\(n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ \Rightarrow n_{Al}=0,2(mol)\Rightarrow m_{Al}=0,2.27=5,4(g)\\ \Rightarrow m_{Cu}=10-5,4=4,6(g)\\ \Rightarrow \%_{Al}=\dfrac{5,4}{10}.100\%=54\%\\ \Rightarrow \%_{Cu}=100\%-54\%=46\%\\ n_{H_2SO_4}=0,3(mol)\Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,3.98}{20\%}=147(g)\)