Đáp án:
\(\begin{array}{l}
b)\\
\% {m_{C{H_3}COOH}} = 72,29\% \\
\% {m_{HCOOH}} = 27,71\% \\
c)\\
{C_M}NaOH = 0,06M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
HCOOH + NaOH \to HCOONa + {H_2}O\\
C{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O\\
b)\\
hh:HCOOH(a\,mol),C{H_3}COOH(b\,mol)\\
\left\{ \begin{array}{l}
46a + 60b = 1,66\\
68a + 82b = 2,32
\end{array} \right.\\
\Rightarrow a = 0,01;b = 0,02\\
\% {m_{C{H_3}COOH}} = \dfrac{{0,02 \times 60}}{{1,66}} \times 100\% = 72,29\% \\
\% {m_{HCOOH}} = 100 - 72,29 = 27,71\% \\
c)\\
{n_{NaOH}} = 0,01 + 0,02 = 0,03\,mol\\
{C_M}NaOH = \dfrac{{0,03}}{{0,5}} = 0,06M
\end{array}\)