Đáp án:
1, Dung dịch \({H_2}S{O_4}\) đặc, nóng
\(\begin{array}{l}
{H_2}S{O_4} + ZnO \to ZnS{O_4} + {H_2}O\\
{H_2}S{O_4} + 2KOH \to {K_2}S{O_4} + 2{H_2}O\\
{H_2}S{O_4} + N{a_2}C{O_3} \to N{a_2}S{O_4} + C{O_2} + {H_2}O\\
2{H_2}S{O_4} + Cu \to C{\rm{uS}}{{\rm{O}}_4} + S{O_2} + 2{H_2}O\\
6{H_2}S{O_4} + 2Fe \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O
\end{array}\)
2,
\(\begin{array}{l}
C{H_3}{\rm{COOH + }}{{\rm{C}}_2}{H_5}OH \to C{H_3}{\rm{COO}}{{\rm{C}}_2}{H_5} + {H_2}O\\
{n_{C{H_3}{\rm{COOH}}}} = 1mol\\
{n_{{{\rm{C}}_2}{H_5}OH}} = 2,2mol\\
{n_{C{H_3}{\rm{COO}}{{\rm{C}}_2}{H_5}}} = {n_{C{H_3}{\rm{COOH}}}} = 1mol \to {m_{C{H_3}{\rm{COO}}{{\rm{C}}_2}{H_5}}} = 88g\\
H = \dfrac{{55}}{{88}} \times 100\% = 62,5\%
\end{array}\)
