$PTPƯ:CaO+2HCl\xrightarrow{} CaCl_2+H_2O$
$\text{Đổi 500 ml = 0,5 lít.}$
$⇒n_{HCl}=0,5.0,4=0,2mol.$
$Theo$ $pt:$ $n_{CaO}=\dfrac{1}{2}n_{HCl}=0,1mol.$
$⇒m_{CaO}=0,1.56=5,6g.$
$⇒m_{Cu}=m_{hỗn\ hợp}-m_{CaO}=14-5,6=8,4g.$
$b,\%m_{CaO}=\dfrac{5,6}{14}.100\%=40\%$
$⇒\%m_{Cu}=\dfrac{8,4}ơ{14}.100\%=60\%$
chúc bạn học tốt!