Đáp án:

a,

$2CH_3COOH+Mg \to (CH_3COO)_2Mg+H_2$ 

b,

$n_{CH_3COOH}=0,2.1=0,2$ mol 

$\Rightarrow n_{H_2}=n_{(CH_3COO)_2Mg}=\dfrac{1}{2}n_{CH_3COOH}=0,1$ mol 

$\Rightarrow V_{H_2}=0,1.22,4=2,24L$ 

c,

$m_{(CH_3COO)_2Mg}=0,1.142=14,2$ gam 

d,

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

$\Rightarrow n_{C_2H_5OH}=n_{CH_3COOH}=0,2$ mol 

$\Rightarrow m_{C_2H_5OH}=0,2.46= 9,2$ gam

Đáp án:

Giải thích các bước giải:

 `a)`

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

`b)`

`n_(CH_3COOH) = 0.2 * 1 = 0.2 (mol)`

Theo phuong trinh : `1/2n_(CH_3COOH) = n_(H_2)= 0.1 (mol)`

`-> V_(H_2) = 0.1 * 24.79 = 2.479 (l)`

`c)`

Theo phuong trinh : `n_(H_2) = n_((CH_3COO)_2Mg) = 0.1 (mol)`

`-> m_((CH_3COO)_2Mg) = 0.1 * 142 = 14.2 (gam)`

`d)`

`2C_2H_5OH + 2O_2 overset(enzym)(->) CH_3COOH + 2H_2O`

Theo phuogn trinh : `n_(CH_3COOH) = n_(C_2H_5OH) = 0.2 (mol)`

`-> m_(C_2H_5OH) = 0.2 *  46 = 9.2 (gam)`