Ta có :
\(nCuCl2=0,4.2=0,8\left(mol\right)\)
\(nNaOH=0,2.2=0,4\left(mol\right)\)
\(PTHH:CuCl2+2NaOH\rightarrow Cu\left(OH\right)2\downarrow+2NaCl\)
\(\Rightarrow nCuCl2\text{dư}=0,8-0,2=0,6\left(mol\right)\)
\(PTHH:Cu\left(OH\right)2\rightarrow CuO+H2O\)
Kết tủa A :Cu(OH)2
Dd B là: CuCl2 ; NaCl
\(\Rightarrow m_{cr}=0,2.80=16\left(g\right)\)
\(\Rightarrow C_M=\frac{0,4+0,6}{0,2+0,2}=2,5M\)