Đáp án:

 b) $\begin{gathered}
  \% {m_{HCOOH}} = 27,71\% ;\% {m_{C{H_3}COOH}} = 72,29\%  \hfill \\
  \% {m_{HCOONa}} = 29,31\% ;\% {m_{C{H_3}COONa}} = 70,69\%  \hfill \\ 
\end{gathered} $

Giải thích các bước giải:

 a) 

$\begin{gathered}
  HCOOH + NaOH \to HCOONa + {H_2}O \hfill \\
  C{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O \hfill \\
  PT{\text{ ion:}} \hfill \\
  HCOOH + O{H^ - } \to HCO{O^ - } + {H_2}O \hfill \\
  C{H_3}COOH + O{H^ - } \to C{H_3}CO{O^ - } + {H_2}O \hfill \\ 
\end{gathered} $

b) Gọi x, y là số mol của $HCOOH;C{H_3}COOH$

Ta có hpt: $\left\{ \begin{gathered}
  46x + 60y = 16,6 \hfill \\
  68x + 82y = 23,2 \hfill \\ 
\end{gathered}  \right. \Rightarrow x = 0,1;y = 0,2$

$\begin{gathered}
  \% {m_{HCOOH}} = \dfrac{{0,1.46}}{{16,6}}.100\%  = 27,71\%  \hfill \\
   \Rightarrow \% {m_{C{H_3}COOH}} = 100 - 27,71 = 72,29\%  \hfill \\
  \% {m_{HCOONa}} = \dfrac{{0,1.68}}{{23,2}}.100\%  = 29,31\%  \hfill \\
   \Rightarrow \% {m_{C{H_3}COONa}} = 100 - 29,31 = 70,69\%  \hfill \\ 
\end{gathered} $