Đáp án + Giải thích các bước giải:

Bài 8:

1/

a) $m_{CH_3COOH} = \dfrac{200 .4,8\%}{100\%} = 9,6g$

$n_{CH_3COOH} = \dfrac{9,6}{60} = 0,16\text{ mol}$

Đổi: $200ml = 0,2l$

$n_{CH_3COOH(I)} = n_{CH_3COOH(II)} = \dfrac{0,16}{2} = 0,08\text{ mol}$

PTHH(I):
$CH_3COOH + NaOH \longrightarrow CH_3COONa + H_2O$

Theo PTHH:
$n_{NaOH} = n_{CH_3COOH} = 0,08\text{ mol}$

$\Rightarrow C_{MNaOH} = \dfrac{0,08}{0,2} = 0,4M$

b) PTHH(II):
$2CH_3COOH + Mg \longrightarrow (CH_3COO)_2Mg + H_2$

Theo PTHH:
$n_{H_2} = \dfrac{1}{2}n_{CH_3COOH} = \dfrac{1}{2} . 0,08 = 0,04\text{ mol}$

$\Rightarrow V_{H_2} = 0,04 . 22,4 = 0,896l$

2/ PTHH:
$C_2H_5OH + O_2 \xrightarrow{\text{ men giấm}} CH_3COOH + H_2O$

$n_{CH_3COOH(lt)} = \dfrac{0,16 . 100\%}{80\%} = 0,2\text{ mol}$

Theo PTHH:
$n_{C_2H_5OH} = n_{CH_3COOH(lt)} = 0,2\text{ mol}$

$\Rightarrow m_{C_2H_5OH} = 0,2 . 46 = 9,2g$

Đáp án + Giải thích các bước giải:

$1\big)$

$a\big)$

$n_{CH_3COOH}=\frac{200.4,8\%}{60}=0,16(mol)$

Phần 1:

$CH_3COOH+NaOH\to CH_3COONa+H_2O$

Theo PT: $n_{NaOH}=n_{CH_3COOH(1P)}=\frac{0,16}{2}=0,08(mol)$

$\to C_{M\,NaOH}=\frac{0,08}{0,2}=0,4M$

$b\big)$

Phần 2:

$Mg+2CH_3COOH\to (CH_3COO)_2Mg+H_2$

Theo PT: $n_{H_2}=\frac{1}{2}n_{CH_3COOH(1P)}=0,04(mol)$

$\to V_{H_2}=0,04.22,4=0,896(l)$

$2\big)$

$n_{CH_3COOH(lt)}=\frac{0,16}{80\%}=0,2(mol)$

$C_2H_5OH+O_2\xrightarrow{men}CH_3COOH+H_2O$

Theo PT: $n_{C_2H_5OH(lt)}=n_{CH_3COOH(lt)}=0,2(mol)$

$\to m_{C_2H_5OH(lt)}=0,2.46=9,2(g)$