Đáp án:
Ta có Cu+HCl(ko pư)
⇒$m_{Cu}$ =12,8(g)
⇒$m_{hh}$ lúc đầu=23,6-12,8=10,8(g)
$m_{HCl}$=91,25×20%=18,25(g)⇒$n_{HCl}$ =18,25÷36,5=0,5(mol)
PTHH: Mg+2HCl→$MgCl_{2}$ +$H_{2}$
x 2x
PTHH: Fe+2HCl→$FeCl_{2}$ +$H_{2}$
y 2y
Hệ$\left \{ {{24x+56y=10,8} \atop {2x+2y=0,5}} \right.$
⇒$\left \{ {{x=0,1} \atop {y=0,15}} \right.$
⇒$m_{Mg}$ =0,1×24=2,4(g)
$m_{Fe}$=0,15×56=8,4(g)
⇒%$m_{Cu}$ =(12,8÷23,6)×100≈54,24%
%$m_{Mg}$=(2,4÷23,6)×100≈10,17%
%$m_{Fe}$=100%-54,24%-10,17%=35,59%