a/
Gọi số mol $CuO$: x
$Al_2O_3$:y
$Fe_2O_3$:z
$m_\text{mỗi phần}=\frac{15,64}{2}=7,82g$
⇒$80x+102y+160z=7,82(1)$
P1:
$mHCl =200.5,2925\%=10,585g$
$nHCl =\frac{10,585}{36,5}=0,29$
$CuO+2HCl \to CuCl_2+H_2O$
$Al_2O_3+6HCl \to 2AlCl_3+3H_2O$
$Fe_2O_3+6HCl \to 2FeCl_3+3H_2O$
⇒$2x+6y+6z=0,29(2)$
P2:
$CuO+H_2 \to Cu+H_2O$
$Fe_2O_3+3H_2 \to 2Fe+3H_2O$
Ta có $mCu+mAl_2O_3+mFe=5,98$
⇒$64x+102y+2z.56=5,98(3)$
$(1)(2)(3)⇒x=0,025 ;y=0,01 ;z=0,03$
$mCuO=0,025.80.2=4g$
$mAl_2O_3=0,01.102.2=2,04g$
$mFe_2O_3=0,03.160.2=9,6g$
b/
$mddsau=200+7,82=207,82g$
$C\%CuCl_2 =\frac{0,025.135}{207,82}.100=1,62\%$
$C\%AlCl_3=\frac{0,02.133,5}{207,82}.100=1,28\%$
$C\%FeCl_3=\frac{0,06.162,5}{207,82}.100=4,69\%$