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: $m_{cr}+m_{dd\ HCL}=m_{dd\ A}+m_{H_2}$
$\Rightarrow 6,88+250=m_{dd\ A}+0,05.2$
$\Rightarrow m_{dd\ A}=256,78\ \text{gam}$
$\Rightarrow \text C\%_{HCl\ du}=\frac{0,06.36,5}{256,78}\cdot 100\%=0,85\%$
$\text C\%_{FeCl_2}=\frac{0,05.127}{256,78}\cdot 100\%=2,47\%$
$\text C\%_{AlCl_3}=\frac{0,08.133,5}{256,78}\cdot 100\%=4,16\%$
