Giải thích các bước giải:

$a)\ Mg+2CH_3COOH\to (CH_3COO)_2Mg+H_2$

$b)$

$n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)$

$n_{CH_3COOH}=2n_{H_2}=0,4(mol)$

$m_{CH_3COOH}=0,4.60=24(g)$

$c)$

$CH_3COOH+C_2H_5OH\to CH_3COOC_2H_5+H_2O$

$n_{CH_3COOC_2H_5}=n_{CH_3COOH}=0,4(mol)$

$m_{CH_3COOC_2H_5}=0,4.88.90\%=31,68(g)$

Đáp án + Giải thích các bước giải:

 `a)` 

`Mg + 2CH_3COOH → (CH_3COO)_2Mg + H_2` 

`b)` 

`n_{H_2} = {4,48}/{22,4} = 0,2(mol)` 

`n_{CH_3COOH} = 2n_{H_2} = 0,4(mol)` 

`m_{CH_3COOH} = 0,4 . 60 = 24(g)` 

`c)` 

`CH_3COOH + C_2H_5OH`$\buildrel{{t°,H_2SO_4}}\over\rightleftharpoons$`CH_3COOC_2H_5 + H_2O` 

`n_{CH_3COOC_2H_5} = n_{CH_3COOH} = 0,4(mol)` 

`m_{CH_3COOC_2H_5} = 0,4 . 88 . 90% = 31,68(g)` 

`#aura`