Đáp án + Giải thích các bước giải:

 a) $m_{CH_3COOH} = \dfrac{C\% . m_{dd}}{100\%} = \dfrac{2,5\% . 300}{100\%} = 7,5(g)$

$n_{CH_3COOH} = \dfrac{m}{M} = \dfrac{7,5}{60} = 0,125(mol)$

PTHH:

$2CH_3COOH + Mg \longrightarrow (CH_3COO)_2Mg + H_2 \uparrow$
Theo PTHH:
$n_{H_2} = \dfrac{1}{2}n_{CH_3COOH} = \dfrac{1}{2} . 0,125 = 0,0625(mol)$
$\Rightarrow V_{H_2} = n . 22,4 = 0,0625 . 22,4 = 1,4l$

b) PTHH:

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

Theo PTHH:
$n_{C_2H_5OH(lt)} = n_{CH_3COOH} = 0,125(mol)$

$\Rightarrow n_{C_2H_5OH(tt)} = 0,125 : 50\% = 0,25(mol)$

$\Rightarrow m_{C_2H_5OH} = n . M = 0,25 . 46 = 11,5g$

$\Rightarrow V_{C_2H_5OH} = \dfrac{m}{D} =\dfrac{11,5}{0,8} = 14,375ml$

$\Rightarrow V_{ddC_2H_5OH} = \dfrac{V_{C_2H_5OH} . 100}{Đ_r} = \dfrac{14,375 . 100}{10} = 143,75ml$

Đáp án + Giải thích các bước giải:

\(\begin{array} {l} a)\\ n_{CH_3COOH}=\dfrac{300.2,5\%}{60}=0,125(mol)\\ Mg+2CH_3COOH\to (CH_3COO)_2Mg+H_2\\ n_{H_2}=\dfrac{1}{2}n_{CH_3COOH}=0,0625(mol)\\ V_{H_2(đktc)}=0,0625.22,4=1,4(l)\\ b)\\ C_2H_5OH+O_2\xrightarrow{men\,giấm}CH_3COOH+H_2O\\ n_{C_2H_5OH(tt)}=n_{CH_3COOH}=0,125(mol)\\ n_{C_2H_5OH(lt)}=\dfrac{0,125}{50\%}=0,25(mol)\\ V_{C_2H_5OH}=\dfrac{0,25.46}{0,8}=14,375(ml)\\ V_{dd\,rượu}=\dfrac{14,375.100}{10}=143,75(ml) \end{array}\)