\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(0.2.......0.3................................0.3\)
\(V_{H_2}=0.3\cdot22.4=6.72\left(l\right)\)
\(C_{M_{H_2SO_4}}=\dfrac{0.3}{0.1}=3\left(M\right)\)