4Al + 3O2 =nhiệt độ=> 2Al2O3
0,1............................................0,05 (mol)
Al2O3 + 6HCl ==> 2AlCl3 + 3H2O
0,05..........0,3................0,1
Ta có: nAl= 2,7/27=0,1 (mol) ==> nAl2O3= (2/4)x0,1= 0,05 (mol) ==> mAl2O3= 0,05x102= 5,1 (g)
==> nHCl= 6nAl2O3= 6x0,05= 0,3 (mol)
==> mHCl= 0,3x36,5=10,95 (g)
==> mddHCl= (mctx100)/C%= (10,95x100)/14,6= 75 (g)
mdd sau phản ứng= mAl2O3 + mdd HCl= 5,1 + 75= 80,1 (g)
mAlCl3= 0,1x133,5= 13,35 (g)
C% AlCl3= (mct/mdd)x100= (13,35/80,1)x100= 16,67%