\(Fe+CuSO_4\rightarrow FeSO_4+Cu\\ n_{CuSO_4\left(bđ\right)}=\dfrac{8\%.100}{160}=0,05\left(mol\right)\\ n_{Fe\left(bđ\right)}=\dfrac{3,8}{56}=\dfrac{19}{280}\left(mol\right)\\ Vì:\dfrac{0,05}{1}< \dfrac{\dfrac{19}{280}}{1}\Rightarrow Fedư\\ Đặt:a=n_{Fe\left(p.ứ\right)}=n_{CuSO_4\left(p.ứ\right)}=n_{Cu}=n_{FeSO_4}\\ \Rightarrow m_{sắt.sau}=m_{sắt.trước}-m_{Fe\left(p.ứ\right)}+m_{Cu}\\ \Leftrightarrow4,04=3,8+8a\\ \Leftrightarrow a=0,03\\ \Rightarrow m_{Cu}=0,03.64=1,92\left(g\right)\)
\(b,C\%_{ddCuSO_4\left(dư\right)}=\dfrac{\left(0,05-0,03\right).160}{0,03.56+100-0,03.64}.100\approx3,208\%\\ C\%_{ddFeSO_4}=\dfrac{0,03.152}{0,03.56+100-0,03.64}.100\approx4,672\%\)
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