Câu 11
-Chat ko tan la Cu-->%Cu=(12,8/23,6)*100=54,2%
Co m(HCl)=(20*91,25)/100=18,25g--> n(HCl)=18,25:36,5=0,5mol
Goi n(Mg)=a; n(Fe)=b
Mg+2HCl->MgCl2+H2
a......2a..........a............a
Fe+2HCl->FeCl2+H2
b......2b..........b.......b.
Khi do ta co he
{24a+56b=10,8
{2a+2b=0,5
->a=0,1; b=0,15
->%Mg=((24*0,1):23,6)*100=10,2%