`a.PTHH:CH_3COOH+KOH->CH_3COOK+H_2O`

`b.n_{KOH}=0,2.1=0,2(mol)`

`n_{CH_3COOH}=n_{CH_3COOK}=n_{KOH}=0,2(mol)`

`m_{CH_3COOH}=0,2.60=12(g)`

`%CH_3COOH=12/{25,8}.100%~~46,5%`

`%C_2H_5OH=100-46,5=53,5%`

`c.m_{CH_3COOK}=0,2.98=19,6(g)`

Đáp án:

Đã giải

Giải thích các bước giải:

`a)`

`CH_3COOH+KOH \to CH_3COOK+H_2O`

`b)`

`n_{KOH}=n_{CH_3COOH}=0,2.1=0,2(mol)`

`=>m_{CH_3COOH}=0,2.60=12(g)`

`=>%m_{CH_3COOH}=(12.100)/(25,8)=46,51(%)`

`=>%m_{CH_3OH}=100-46,51=53,49(%)`

`c)`

`m_{CH_3COOK}=0,2.98=19,6(g)`