\(\begin{array}{l}
3)\\
a)N{a_2}O + 2{H_2}O \to 2NaOH\\
nN{a_2}O = \dfrac{{15,5}}{{62}} = 0,25\,mol\\
= > nNaOH = 0,25 \times 2 = 0,5\,mol\\
CMNaOH = \dfrac{{0,5}}{{0,5}} = 1M\\
b)\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
n{H_2}S{O_4} = \dfrac{{0,5}}{2} = 0,25\,mol\\
m{\rm{dd}}{H_2}S{O_4} = \dfrac{{0,25 \times 98}}{{20\% }} = 122,5g\\
V{H_2}S{O_4} = \dfrac{{122,5}}{{1,4}} = 87,5ml\\
4)\\
2Fe{(OH)_3} \to F{e_2}{O_3} + 3{H_2}O\\
Zn{(OH)_2} + {H_2}S{O_4} \to ZnS{O_4} + 2{H_2}O\\
NaOH + HCl \to NaCl + {H_2}O\\
2NaOH + C{O_2} \to N{a_2}C{O_3} + {H_2}O\\
{H_2}S{O_4} + 2NaOH \to N{a_2}S{O_4} + 2{H_2}O
\end{array}\)