VddHCl=340ml=0,34l
CM=3M
=>nddHCl=V.CM=1,02mol
Fe+2HCl->FeCl2+H2
x 2x x x
Mg+2HCl->MgCl2+H2
y 2y y y
m muối khan =mFeCl2+mMgCl2 =57,09 g
57,09= 127x+95y
2x+2y=1,02
=>x=0,27
y=0.24
% mFe= (0,27.56)\(0,27.56+0,24.24) . 100%~~72,4%
=>%mMg=100-72,4=27,6%