\(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1(mol)\\ Al_2O_3+3H_2SO_4\to Al_2(SO_4)_3+3H_2O\\ \Rightarrow n_{H_2SO_4}=0,3(mol)n_{Al_2(SO_4)_3}=0,1(mol)\\ \Rightarrow C\%_{H_2SO_4}=\dfrac{0,3.98}{300}.100\%=9,8\%\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{0,1.342}{10,2+300}.100\%=11,03\%\)