\(n_{CuSO_4}=\dfrac{200.12\%}{160}=0,15\left(mol\right)\)
PTHH: \(Zn+CuSO_4\rightarrow ZnSO_4+Cu\downarrow\)
0,15<--0,15------>0,15----->0,15
`=> m_{Zn} = 0,15.65 = 9,75 (g)`
Ta có: \(m_{ddspư}=200+9,75-0,15.64=200,15\left(g\right)\)
`=>` \(C\%_{ZnSO_4}=\dfrac{0,15.161}{200,15}.100\%=12,07\%\)