\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{H_2SO_4}=1,5\times0,2=0,3\left(mol\right)\)
PTHH:\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
Theo PTHH ta có :
\(\dfrac{n_{Al}}{2}=\dfrac{n_{H_2SO_4}}{3}\)\(\Rightarrow\) \(H_2SO_4\)và \(Al\) phản ứng hết hoàn toàn
\(n_{H2}=\dfrac{3n_{AL}}{2}=\dfrac{3.0,2}{2}=0,3\left(mol\right)\)
Thể tích khí :\(V_{H_2}=22,4\times0,3=6,72\left(l\right)\)