PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
Ta có: \(n_{H_2}=\frac{13,44}{22,4}=0,6\left(mol\right)\)
\(\Rightarrow n_{Al}=0,4mol\) \(\Rightarrow m_{Al}=0,4\cdot27=10,8\left(g\right)\)
\(\Rightarrow\%m_{Al}=\frac{10,8}{12}\cdot100=90\%\)
\(\Rightarrow\%m_{Ag}=10\%\)