Bảo toàn e:
\(3n_{Al}=2n_{H_2}\)
\(\Rightarrow n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}.0,6=0,4\left(mol\right)\)
\(\Rightarrow m_{Al}=10,8\left(g\right)\)
\(\Rightarrow\%m_{Al}=\dfrac{10,8}{21,6}.100\%=50\%\)
\(\Rightarrow\%m_{Ag}=50\%\)
\(n_{H_2}=\dfrac{13,44}{22,4}=0,6(mol)\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ \Rightarrow n_{Al}=\dfrac{2}{3}n_{H_2}=0,4(mol)\\ \Rightarrow m_{Al}=0,4.27=10,8(g)\\ \Rightarrow \%_{Al}=\dfrac{10,8}{21,6}.100\%=50\%\\ \Rightarrow \%_{Ag}=100\%-50\%=50\%\)