Đáp án:

\(\begin{array}{l}
a)\\
{m_{Cu{{(OH)}_2}}} = 19,6g\\
b)\\
{C_\% }NaOH = 1,81\% \\
{C_\% }N{a_2}S{O_4} = 12,89\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
1)\\
a)\\
{n_{CuS{O_4}}} = \dfrac{{160 \times 20\% }}{{160}} = 0,2\,mol\\
{n_{NaOH}} = \dfrac{{80 \times 25\% }}{{40}} = 0,5\,mol\\
CuS{O_4} + 2NaOH \to Cu{(OH)_2} + N{a_2}S{O_4}\\
{n_{CuS{O_4}}} < \dfrac{{{n_{NaOH}}}}{2} \Rightarrow NaOH \text{ dư }\\
{n_{Cu{{(OH)}_2}}} = {n_{CuS{O_4}}} = 0,2\,mol\\
{m_{Cu{{(OH)}_2}}} = 0,2 \times 98 = 19,6g\\
b)\\
{n_{NaOH}} \text{ dư } = 0,5 - 0,2 \times 2 = 0,1\,mol\\
{n_{N{a_2}S{O_4}}} = {n_{CuS{O_4}}} = 0,2\,mol\\
{m_{{\rm{dd}}spu}} = 160 + 80 - 19,6 = 220,4g\\
{C_\% }NaOH \text{ dư }= \dfrac{{0,1 \times 40}}{{220,4}} \times 100\%  = 1,81\% \\
{C_\% }N{a_2}S{O_4} = \dfrac{{0,2 \times 142}}{{220,4}} \times 100\%  = 12,89\% 
\end{array}\)