Đáp án:
\(\begin{array}{l}
a)\\
{m_{CuO}} = 40g\\
b)\\
C{\% _{NaCl}} = 19,435\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
CuC{l_2} + 2NaOH \to 2NaCl + Cu{(OH)_2}\\
Cu{(OH)_2} \to CuO + {H_2}O\\
{m_{{\rm{dd}}NaOH}} = \dfrac{{200 \times 20}}{{100}} = 40g\\
{n_{NaOH}} = \dfrac{m}{M} = \dfrac{{40}}{{40}} = 1mol\\
{n_{Cu{{(OH)}_2}}} = \frac{{{n_{NaOH}}}}{2} = 0,5mol\\
{n_{CuO}} = {n_{Cu{{(OH)}_2}}} = 0,5mol\\
{m_{CuO}} = n \times M = 0,5 \times 80 = 40g\\
b)\\
{m_{Cu{{(OH)}_2}}} = n \times M = 0,5 \times 98 = 49g\\
{m_{{\rm{dd}}spu}} = 150 + 200 - 49 = 301g\\
{n_{NaCl}} = {n_{NaOH}} = 1mol\\
{m_{NaCl}} = n \times M = 1 \times 58,5 = 58,5g\\
C{\% _{NaCl}} = \dfrac{{{m_{NaCl}}}}{{{m_{{\rm{dd}}spu}}}} \times 100\% = \dfrac{{58,5}}{{301}} \times 100\% = 19,435\%
\end{array}\)
