Đáp án:
\(\begin{array}{l}
{m_{NaOH}} = 4g\\
{m_{NaCl}} = 23,4g\\
{m_{Cu{{(OH)}_2}}} = 19,6g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
CuC{l_2} + 2NaOH \to Cu{(OH)_2} + 2NaCl\\
b)\\
{n_{NaOH}} = \dfrac{m}{M} = \dfrac{{20}}{{40}} = 0,5\,mol\\
\dfrac{{0,2}}{1} < \dfrac{{0,5}}{2} \Rightarrow \text{ NaOH dư}\\
{n_{NaOH}} \text{ dư}= 0,5 - 0,2 \times 2 = 0,1\,mol\\
{n_{NaCl}} = 2{n_{CuC{l_2}}} = 0,4\,mol\\
{n_{Cu{{(OH)}_2}}} = {n_{CuC{l_2}}} = 0,2\,mol\\
{m_{NaOH}}\text{ dư} = 0,1 \times 40 = 4g\\
{m_{NaCl}} = 0,4 \times 58,5 = 23,4g\\
{m_{Cu{{(OH)}_2}}} = 0,2 \times 98 = 19,6g
\end{array}\)
