$n_{HCl}=1(mol)$
Đặt $2x$, $x$ là số mol $AO, B_2O_3$
$AO+2HCl\to ACl_2+H_2O$
$B_2O_3+6HCl\to 2BCl_3+3H_2O$
$\to 2x.2+x.6=1$
$\to x=0,1$
$\to n_{AO}=0,2(mol); n_{B_2O_3}=0,1(mol)$
Ta có: $0,2(M_A+16)+0,1(2M_B+16.3)=18,2$
$\to M_A+M_B=51$
$\to M_A=24(Mg); M_B=27(Al)$
Vậy hai oxit là $MgO, Al_2O_3$