\(n_{Al}=\dfrac{5,4}{27}=0,2mol\\ n_{H_2Ô_4SO_4}=\dfrac{392.10\%}{100\%.98}=0,4mol\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ \Rightarrow\dfrac{0,2}{2}< \dfrac{0,4}{3}\Rightarrow H_2SO_4.dư\\ n_{H_2}=n_{Al}=0,2mol\\ V_{H_2}=0,2.24,79=4,958l\)