a) 2Al + 6HCl --> 2AlCl3 +3H2(1)
Al2O3 +6HCl --> 2AlCl3 + 3H2O (2)
nH2=0,15(mol)
theo (1) : nAl=2/3nH2=0,1(mol)
=>mAl=2,7(g)
=>mAl2O3=7,8-2,7=5,1(g)
b) nAl2O3=0,05(mol)
theo (1) : nHCl(1)=3nAl=0,3(mol)
theo (2) : nHCl(2) =6nAl2O3=0,3(mol)
=>\(\Sigma nHCl=0,6\left(mol\right)\)
=>VHCl(cần dùng)=0,6/3=0,2(l)