Đáp án:

a,

$K_2CO_3+2CH_3COOH\to 2CH_3COOK+CO_2+H_2O$ 

b,

$n_{CO_2}=\dfrac{0,84}{22,4}=0,0375$ mol 

Theo PTHH:

$n_{K_2CO_3}=n_{CO_2}=0,0375$ mol 

$\Rightarrow m=m_{K_2CO_3}=0,0375.138=5,175$ gam 

Theo PTHH:

$n_{CH_3COOK}=2n_{CO_2}=0,0375.2=0,075$ mol 

$\Rightarrow m_{CH_3COOK}=0,075.98=7,35$ gam 

c,

$m_{\text{dd sau}}=5,175+80-0,0375.44=83,525$ gam 

$\Rightarrow C\%_{CH_3COOK}=\dfrac{7,35.100\%}{83,525}=8,8\%$