a,
$MgO+2HCl\to MgCl_2+H_2O$
$Al_2O_3+6HCl\to 2AlCl_3+3H_2O$
b,
Gọi x, y là số mol $MgO$, $Al_2O_3$
$\Rightarrow 40x+102y=14,2$ (1)
$n_{HCl}=\dfrac{29,2}{36,5}=0,8(mol)$
$\Rightarrow 2x+6y=0,8$ (2)
Từ (1)(2) suy ra $x=y=0,1$
$\%m_{MgO}=\dfrac{0,1.40.100}{14,2}=28,17\%$
$\%m_{Al_2O_3}=71,83\%$