V H2=13,44 lít nha
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
Ta có:\(n_{H2}=\frac{13,44}{22,4}=0,6\left(mol\right)\)
\(n_{Al}=\frac{2}{3}n_{H2}=0,4\left(mol\right)\)
\(\Rightarrow m_{Al}=0,4.27=10,8\left(g\right)\)
\(4Al+3O_2\rightarrow2Al_2O_3\)
Ta có: \(n_{O2}=\frac{3}{4}n_{Al}=\frac{3}{4}.0,4=0,3\left(mol\right)\Rightarrow V_{O2}=0,3.22,4=6,72\left(l\right)\)