$ n_{Zn} = \dfrac{m}{M} = \dfrac{9,75}{65} = 0,15 (mol)$

$ Zn + 2CH_3COOH \rightarrow (CH_3COO)_2Zn + H_2↑ $ 

a,

$ n_{H_2} = n_{Zn} = 0,15 (mol) $ 

$ \rightarrow V_{H_2} = n × 22,4 = 0,15 × 22,4 = 3,36 (l) $ 

b,

$ n_{ (CH_3COO)_2Zn} = n_{Zn} = 0,15 (mol) $ 

$ \rightarrow m_{ (CH_3COO)_2Zn} = n × M = 0,15 × 183 = 27,45 (g) $ 

c,

$ n_{CH_3COOH} = 2 × n_{Zn} = 2 × 0,15 = 0,3 (mol) $ 

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

$ n_{C_2H_5OH} = n_{CH_3COOH} = 0,3 (mol) $ 

$ \rightarrow m_{ctC_2H_5OH} = n × M = 0,3 × 46 = 13,8 (g) $

$ \rightarrow m_{ddC_2H_5OH} = \dfrac{13,8}{20} × 100 = 69 (g) $ 

Mà hiệu suất chỉ đạt $80\%$ 

$ \rightarrow m_{ttC_2H_5OH} = \dfrac{ H × m_{lt}}{100\%} = \dfrac{ 80 × 69 }{100} = 55,2 (g) $ 

PTHH:`2CH_3COOH+Zn→(CH_3COO)_2Zn+H_2↑` (1)

`a)`

Ta có: `m_(Zn)=9,75/65=0,15`(mol)

Theo `PTHH`,ta thấy: `n_(H_2)=n_Zn=0,15(mol)`

`⇒V_(H_2)=0,15.22,4=3,36`(l)

`b)`

Theo `PTHH`,ta thấy: `n_(H_2)=n_((CH_3COO)_2Zn)=0,15`(mol)

`⇒m_((CH_3COO)_2Zn)=0,15.(12+3+12+16.2)=8,85`(g)

`c)`

`PTHH: C_2H_5OH  \overset{Men -giấm}\to CH_3COOH+H_2O`(2)

Theo `PTHH`(1),ta thấy: `n_(CH_3COOH)=2n_(H_2)=2.0,15=0,3`(mol)

Theo `PTHH(2)`,ta thấy: `n_(CH_3COOH)=n_(C_2H_5OH)=0,3`(mol)

`⇒m_(ctC_2H_5OH)=n.M=0,3.46=13,8 (g)`

Mà: `%C_2H_5OH=(m_(ctC_2H_5OH))/(m_(ddC_2H_5OH)).100`

`⇒m_(ddC_2H_5OH)=100.(13,8)/20=69`(g)

Lại có hiệu suất đạt `80%` nên,ta có:

$⇒ m_{ttC_2H_5OH}=\dfrac{ H × m_{ltC_2H_5OH}}{100\%}= \dfrac{ 80 × 69 }{100}=55,2$ (g)

Xin hay nhất =_=