\(n_{NaOH}=2\cdot0,4=0,8\left(mol\right)\\ n_{CuCl_2}=\dfrac{67,5}{135}=0,5\left(mol\right)\\ PTHH:2NaOH+CuCl_2\rightarrow2NaCl+Cu\left(OH\right)_2\)
Vì \(\dfrac{n_{NaOH}}{2}< \dfrac{n_{CuCl_2}}{1}\) nên \(CuCl_2\) dư
\(\Rightarrow n_{Cu\left(OH\right)_2}=\dfrac{1}{2}n_{NaOH}=0,4\left(mol\right)\\ \Rightarrow m_{Cu\left(OH\right)_2}=0,4\cdot98=39,2\left(g\right)\)