\(a.Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ n_{Zn}=n_{H_2}=0,2\left(mol\right)\\ \Rightarrow\%m_{Zn}=\dfrac{0,2.65}{25.8}.100=50,39\%\\ \%m_{Cu}=100-50,39=49,61\%\\ b.m_{ddsaupu}=0,2.65+200-0,2.2=212,6\left(g\right)\\ n_{ZnSO_4}=n_{H_2}=0,2\left(mol\right)\\ C\%_{ZnSO_4}=\dfrac{0,2.161}{212,6}.100=15,15\%\)