$n_{HCl}$ `=0,2 . 2 = 0,4` mol
Đặt $n_{CuO}$ `=x` mol , $n_{MgO}$ `=y` mol
`PTHH : CuO + 2HCl -> CuCl_2 + H_2O`
Pứ :        `x -> 2x -> x -> x`
`PTHH : MgO + 2HCl -> MgCl_2 + H_2O`
Pứ :          `y -> 2y -> y -> y`

 Theo pt có hệ
$\begin{cases} 2x+ 2y = 0,4\\80x+40y = 12\\ \end{cases}$
`<=> x = y = 0,1` mol
`%`$m_{CuO}$ `= `$\dfrac{0,1.80}{12}$ `.100% = 66,67%`
`->%`$m_{MgO}$ `=100% - 66,67% =33,33%`
$m_{CuCl_2}$ `=0,1 . 135 = 13,5g`
$m_{MgCl_2}$ `=0,1 . 95=9,5g`