$ m_{CuSO_4} = 10\% × 20 = 2 (g) $
$ n_{CuSO_4} = \dfrac{m}{M} = \dfrac{2}{160} = \dfrac{1}{80} (mol) $
$ Zn + CuSO_4 \to Cu ↓ + ZnSO_4 $
Theo phương trình :
$ n_{Zn} = n_{CuSO_4} = \dfrac{1}{80} (mol) $
$ \to m_{Zn} = n × M = \dfrac{1}{80} × 65 = 0,8125 (g) $
$ n_{ZnSO_4} = n_{CuSO_4} = \dfrac{1}{80} (mol) $
$ \to m_{ZnSO_4} = n × M = \dfrac{1}{80} × 161 = 2,0125 (g) $
$ m_{ddspứ } = m_{Zn} + m_{ddCuSO_4} - m_{Cu} = 0,8125 + 20 - \dfrac{1}{80} × 64 = 20,0125 (g) $
$ \to C\%_{ZnSO_4} = \dfrac{m_{ct}}{m_{dd}} × 100\% = \dfrac{ 2,0125 }{ 20,0125 } × 100 ≈ 10,056\% $