a)

$mNaOH$ = $30$ × $20%$ = $6$ g

⇒ $nNaOH$ = $\frac{6}{40}$ = $0,15$ $mol$

⇒ $nCH_3COOH$ = $0,15$ $mol$

⇒ $CmCH_3COOH$ = $\frac{0,15}{0,5}$ = $0,3$ M

$CH_3COOH$ + $NaOH$ → $CH_3COONa$ +$H_2O$

b)

  $nNa_2CO_3$ = $0,2$ × $0,5$ = $0,1$ mol

Nhận thấy : $nNa_2CO_3$ > $\frac{nCH_3COOH}{2}$

⇒ $CH_3COOH$ hết

          $2CH_3COOH$ + $Na_2CO_3$ → $2CH_3COONa $+ $H_2O$ +$ CO_2 $       

⇒ $nCO_2$ = $\frac{nCH_3COOH}{2}$ = $\frac{0,15}{2}$ = $0,075$ $mol$

⇒ $VCO_2$ = $0,075$ ×$22,4$ = $1.68$ l

Đáp án:

$a\big)C_{M\,CH_3COOH}=0,3M$

$b\big)V_{CO_2(đktc)}=1,68(l)$

Giải thích các bước giải:

$a\big)$

Đổi $500ml=0,5l$

$m_{NaOH}=30.20\%=6(g)$

$\to n_{NaOH}=\frac{6}{40}=0,15(mol)$

$CH_3COOH+NaOH\to CH_3COONa+H_2O$

Theo PT: $n_{CH_3COOH}=n_{NaOH}=0,15(mol)$

$\to C_{M\,CH_3COOH}=\frac{0,15}{0,5}=0,3M$

$b\big)$

Đổi $200ml=0,2l$

$\to n_{Na_2CO_3}=0,2.0,5=0,1(mol)$

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

Vì $n_{Na_2CO_3}>\frac{n_{CH_3COOH}}{2}\to Na_2CO_3$ dư

Theo PT: $n_{CO_2}=\frac{1}{2}n_{CH_3COOH}=0,075(mol)$

$\to V_{CO_2(đktc)}=0,075.22,4=1,68(l)$