Giải:
`ZnO+2HCl->ZnCl_2+H_2O`
`CuO+2HCl->CuCl_2+H_2O`
`n_{ZnCl_2}=amol;n_{CuCl_2}=bmol`
$m_{muoi\ khan}=136a+135b=20,35g$
$m_{hon\ hop}=81x+80y=12,1$
`x=0,1mol;y=0,05mol`
$Theo\ phuong\ trinh: n_{HCl}=(0,1+0,05).2=0,3mol$
`CM_{HCl}=(0,3)/(0,2)=1,5M``->a=1,5M`