Đáp án:
\(\begin{array}{l}
a)\\
\% C{H_3}COOH = 56,6\% \\
\% HCOOH = 43,4\% \\
b)\\
m = 15g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
C{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O\\
HCOOH + NaOH \to HCOONa + {H_2}O\\
{n_{NaOH}} = 0,2 \times 1 = 0,2mol\\
hh:C{H_3}COOH(a\,mol),HCOOH(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,2\\
60a + 46b = 10,6
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,1\\
{m_{C{H_3}COOH}} = 0,1 \times 60 = 6g\\
\% C{H_3}COOH = \dfrac{6}{{10,6}} \times 100\% = 56,6\% \\
\% HCOOH = 100 - 56,6 = 43,4\% \\
b)\\
{n_{C{H_3}COONa}} = {n_{C{H_3}COOH}} = 0,1mol\\
{m_{C{H_3}COONa}} = 0,1 \times 82 = 8,2g\\
{n_{HCOONa}} = {n_{HCOOH}} = 0,1mol\\
{m_{HCOONa}} = 0,1 \times 68 = 6,8g\\
m = 8,2 + 6,8 = 15g
\end{array}\)
