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\(\begin{array}{l}
4)\\
Zn + C{l_2} \to ZnC{l_2}\\
{n_{C{l_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{n_{Zn}} = {n_{C{l_2}}} = 0,1mol\\
{m_{Zn}} = 0,1 \times 65 = 6,5g\\
{n_{ZnC{l_2}}} = {n_{C{l_2}}} = 0,1mol\\
{m_{ZnC{l_2}}} = 0,1 \times 136 = 13,6g\\
5)\\
3NaOH + {H_3}P{O_4} \to N{a_3}P{O_4} + 3{H_2}O\\
{m_{NaOH}} = \dfrac{{200 \times 4}}{{100}} = 8g\\
{n_{NaOH}} = \dfrac{8}{{40}} = 0,2mol\\
{n_{{H_3}P{O_4}}} = \dfrac{{{n_{NaOH}}}}{3} = \frac{1}{{15}}mol\\
{m_{{H_3}P{O_4}}} = \dfrac{1}{{15}} \times 98 = 6,53g\\
C{\% _{{H_3}P{O_4}}} = \dfrac{{6,53}}{{100}} \times 100\%  = 6,53\% \\
{n_{N{a_3}P{O_4}}} = \dfrac{{{n_{NaOH}}}}{3} = \dfrac{1}{{15}}mol\\
{m_{N{a_3}P{O_4}}} = \dfrac{1}{{15}} \times 164 = 10,93g\\
C{\% _{N{a_3}P{O_4}}} = \dfrac{{10,93}}{{200 + 100}} \times 100\%  = 3,64\% \\
6)\\
2KOH + {H_2}S{O_4} \to {K_2}S{O_4} + 2{H_2}O\\
{n_{KOH}} = 0,1 \times 2 = 0,2mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{{n_{KOH}}}}{2} = 0,1mol\\
{C_{{M_{{H_2}S{O_4}}}}} = \dfrac{{0,1}}{{0,2}} = 0,5mol\\
{n_{{K_2}S{O_4}}} = \dfrac{{{n_{KOH}}}}{2} = 0,1mol\\
{C_{{M_{{K_2}S{O_4}}}}} = \dfrac{{0,1}}{{0,1 + 0,2}} = \dfrac{1}{3}M
\end{array}\)