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Giải thích các bước giải:

\(\begin{array}{l}
1.\\
KOH + HCl \to KCl + {H_2}O\\
{n_{HCl}} = 0,375mol\\
a)\\
{n_{KOH}} = {n_{HCl}} = 0,375mol\\
 \to {V_{KOH}} = \dfrac{{0,375}}{2} = 0,1875l\\
b)\\
{n_{KCl}} = {n_{HCl}} = 0,375mol\\
 \to C{M_{KCl}} = \dfrac{{0,375}}{{0,1875 + 0,25}} = 0,86M\\
c)\\
NaOH + HCl \to NaCl + {H_2}O\\
{n_{NaOH}} = {n_{HCl}} = 0,375mol \to {m_{NaOH}} = 15g\\
 \to {m_{{\rm{dd}}NaOH}} = \dfrac{{15}}{{10\% }} \times 100\%  = 150g\\
2.\\
a)Zn + C{\rm{uS}}{O_4} \to ZnS{O_4} + Cu\\
b)\\
{m_{{\rm{CuS}}{O_4}}} = \dfrac{{32 \times 10\% }}{{100\% }} = 3,2g\\
 \to {n_{{\rm{CuS}}{O_4}}} = 0,02mol\\
 \to {n_{Zn}} = {n_{{\rm{CuS}}{O_4}}} = 0,02mol \to {m_{Zn}} = 1,3g\\
c)\\
{n_{ZnS{O_4}}} = {n_{{\rm{CuS}}{O_4}}} = 0,02mol \to {m_{ZnS{O_4}}} = 3,22g\\
{n_{Cu}} = {n_{{\rm{CuS}}{O_4}}} = 0,02mol \to {m_{Cu}} = 1,28g\\
{m_{{\rm{dd}}}} = {m_{Zn}} + {m_{{\rm{dd}}}}_{{\rm{CuS}}{O_4}} - {m_{Cu}} = 32,02g\\
 \to C{\% _{ZnS{O_4}}} = \dfrac{{3,22}}{{32,02}} \times 100\%  = 10,06\% \\
3.\\
2KOH + {H_2}S{O_4} \to {K_2}S{O_4} + 2{H_2}O\\
{m_{{H_2}S{O_4}}} = \dfrac{{200 \times 14,7\% }}{{100\% }} = 29,4g\\
 \to {n_{{H_2}S{O_4}}} = 0,3mol\\
a)\\
{n_{KOH}} = 2{n_{{H_2}S{O_4}}} = 0,6mol \to {m_{KOH}} = 33,6g\\
 \to {m_{{\rm{dd}}KOH}} = \dfrac{{33,6}}{{5,6\% }} \times 100\%  = 600g\\
 \to {V_{{\rm{dd}}KOH}} = \dfrac{{600}}{{10,45}} = 57,42ml = 0,06l\\
b)\\
{n_{{K_2}S{O_4}}} = {n_{{H_2}S{O_4}}} = 0,3mol \to {m_{{K_2}S{O_4}}} = 52,2g\\
 \to {m_{{\rm{dd}}}} = {m_{{\rm{dd}}KOH}} + {m_{{\rm{dd}}{H_2}S{O_4}}} = 800g\\
 \to C{\% _{{K_2}S{O_4}}} = \dfrac{{52,2}}{{800}} \times 100\%  = 6,525\% 
\end{array}\)