`#boiyg`

`a.`

PTHH : `CH_3COOH+NaOH->CH_3COONa+H_2O`

`n_{NaOH}=0,1.2=0,2` `mol`

Theo phương trình : `n_{CH_3COONa}=n_{NaOH}=0,2` `mol`

`=>m_{CH_3COONa}=0,2.82=16,4` `g`

`b.`

Theo phương trình : `n_{CH_3COOH}=n_{NaOH}=0,2` `mol`

`=>C_MCH_3COOH=(0,2)/(0,3)=(2)/(3)` `M`

`c.`

`C_MCH_3COONa=(0,2)/(0,1+0,3)=0,5` `M`

Đáp án:

a,

$CH_3COOH+NaOH\to CH_3COONa+H_2O$

$n_{NaOH}=0,1.2=0,2$ mol 

Theo PTHH:

$n_{CH_3COONa}=n_{CH_3COOH}=n_{NaOH}=0,2$ mol 

$\Rightarrow m_{CH_3COONa}=0,2.82=16,4$ gam 

b,

$V_{dd CH_3COOH}=300 mL=0,3L$

$C_{M_{CH_3COOH}}=\dfrac{0,2}{0,3}=\dfrac{2}{3}M$ 

c,

$V_{\text{dd sau}}=300+100=400mL=0,4L$ 

$C_{M_{CH_3COONa}}=\dfrac{0,2}{0,4}=0,5M$