a)
2Al+6HCl\(\rightarrow\)2AlCl3+3H2(1)
Al2O3+6HCl\(\rightarrow\)2AlCl3+3H2O(2)
nH2=\(\frac{13,44}{22,4}\)=0,6(mol)
\(\rightarrow\)nAl=\(\frac{\text{0,6.2}}{3}\)=0,4(mol)
nHCl=0,5.3,6=1,8(mol)
nHCl(2)=nHCl-nHCl(1)=1,8-0,6.2=0,6(mol)
\(\rightarrow\)nAl2O3=\(\frac{0,6}{6}\)=0,1(mol)
mAl=0,4.27=10,8(g)
mAl2O3=0,1.102=10,2(g)
b)
nAlCl3=nAlCl3(1)+nAlCl3(2)=0,4+0,1.2=0,6(mol)
m=mAlCl3=0,6.133,5=80,1(g)