\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ a,Zn+H_2SO_4\to ZnSO_4+H_2\\ b,n_{Zn}=0,1(mol)\Rightarrow m_{Zn}=0,1.65=6,5(g)\\ \Rightarrow \%_{Zn}=\dfrac{6,5}{10,5}.100\%=61,9\%\\ \Rightarrow \%_{Cu}=100\%-61,9\%=38,1\%\\ c,n_{H_2SO_4}=0,1(mol)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,1}{0,5}=0,2M\)