Giải thích các bước giải:

\(\begin{array}{l}
C{H_3}{\rm{COO}}H + {C_2}{H_5}OH \to C{H_3}{\rm{COO}}{C_2}{H_5} + {H_2}O\\
a)\\
{n_{C{H_3}{\rm{COO}}H}} = 0,75mol\\
{n_{{C_2}{H_5}OH}} = 1,63\\
 \to {n_{{C_2}{H_5}OH}} > {n_{C{H_3}{\rm{COO}}H}}\\
 \to {n_{{C_2}{H_5}OH}}dư\\
 \to {n_{C{H_3}{\rm{COO}}{C_2}{H_5}}} = {n_{C{H_3}{\rm{COO}}H}} = 0,75mol\\
 \to {m_{C{H_3}{\rm{COO}}{C_2}{H_5}}} = 0,75 \times 88 = 66g\\
 \to H = \dfrac{{41,25}}{{66}} \times 100\%  = 62,5\% \\
b)\\
C{H_3}{\rm{COO}}H + NaHC{O_3} \to C{H_3}{\rm{COO}}Na + {H_2}O + C{O_2}\\
{n_{NaHC{O_3}}} = {n_{C{H_3}{\rm{COO}}H}} = 0,75mol\\
 \to {m_{NaHC{O_3}}} = 0,75 \times 84 = 63g\\
 \to {m_{{\rm{dd}}NaHC{O_3}}} = \dfrac{{63}}{{6,3\% }} \times 100\%  = 1000g\\
{n_{C{H_3}{\rm{COONa}}}} = {n_{C{O_2}}} = {n_{C{H_3}{\rm{COO}}H}} = 0,75mol\\
 \to {m_{C{H_3}{\rm{COONa}}}} = 0,75 \times 82 = 61,5g\\
 \to {m_{{\rm{dd}}C{H_3}{\rm{COONa}}}} = {m_{C{H_3}{\rm{COO}}H}} + {m_{{\rm{dd}}NaHC{O_3}}} - {m_{C{O_2}}}\\
 \to {m_{{\rm{dd}}C{H_3}{\rm{COONa}}}} = 45 + 1000 - 0,75 \times 44 = 1012g\\
 \to C{\% _{{\rm{dd}}C{H_3}{\rm{COONa}}}} = \dfrac{{61,5}}{{1012}} \times 110\%  = 6,08\% 
\end{array}\)