\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
\(n_{H_2}=\frac{13,44}{22.4}=0,6\left(mol\right)\)
a) Theo PT => \(m_{Al}=\frac{0,6.2}{3}.27=10,8\left(g\right)\)
\(\%m_{Al}=\frac{10,8}{12}.100=90\%\)
=> %mAg=100-90=10%
b)Theo PT: \(\Rightarrow m_{H_2SO_4}=0,6.98=58,8\left(g\right)\)
\(\Rightarrow C\%_{H_2SO_4}=\frac{58,8}{300}.100=19,6\%\)