$Zn +2CH_3COOH \xrightarrow{} (CH_3COO)_2Zn +H_2$ 

$n_{Zn}=\dfrac{13}{65}=0,2(mol)$ 

theo pt 

$n_{H_2}=n_{Zn}=0,2(mol)$ 

`->`$V_{H_2}=0,2.22,4=4,48(lít)$ 

`b)` 

$CH_3COOH +C_2H_5OH \xrightarrow{to, H^+} CH_3COOC_2H_5 +H_2O$ 

theo pt 

$n_{CH_3COOC_2H_5}=n_{CH_3COOH}=2n_{Zn}=0,4(mol)$ 

`->`$m_{CH_3COOC_2H_5}=0,4.88=35,2(g)$ 

mà ta có `H=90%` 

suy ra `m_{CH_3COOC_2H_5(thực-tế )}=35,2.90%=31,68(g)`

$\rm n_{Zn} = \dfrac{13}{65} = 0,2$ mol

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

a)$\rm n_{H_2} = n_{Zn} = 0,2$ mol

$\rm \Rightarrow V_{H_2} = 0,2 . 22,4 = 4,48l$

b)

$\rm n_{CH_3COOH} = 2n_{Zn} = 2 . 0,2 = 0,4$ mol

$\rm CH_3COOH + C_2H_5OH \buildrel{{t^o, H_2SO_4 đặc}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$

$\rm n_{CH_3COOC_2H_5(lí thuyết)} = n_{CH_3COOH} = 0,4$ mol

$\rm \Rightarrow n_{CH_3COOC_2H_5(thực tế)} = \dfrac{0,4 . 90}{100} = 0,36$ mol

$\rm \Rightarrow m_{CH_3COOC_2H_5(thực tế)} = 0,36 . 88 = 31,68g$