Em tham khảo nha :

\(\begin{array}{l}
a)\\
CuC{l_2} + 2NaOH \to Cu{(OH)_2} + 2NaCl\\
{m_{NaOH}} = \dfrac{{200 \times 10}}{{100}} = 20g\\
{n_{NaOH}} = \dfrac{{20}}{{40}} = 0,5mol\\
\dfrac{{0,3}}{1} > \dfrac{{0,5}}{2} \Rightarrow CuC{l_2}\text{ dư}\\
{n_{Cu{{(OH)}_2}}} = \dfrac{{{n_{NaOH}}}}{2} = 0,25mol\\
Cu{(OH)_2} \to CuO + {H_2}O\\
{n_{CuO}} = {n_{Cu{{(OH)}_2}}} = 0,25mol\\
{m_{CuO}} = 0,25 \times 80 = 20g\\
b)\\
{n_{CuC{l_2}d}} = 0,3 - \dfrac{{0,5}}{2} = 0,05mol\\
{m_{CuC{l_2}d}} = 0,05 \times 135 = 6,75g\\
{n_{NaCl}} = {n_{NaOH}} = 0,5mol\\
{m_{NaCl}} = 0,5 \times 58,5 = 29,25g\\
c)\\
{m_{Cu{{(OH)}_2}}} = 0,25 \times 98 = 24,5g\\
{m_{{\rm{dd}}spu}} = 100 + 200 - 24,5 = 275,5g\\
C{\% _{CuC{l_2}}} = \dfrac{{6,75}}{{275,5}} \times 100\%  = 2,45\% \\
C{\% _{NaCl}} = \dfrac{{29,25}}{{275,5}} \times 100\%  = 10,62\% 
\end{array}\)