Giải thích các bước giải:

$a)$

Đổi $200ml=0,2l$

$n_{NaOH}=0,2.1=0,2(mol)$

$CH_3COOH+NaOH\to CH_3COONa+H_2O$

$n_{CH_3CHOOH}=n_{NaOH}=0,2(mol)$

$m_{CH_3COOH}=0,2.60=12(g)$

$b)$

$n_{C_2H_5OH}=\dfrac{23}{46}=0,5(mol)$

$n_{C_2H_5OH}>n_{CH_3COOH}\to H$ tính theo $CH_3COOH$

$n_{CH_3COOC_2H_5}=n_{CH_3COOH}=0,2(mol)$

$m_{CH_3COOC_2H_5\ (lt)}=0,2.88=17,8(g)$

$H=\dfrac{m_{tt}}{m_{lt}}=\dfrac{10,56}{17,6}.100=60\%$

`a)`

`CH_3COOH + NaOH → CH_3COONa + H_2O`

`n_{NaOH} = {200}/{1000} .1 = 0,2(mol)`

Theo PT: `n_{CH_3COOH} = n_{NaOH} = 0,2(mol)`

`a = 0,2 . 60 = 12(g)`

`b)`

`CH_3COOH + C_2H_5OH → CH_3COOC_2H_5 + H_2O`

`n_{C_2H_5OH} = {23}/{46} = 0,5(mol)`

`n_{CH_3COOH} < n_{C_2H_5OH} → C_2H_5OH` dư

Theo PT: `n_{text{este}} = n_{CH_3COOH} = 0,2(mol)`

`H% = {10,56}/{0,2. 88} .100 = 60%`