Đáp án:

\(\begin{array}{l}
b)\\
m = 14,7g\\
a = 12g\\
c)\\
{C_M}NaCl = 0,43M\\
{C_M}NaOH \text{ dư } = 0,14M
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
CuC{l_2} + 2NaOH \to Cu{(OH)_2} + 2NaCl\\
Cu{(OH)_2} \xrightarrow{t^0} CuO + {H_2}O\\
b)\\
{n_{CuC{l_2}}} = 0,3 \times 0,5 = 0,15\,mol\\
{n_{NaOH}} = 0,4 \times 1 = 0,4\,mol\\
{n_{CuC{l_2}}} < \dfrac{{{n_{NaOH}}}}{2} \Rightarrow NaOH \text{ dư }\\
{n_{Cu{{(OH)}_2}}} = {n_{CuC{l_2}}} = 0,15\,mol\\
m = 0,15 \times 98 = 14,7g\\
{n_{CuO}} = {n_{Cu{{(OH)}_2}}} = 0,15\,mol\\
{m_{CuO}} = 0,15 \times 80 = 12g\\
c)\\
{n_{NaCl}} = 2{n_{CuC{l_2}}} = 0,3\,mol\\
{n_{NaOH}} \text{ dư } = 0,4 - 0,15 \times 2 = 0,1\,mol\\
{V_{{\rm{dd}}spu}} = 0,3 + 0,4 = 0,7l\\
{C_M}NaCl = \dfrac{{0,3}}{{0,7}} = 0,43M\\
{C_M}NaOH \text{ dư }= \dfrac{{0,1}}{{0,7}} = 0,14M
\end{array}\)