\(n_{Zn}=\dfrac{m}{M}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(Zn+CuSO_4\rightarrow ZnSO_4+Cu\)
Theo PTHH : \(n_{ZnSO_4}=n_{Cu}=n_{Zn}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{ZnSO_4}=n.M=0,1.161=16,1\left(g\right)\\m_{Cu}=n.M=0,1.64=6,4\left(g\right)\end{matrix}\right.\)
nZn = 0,1 mol
Zn + CuSO4 → ZnSO4 + Cu
0,1.........................0,1......0,1
⇒ mZnSO4 = 0,1.161 = 16,1 (g)
⇒ mCu = 0,1.64 = 6,4 (g)