PTHH: \(Ba\left(OH\right)_2+CuSO_4\rightarrow Cu\left(OH\right)_2\downarrow+BaSO_4\downarrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Ba\left(OH\right)_2}=\dfrac{200\cdot17,1\%}{171}=0,2\left(mol\right)\\n_{CuSO_4}=\dfrac{500\cdot8\%}{160}=0,25\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) CuSO4 dư
\(\Rightarrow\left\{{}\begin{matrix}n_{Cu\left(OH\right)_2}=0,2\left(mol\right)=n_{BaSO_4}\\n_{CuSO_4\left(dư\right)}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{kết.tủa}=0,2\cdot98+0,2\cdot233=66,2\left(g\right)\\m_{CuSO_4\left(dư\right)}=0,05\cdot160=8\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{ddBa\left(OH\right)_2}+m_{ddCuSO_4}-m_{kết.tủa}=633,8\left(g\right)\)
\(\Rightarrow C\%_{CuSO_4\left(dư\right)}=\dfrac{8}{633,8}\cdot100\%\approx1,26\%\)
