Đáp án+Giải thích các bước giải:
a.b.

` n_{CH_3COOH} = {400.4,8%}/60 = 0,32(mol) `

` Mg + 2CH_3COOH --> (CH_3COO)_2Mg + H_2`

                  0,32             ----->                  0,16     ( mol )

` V_{H_2} =0,16.22,4=3,584(lít)`  

c.

` n_{CH_3COOH} = 300/60=5(mol)`

` C_2H_5OH + O_2 --> CH_3COOH + H_2O`

       0,32           <----- 0,32                         ( mol )

`m_{C_2H_5OH}={0,32.46}/{80%}=18,4(g)`

Đáp án + Giải thích các bước giải:

 a) PTHH:
$2CH_3COOH + Mg \longrightarrow (CH_3COO)_2Mg + H_2 \uparrow$

b) $m_{CH_3COOH} = \dfrac{C\% . m_{dd}}{100\%} = \dfrac{4,8\% . 400}{100\%} = 19,2g$

$\Rightarrow n_{CH_3COOH} = \dfrac{m}{M}= \dfrac{19,2}{60} = 0,32\text{ mol}$

Theo PTHH:
$n_{H_2} = \dfrac{1}{2}n_{CH_3COOH} = \dfrac{1}{2} . 0,32 = 0,16\text{ mol}$
$\Rightarrow V_{H_2} = n . 22,4 = 0,16 . 22,4 = 3,584l$

c) $m_{CH_3COOH} = \dfrac{C\% . m_{dd}}{100\%} = \dfrac{4,8\% . 300}{100\%} = 14,4g$

$\Rightarrow n_{CH_3COOH} = \dfrac{m}{M} = \dfrac{14,4}{60} = 0,24\text{ mol}$

PTHH:
$C_2H_5OH + O_2 \xrightarrow{\text{men giấm}} CH_3COOH + H_2O$

Theo PTHH:
$n_{C_2H_5OH(tt)} = n_{CH_3COOH} = 0,24\text{ mol}$
$\Rightarrow n_{C_2H_5OH(lt)} = 0,24 : 80\% = 0,3\text{ mol}$

$\Rightarrow m_{C_2H_5OH} =n .M = 0,3 . 46 = 13,8g$