Đáp án + Giải thích các bước giải:

$a\big)$

Đặt $n_{CH_3COOH}=x(mol);n_{HCOOH}=y(mol)$

$\to 60x+46y=10,6(1)$

$CH_3COOH+NaOH\to CH_3COONa+H_2O$

$HCOOH+NaOH\to HCOONa+H_2O$

Theo PT: $x+y=n_{NaOH}=\frac{200}{1000}.1=0,2(2)$

Từ $(1)(2)\to x=y=0,1(mol)$

$\%m_{CH_3COOH}=\frac{0,1.60}{10,6}.100\%\approx 56,6\%$

$\%m_{HCOOH}=100-56,6=43,4\%$

$b\big)$

Theo PT: $n_{CH_3COONa}=0,1(mol);n_{HCOONa}=0,1(mol)$

$\to m_{CH_3COONa}=0,1.82=8,2(g)$

$\to m_{HCOONa}=0,1.68=6,8(g)$

Em tham khảo nha :

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