\(n_{H_2}=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(0.3.............................0.3..........0.45\)
\(m_{Al}=0.3\cdot27=8.1\left(g\right)\)
\(m_{Al_2\left(SO_4\right)_3}=0.3\cdot342=102.6\left(g\right)\)