$\rm PTHH: 2CH_3COOH +Na_2CO_3 \xrightarrow{} 2CH_3COONa +CO_2 +H_2O$ 

$\rm n_{CO_2}=\dfrac{1,12}{22,4}=0,05(mol)$

theo pthh 

b)

$\rm n_{CH_3COOH}=2n_{CO_2}=0,1(mol)$ 

`->`$\rm C_{M_{CH_3COOH}}=\dfrac{0,1}{0,2}=0,5M$ 

c)

$\rm m_{CH_3COOH}=0,1.60=6(g)$ 

$\rm m_{(giấm-ăn)}=\dfrac{6.100}{4}=150(g)$ 

Đáp án:

$\rm a)$

Phản ứng xảy ra:

$\rm 2CH_3COOH+Na_2CO_3\to 2CH_3COONa+CO_2+H_2O$

$\rm b)\\n_{CO_2}=\dfrac{11,2}{22,4}=0,5\ (mol)$

Theo phản ứng:

$\rm n_{CH_3COOH}=2n_{CO_2}=1\ (mol)$

$\rm C_{M\ CH_3COOH}=\dfrac{1}{0,2}=5\ (M)$

$\rm c)$

$\rm m_{dd\ CH_3COOH}=\dfrac{1×60}{4\%}=1500\ (g)$