Đáp án:

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

Phương trình phản ứng:

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

Theo phương trình:

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

$\rm V_{H_2}=0,2×22,4=4,48\ (L)$

$\rm b)$

Theo phương trình:

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

$\rm m_{dd\ CH_3COOH}=\dfrac{0,4×60}{12\%}=200\ (g)$

$\rm n_{Zn}=\dfrac{13}{65}=0,2(mol)$ 

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

theo pthh 

$n_{H_2}=n_{Zn}=0,2(mol)$ 

$n_{CH_3COOH}=2n_{Zn}=0,4(mol)$ 

`a)` 

$V_{H_2}=0,2.22,4=4,48(l)$ 

`b)$ 

$m_{CH_3COOH}=0,4.60=24(g)$ 

`->`$m_{dd \ CH_3COOH}=\dfrac{24}{12\%}=200(g)$