Giải thích các bước giải:

\(\begin{array}{l}
5.\\
{\rm{a)CuS}}{O_4} + 2NaOH \to Cu{(OH)_2} + N{a_2}S{O_4}\\
b)\\
{n_{CuS{O_4}}} = 0,25mol\\
{n_{NaOH}} = 0,6mol\\
 \to {n_{NaOH}} > {n_{CuS{O_4}}} \to {n_{NaOH}}dư\\
 \to {n_{Cu{{(OH)}_2}}} = {n_{CuS{O_4}}} = 0,25mol\\
 \to {m_{Cu{{(OH)}_2}}} = 24,5g\\
c)\\
{n_{N{a_2}S{O_4}}} = {n_{CuS{O_4}}} = 0,25mol\\
{n_{NaOH(pt)}} = 2{n_{CuS{O_4}}} = 0,5mol\\
 \to {n_{NaOH(dư)}} = 0,1mol\\
 \to C{M_{N{a_2}S{O_4}}} = \dfrac{{0,25}}{{0,2 + 0,3}} = 0,5M\\
 \to C{M_{NaOH(dư)}} = \dfrac{{0,1}}{{0,2 + 0,3}} = 0,2M\\
8.\\
a)\\
CuC{l_2} + 2NaOH \to Cu{(OH)_2} + 2NaCl\\
Cu{(OH)_2} \to CuO + {H_2}O\\
b)\\
{n_{NaOH}} = 0,5mol\\
 \to {n_{NaOH}} > {n_{CuC{l_2}}} \to {n_{NaOH}}dư\\
 \to {n_{Cu{{(OH)}_2}}} = {n_{CuC{l_2}}} = 0,2mol\\
 \to {n_{CuO}} = {n_{Cu{{(OH)}_2}}} = 0,2mol\\
 \to {m_{CuO}} = 16g\\
c)\\
{n_{NaCl}} = 2{n_{CuC{l_2}}} = 0,4mol\\
 \to {m_{NaCl}} = 23,4g\\
{n_{NaOH(pt)}} = 2{n_{CuC{l_2}}} = 0,4mol\\
 \to {n_{NaOH(dư)}} = 0,1mol\\
 \to {m_{NaOH(dư)}} = 4g
\end{array}\)