a)
2Al+6HCl->2AlCl3+3H2
Fe+2HCl->FeCl2+H2
gọi a là nAl b là nFe
nHCl=200x11,68%/36,5=0,64 mol
ta có
3a+2b=0,64
133,5a+127b=29,24
=>a=0,2 b=0,02
mhh=0,2x27+0,02x56=6,52 g
b)
mdd spu=6,52+200-0,32x2=205,88 g
C%AlCl3=0,2x133,5/205,88x100%=12,97%
C%FeCl2=0,02x127/205,88x100%=1,23%