a)\(4Al+3O2-->2Al2O3\)
\(n_{O2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(n_{Al}=\frac{5,4}{27}=0,2\left(mol\right)\)
Lập tỉ lệ
\(n_{O2}\left(\frac{0,15}{3}\right)=n_{Al}\left(\frac{0,2}{4}\right)\)
=. pư xảy ra hoàn toàn =>k dư
Vậy \(\%m_{Al2O3}=100\%\)
b) \(Al2O3+6HCl-->2AlCl3+3H2O\)
\(n_{Al2O3}=\frac{1}{2}n_{Al}=0,1\left(mol\right)\)
\(n_{HCl}=6n_{Al2O3}=0,6\left(mol\right)\)
\(V_{HCl}=\frac{0,6}{0,4}=1,5\left(M\right)\)