Gọi số mol $MgO$:x
$CuO$: y
Ta có $mMgO+mCuO=52$
⇒$40x+80y=52(1)$
$MgO+2HCl \to MgCl_2+H_2O$
$CuO+2HCl \to CuCl_2+H_2O$
$500ml=0,5lit$
$nHCl=0,5.3,2=1,6$
⇒$2x+2y=1,6(2)$
(1)(2)⇒$\left \{ {{x=0,3} \atop {y=0,5}} \right.$
$mHCl=1,6.36,5=58,4g$
$mddsau=mOxit+mHCl=52+58,4=110,4g$
$C\%MgCl_2=\frac{0,3.95}{110,4}.100=25,82\%$
$C\%CuCl_2=\frac{0,5.135}{110,4}.100=61,14\%$
$CMMgCl_2=\frac{0,3}{0,5}=0,6M$
$CMCuCl_2=\frac{0,5}{0,5}=1M$