`n_{CH_3COOH}={200.5}/{100.60}=1/6(mol)`

`PTHH:2CH_3COOH+Na_2CO_3->2CH_3COONa+CO_2+H_2O`

`(mol)`   `1/6->`                   `1/12`              `1/6`                   `1/12`

`V_{CO_2}=1/12 .22,4~~1,87(l)`

`m_{dd Na_2CO_3}={1/12.106.100}/10 ~~88,33(g)`

`m_{dd sau}=200+88,33-1/12 .44~~284,66(g)`

`C% CH_3COONa ={1/6 .82.100%}/{284,66} ~~4,8%`

Theo đề

$\rm n_{CH_3COOH} = \dfrac{200 . 5}{100 . 60} = \dfrac{1}{6}$ mol

Phản ứng xảy ra:

$\rm Na_2CO_3 + 2CH_3COOH \longrightarrow 2CH_3COONa + CO_2 + H_2O$

Theo PTHH:

$\rm n_{CO_2} = \dfrac{1}{2}n_{CH_3COOH} = \dfrac{1}{2} . \dfrac{1}{6} = \dfrac{1}{12}$ mol

Thể tích khí $\rm CO_2$ sinh ra:

$\rm V_{CO_2} = \dfrac{1}{12} . 22,4 \approx 1,867L$

Theo PTHH:

$\rm n_{Na_2CO_3} = \dfrac{1}{2}n_{CH_3COOH} = \dfrac{1}{2}. \dfrac{1}{6} = \dfrac{1}{12}$ mol

Khối lượng dd $\rm Na_2CO_3$ cần:

$\rm m_{dd Na_2CO_3} = \dfrac{ \dfrac{1}{12} . 106 . 100}{10} = \dfrac{265}{3}g$

Bảo toàn khối lượng:

$\rm m_{dd sau} = 200 + \dfrac{265}{3} - \dfrac{1}{12} . 44 = \dfrac{854}{3}g$

Theo PTHH:

$\rm n_{CH_3COONa} = n_{CH_3COOH} = \dfrac{1}{6}$ mol

Nồng độ phần trăm dung dịch sau là

$\rm C$`%_{text[dd sau]}`$\rm = \dfrac{ \dfrac{1}{6}. 82}{ \dfrac{854}{3}} .100$`%`$\rm \approx 4,801$`%`