\(n_{Zn}=\dfrac{6,5}{65}=0,1(mol)\\ m_{dd_{CuSO_4}}=1,12.200=224(g)\\ \Rightarrow n_{CuSO_4}=\dfrac{224.10\%}{100\%.160}=0,14(mol)\\ PTHH:Zn+CuSO_4\to ZnSO_4+Cu\)
Vì \(\dfrac{n_{Zn}}{1}<\dfrac{n_{CuSO_4}}{1}\) nên \(CuSO_4\) dư
\(a,n_{Cu}=n_{Zn}=0,1(mol)\\ \Rightarrow m_{Cu}=0,1.64=6,4(g)\\ b,n_{ZnSO_4}=n_{Zn}=0,1(mol)\\ \Rightarrow C_{M_{ZnSO_4}}=\dfrac{0,1}{0,2}=0,5M\\ C\%_{ZnSO_4}=\dfrac{0,1.161}{6,5+224-6,4}.100\%=7,18\%\)