`1.`

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

`m_{CH_3COOH}=180.20%=36(g)`

`m_{NaOH}=200.20%=40(g)`

`n_{NaOH}={40}/{40}=1 (mol)`

`n_{CH_3COOH}={36}/{60}=0,6(mol)`

Vì `n_{NaOH}>n_{CH_3COOH}=> ` Tính theo `n_{CH_3COOOH}`

Theo pt : `n_{CH_3COOOH}=n_{CH_3COOONa}=0,6 (mol)`

`=>m_{CH_3COONa=0,6.82=49,2(g) `

`2.`

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

`m_{CH_3COOH}=60.20%=12 (g)`

`=>n _{CH_3COOH}=12/ {60}=0,2 (mol )`

`m_{NaOH}=20.30% =6 (g)`

`=> n_{NaOH}=6/{40}=0,15(mol)`

Vì `n_{NaOH}<n_{CH_3COOH}=> ` Tính theo `NaOH`

`=>n_{NaOH}=n_{CH_3COONa}=0,15(mol )`

`m _{CH_3COONa} ` theo lí thuyết ` =0,15.82=12,3(g)`

`H%={9,84}/{12,3} .100%=80%` 

`1)`

`n_{CH_3COOH}={180.20\%}/{60}=0,6(mol)`

`n_{NaOH}={200.20\%}/{40}=1(mol)`

`CH_3COOH+NaOH->CH_3COONa+H_2O`

`0,6<1->NaOH` dư

`->n_{CH_3COONa}=n_{CH_3COOH}=0,6(mol)`

`->m_{CH_3COONa}=0,6.82=49,2(g)`

`2)`

`n_{CH_3COOH\ l t}={60.20\%}/{60}=0,2(mol)`

`n_{NaOH\ l t}={20.30\%}/{40}=0,15(mol)`

`CH_3COOH+NaOH->CH_3COONa+H_2O`

`0,2>0,15->H` theo NaOH

`->n_{CH_3COONa\ l t}=n_{NaOH\ l t}=0,15(mol)`

`->H={9,84}/{0,15.82}.100\%=80\%`