Lời giải:

$\rm a)\\n_{CH_3COOH}=\dfrac{100\cdot 12\%}{60}=0,2\ (mol)$

Phản ứng xảy ra:

$\rm CH_3COOH+NaHCO_3\to CH_3COONa+CO_2+H_2O$

Theo phản ứng:

$\rm n_{NaHCO_3}=n_{CH_3COOH}=0,2\ (mol)$

$\rm C\%_{NaHCO_3}=\dfrac{0,2\cdot 84}{200}\cdot 100\%=8,4\%$

$\rm b)$

Theo phản ứng:

$\rm n_{CH_3COONa}=n_{CO_2}=n_{CH_3COOH}=0,2\ (mol)$

$\rm C\%_{CH_3COONa}=\dfrac{0,2\cdot 82}{100+200-0,2\cdot 44}\cdot 100\%=5,63\%$

`+)m_{CH_3COOH}=100.12\%=12(g)`

`n_{CH_3COOH}=(12)/(60)=0,2(mol)`

`a,` PTHH:

`CH_3COOH+NaHCO_3\to CH_3COONa+CO_2\uparrow+H_2O`

      `0,2`        `\to 0,2`           `\to 0,2`          `\to 0,2(mol)`

`=>m_{NaHCO_3}=0,2.84=16,8(g)`

`C\%_{NaHCO_3}=(16,8)/(200).100\%=8,4\%`

`b,` Dung dịch muối thu được là `CH_3COONa`

`m_{CH_3COONa}=0,2.82=16,4(g)`

`m_{\text{dung dịch sau p/ứ}}=m_{\text{dd }CH_3COOH}+m_{\text{dd } NaHCO_3}-m_{\uparrow}`

`=100+200-0,2.44=291,2(g)`

`C\%_{CH_3COONa}=(16,4)/(291,2).100\%≈5,63\%`