$nAl=\frac{2,7}{27}=0,1(mol)$a)$4Al+3O2\rightarrow 2Al2O3$(1)
$Al2O3+6HCl\rightarrow 2AlCl3+3H2O$(2)
b)Từ pt(1)$\Rightarrow$nO2=0,075(mol)
$\Rightarrow$V O2=0,075.22,4=1,68(l)
c)Từ pt(2)$\Rightarrow$nHCl=6n Al2O3=0,3(mol)
$\Rightarrow$mHCl=0,3.36,5=10,95(g)
$\Rightarrow$m dd HCl=$\frac{10,95.100}{14,6}$=75(g)