mkl ko tan =mCu = 1,86(g)
nh2= 0,135 mol
Fe + 2HCl--> FeCl2 + H2
x---------------------------x
2Al + 6HCl---> 2AlCl3 + 3H2
y-----------------------------3/2y
ta có hpt
56x+27y = 6-1.86=4,14
x+3/2y=0,135
=> x=0,045 => mFe=2.52(g)
y=0,06 => mAl=1,62(g)
mCu=1,86(g)