Đáp án:

\(\begin{array}{l}
b)\\
{m_{CuO}} = 16g\\
c)\\
C{\% _{{K_2}S{O_4}}} = 5,11\% \\
C{\% _{CuS{O_4}}} = 2,35\% 
\end{array}\)

Giải thích các bước giải:

 300g dung dịch $CuSO_4$ 16%

\(\begin{array}{l}
a)\\
2KOH + CuS{O_4} \to {K_2}S{O_4} + Cu{(OH)_2}\\
Cu{(OH)_2} \to CuO + {H_2}O\\
b)\\
{m_{KOH}} = \dfrac{{400 \times 5,6}}{{100}} = 22,4g\\
{n_{KOH}} = \dfrac{m}{M} = \dfrac{{22,4}}{{56}} = 0,4mol\\
{m_{CuS{O_4}}} = \dfrac{{300 \times 16}}{{100}} = 48g\\
{n_{CuS{O_4}}} = \dfrac{m}{M} = \dfrac{{48}}{{160}} = 0,3mol\\
\dfrac{{0,4}}{2} < \dfrac{{0,3}}{1} \Rightarrow CuS{O_4}\text{ dư}\\
{n_{Cu{{(OH)}_2}}} = \dfrac{{{n_{KOH}}}}{2} = 0,2mol\\
 \Rightarrow {n_{CuO}} = {n_{Cu{{(OH)}_2}}} = 0,2mol\\
{m_{CuO}} = n \times M = 0,2 \times 80 = 16g\\
c)\\
{n_{{K_2}S{O_4}}} = \dfrac{{{n_{KOH}}}}{2} = 0,2mol\\
{m_{{K_2}S{O_4}}} = n \times M = 0,2 \times 174 = 34,8g\\
{n_{CuS{O_4}d}} = {n_{CuS{O_4}}} - \dfrac{{{n_{KOH}}}}{2} = 0,1mol\\
{m_{CuS{O_4}d}} = n \times M = 0,1 \times 160 = 16g\\
{m_{Cu{{(OH)}_2}}} = n \times M = 0,2 \times 98 = 19,6g\\
{m_{ddspu}} = 400 + 300 - 19,6 = 680,4g\\
C{\% _{{K_2}S{O_4}}} = \dfrac{{34,8}}{{680,4}} \times 100\%  = 5,11\% \\
C{\% _{CuS{O_4}}} = \dfrac{{16}}{{680,4}} \times 100\%  = 2,35\% 
\end{array}\)