Gọi a, b, c là mol $MgO$, $Fe_2O_3$, $CuO$ trong 24g A.
$\Rightarrow 40a+160b+80c=24$ (1)
$n_{HCl}=0,6.1,5=0,9 mol$
$MgO+2HCl\to MgCl_2+H_2O$
$Fe_2O_3+6HCl\to 2FeCl_3+3H_2O$
$CuO+2HCl\to CuCl_2+H_2O$
$\Rightarrow 2a+6b+2c=0,9$ (2)
Trong 12g A có $0,5a$, $0,5b$, $0,5c$ mol mỗi chất.
$Fe_2O_3+3CO\to 2Fe+3CO_2$
$CuO+CO\to CO_2+Cu$
$\Rightarrow n_{Fe}=b$, $n_{Cu}=0,5c$
$\Rightarrow 40.0,5a+56b+64.0,5c=10$ (3)
(1)(2)(3)$\Rightarrow a=0,2; b=0,05; c=0,1$
$\%m_{MgO}=\frac{0,2.40.100}{24}=33,33\%$
$\%m_{Fe_2O_3}=\frac{0,05.160.100}{24}=33,33\%$
$\%m_{CuO}=33,34\%$