Đáp án:

 `C% = 6,25%`

Giải thích các bước giải:

 `2CH_3COOH + Na_2CO_3 → 2CH_3COONa + CO_2 + H_2O`

`n_{CH_3COOH} = {300 . 6%}/{60} = 0,3(mol)` 

`n_{Na_2CO_3} = 1/2n_{CH_3COOH} = 0,15(mol)` 

`n_{CO_2} = n_{Na_2CO_3} =0,15(mol)` 

`n_{CH_3COONa} = n_{CH_3COOH} = 0,3(mol)` 

`mdd = 300 + {0,15 . 106}/{15,9%} - 0,15 . 44 = 393,4(g)`

`C%_{CH_3COONa} = {0,3 . 82}/{393,4} . 100 = 6,25(%)` 

`#aura`

Em tham khảo nha :

\(\begin{array}{l}
2C{H_3}COOH + N{a_2}C{O_3} \to 2C{H_3}COONa + C{O_2} + {H_2}O\\
{n_{C{H_3}COOH}} = \dfrac{{300 \times 6\% }}{{60}} = 0,3mol\\
{n_{N{a_2}C{O_3}}} = \dfrac{{{n_{C{H_3}COOH}}}}{2} = 0,15mol\\
{m_{N{a_2}C{O_3}}} = 0,15 \times 106 = 15,9g\\
{m_{ddN{a_2}C{O_3}}} = \dfrac{{15,9 \times 100}}{{15,9}} = 100g\\
{n_{C{O_2}}} = \frac{{{n_{C{H_3}COOH}}}}{2} = 0,15mol\\
{n_{C{H_3}COONa}} = {n_{C{H_3}COOH}} = 0,3mol\\
{m_{C{H_3}COONa}} = 0,3 \times 82 = 24,6g\\
C{\% _{C{H_3}COONa}} = \dfrac{{24,6}}{{300 + 100 - 0,15 \times 44}} \times 100\%  = 6,25\% 
\end{array}\)