nAl = 0.2 mol
mHCl = 25.55g
nHCl = 0.7 mol
2Al + 6HCl --> 2AlCl3 + 3H2
Bđ:0.2____0.7
Pư: 0.2____0.6____________0.3
Kt: 0______0.1____________0.3
mHCl dư = 0.1*36.5 = 3.65 g
RxOy + yH2 -to-> xR + yH2O
0.3/y___0.3
M = 16/0.3/y =160/3 y
<=> Rx + 16y = 160/3 y
<=> Rx = 112/3 y
=> x = 2 , y = 3
=> R = 56
Vậy: CTHH là : Fe2O3
