$n_{Al2O3}$ = 0,1 mol
$m_{HCl}$ = $\frac{400.9,125}{100}$ = 36,5g =>$n_{HCl}$ = 1 mol
Al2O3 + 6HCl ---> 2AlCl3 + 3H2O
a. Ta thấy: $\frac{nAl2O3}{1}$<$\frac{nHCl}{6}$
=> HCl dư, tính theo Al2O3
$n_{AlCl3}$=$n_{Al2O3}$.2=0,1.2=0,2 mol
=>$m_{AlCl3}$=0,2.133,5=26,7g
$n_{HCl pư}$=6$n_{Al2O3}$=0,1.6=0,6 mol
=> $n_{HCl}$ dư=$n_{HCl}$ bđ - $n_{HCl}$ pư=1-0,6=0,4 mol
=> $m_{HCl}$ sau pư=0,4.36,5=14,6g
b.
C%$_{HCl}$ sau pư= $\frac{14,6}{14,6+26,7}$ . 100%≈35,351%
C%$_{AlCl3}$= $\frac{26,7}{14,6+26,7}$ . 100%≈64,649%