42mol

→nNaOH td với HCl=0,48-0,42=0,06mol

nHCl tổng=0,34+0,06=0,4mol

mHCl=0,4.36,5=14,6g

C%HCl=14,6/250.100%=5,84%

mdd sau pứ=6,88+250-0,05.2=256,78g

C%HCl dư=0,06.36,5/256,78.100%=0,85%

C%FeCl2=0,05.127/256,78.100%=2,47%

C%AlCl3=0,08.133,5/256,78.100%=4,16%