Đáp án:

\(\begin{array}{l}
2.\\
{m_{CuO}} = 16g\\
3.\\
C{\% _{NaOH(dư)}} = 2,5\% \\
C{\% _{N{a_2}S{O_4}}} = 17,7\% 
\end{array}\) 

Giải thích các bước giải:

\(\begin{array}{l}
1.\\
2NaOH + C{\rm{uS}}{{\rm{O}}_4} \to N{a_2}S{O_4} + Cu{(OH)_2}\\
Cu{(OH)_2} \to CuO + {H_2}O\\
2.\\
{n_{NaOH}} = 0,5mol\\
{m_{{\rm{CuS}}{O_4}}} = \dfrac{{160 \times 20\% }}{{100\% }} = 32g\\
 \to {n_{{\rm{CuS}}{O_4}}} = 0,2mol\\
 \to {n_{{\rm{CuS}}{O_4}}} < {n_{NaOH}} \to {n_{NaOH}}dư\\
 \to {n_{Cu{{(OH)}_2}}} = {n_{{\rm{CuS}}{O_4}}} = 0,2mol\\
 \to {n_{CuO}} = {n_{Cu{{(OH)}_2}}} = 0,2mol\\
 \to {m_{CuO}} = 16g\\
3.\\
{n_{N{a_2}S{O_4}}} = {n_{{\rm{CuS}}{O_4}}} = 0,2mol \to {m_{N{a_2}S{O_4}}} = 28,4g\\
{n_{NaOH(pt)}} = 2{n_{{\rm{CuS}}{O_4}}} = 0,4mol\\
 \to {n_{NaOH(dư)}} = 0,1mol \to {m_{NaOH(dư)}} = 4g\\
{m_{{\rm{dd}}}} = {m_{NaOH}} + {m_{{\rm{dd}}CuS{O_4}}} - {m_{Cu{{(OH)}_2}}} = 160,4g\\
 \to C{\% _{NaOH(dư)}} = \dfrac{4}{{160,4}} \times 100\%  = 2,5\% \\
 \to C{\% _{N{a_2}S{O_4}}} = \dfrac{{28,4}}{{160,4}} \times 100\%  = 17,7\% 
\end{array}\)

`n_(NaOH)=\frac{20}{40}=0,5(mol)`

`n_(CuSO_4)=\frac{160.20%}{160}=0,2(mol)`

`CuSO_4+2NaOH->Na_2SO_4+Cu(OH)_2`

$Cu(OH)_2\xrightarrow{t^o}CuO+H_2O$

`b,`

Theo `PT`

`n_(CuO)=n_(CuSO_4)=0,2(mol)`

`=>m_(CuO)=0,2.80=16(g)`

`c,`

`m_(dd)=20+160-0,2.98=160,4(g)`

`C%_(Na_2SO_4)=\frac{0,2.142}{160,4}.100=17,7%`

`C%_(NaOH)=\frac{0,1.40}{160,4}.100=2,45%`