\(n_{Al}=\dfrac{4,05}{27}=0,15\left(mol\right)\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ n_{H_2SO_4}=\dfrac{3}{2}.0,15=0,225\left(mol\right)\\ V_{ddH_2SO_4}=\dfrac{0,225}{2,5}=0,09\left(l\right)=90\left(ml\right)\\ Vậy:V=90\left(ml\right)\\ n_{Al_2\left(SO_4\right)_3}=\dfrac{0,15}{2}=0,075\left(mol\right)\\ m_{Al_2\left(SO_4\right)_3}=342.0,075=25,65\left(g\right)\)