$a)ZnO+2HCl→ZnCl_2+H_2O$
$Al_2O_3+6HCl→2AlCl_3+3H_2O$
$n_{HCl}=0,25.2=0,5mol$
Gọi $ZnO=a,Al_2O_3=b$
$⇒$$\left \{ {{81a+102b=13,2} \atop {2a+6b=0,5}} \right.$
⇔$\left \{ {{a=0,1} \atop {b=0,05}} \right.$
$⇒m_{ZnO}=0,1.81=8,1g$
%$ZnO=8,1/13,2.100=61,36$%
%$Al_2O_3=100-61,36=38,64$%
$c)m_{ZnCl_2}=0,1.136=13,6g$
$m_{AlCl_3}=0,1.133,5=13,35g$