a) PTHH: \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\downarrow\)
b) Ta có: \(\left\{{}\begin{matrix}n_{CuCl_2}=0,2mol\\n_{NaOH}=\frac{20}{40}=0,5\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\frac{0,2}{1}< \frac{0,5}{2}\) \(\Rightarrow\) NaOH còn dư, CuCl2 phản ứng hết
\(\Rightarrow n_{Cu\left(OH\right)_2}=0,2mol\) \(\Rightarrow m_{Cu\left(OH\right)_2}=0,2\cdot98=19,6\left(g\right)\)
c) Theo PTHH: \(\left\{{}\begin{matrix}n_{NaCl}=0,4mol\\n_{NaOH\left(dư\right)}=0,1mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\\m_{NaOH\left(dư\right)}=0,1\cdot40=4\left(g\right)\end{matrix}\right.\)