Đáp án:

\(\begin{array}{l}
2)\\
{m_{CuO}} = 16g\\
3)\\
C{\% _{N{a_2}S{O_4}}} = 17,7\% \\
C{\% _{NaOH}} = 2,49\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
1)\\
2NaOH + CuS{O_4} \to Cu{(OH)_2} + N{a_2}S{O_4}\\
Cu{(OH)_2} \to CuO + {H_2}O\\
2)\\
{n_{NaOH}} = \dfrac{m}{M} = \dfrac{{20}}{{40}} = 0,5mol\\
{m_{CuS{O_4}}} = \dfrac{{{m_{{\rm{dd}}CuS{O_4}}} \times 20\% }}{{100\% }} = \dfrac{{160 \times 20}}{{100}} = 32g\\
{n_{CuS{O_4}}} = \dfrac{m}{M} = \dfrac{{32}}{{160}} = 0,2mol\\
\dfrac{{0,5}}{2} > \dfrac{{0,2}}{1} \Rightarrow NaOH\text{ dư}\\
 \Rightarrow {n_{Cu{{(OH)}_2}}} = {n_{CuS{O_4}}} = 0,2mol\\
 \Rightarrow {n_{CuO}} = {n_{Cu{{(OH)}_2}}} = 0,2mol\\
{m_{CuO}} = n \times M = 0,2 \times 80 = 16g\\
3)\\
{m_{Cu{{(OH)}_2}}} = n \times M = 0,2 \times 98 = 19,6g\\
{m_{{\rm{dd}}spu}} = {m_{NaOH}} + {m_{{\rm{dd}}CuS{O_4}}} - {m_{Cu{{(OH)}_2}}}\\
\,\,\,\,\,\,\,\,\,\,\,\, = 20 + 160 - 19,6 = 160,4g\\
{n_{N{a_2}S{O_4}}} = {n_{CuS{O_4}}} = 0,2mol\\
{m_{N{a_2}S{O_4}}} = n \times M = 0,2 \times 142 = 28,4g\\
C{\% _{N{a_2}S{O_4}}} = \dfrac{{{m_{N{a_2}S{O_4}}}}}{{{m_{{\rm{ddspu}}}}}} \times 100\%  = \dfrac{{28,4}}{{160,4}} \times 100\%  = 17,7\% \\
{n_{NaO{H_d}}} = {n_{NaOH}} - 2{n_{CuS{O_4}}} = 0,5 - 2 \times 0,2 = 0,1mol\\
{m_{NaO{H_d}}} = n \times M = 0,1 \times 40 = 4g\\
C{\% _{NaOH}} = \dfrac{{{m_{NaOHd}}}}{{{m_{{\rm{ddspu}}}}}} \times 100\%  = \dfrac{4}{{160,4}} \times 100\%  = 2,49\% 
\end{array}\)