Đáp án + Giải thích các bước giải:

 a) PTHH:
$C_2H_5OH + 3O_2 \xrightarrow{t^o} 2CO_2 + 3H_2O$

$CO_2 + Ca(OH)_2 \longrightarrow CaCO_3 \downarrow + H_2O$

$n_{CaCO_3} = \dfrac{m}{M} = \dfrac{100}{100} = 1\text{ mol}$

Theo PTHH:
$n_{CaCO_3} = n_{CO_2} = \dfrac{2}{3}n_{O_2} = 1\text{ mol}$
$\Rightarrow n_{O_2} = \dfrac{3}{2}n_{CaCO_3} = \dfrac{3}{2} . 1 = 1,5\text{ mol}$

$\Rightarrow V_{O_2} = n . 22,4 = 1,5 . 22,4 = 33,6l$

$V_{kk} = V_{O_2} : 20\% = 33,6 : 20\% = 168l$
b) Theo PTHH:
$n_{C_2H_5OH} = \dfrac{1}{3}n_{O_2} = \dfrac{1}{3} . 1,5 = 0,5\text{ mol}$
$\Rightarrow m_{C_2H_5OH} = n . M = 0,5 . 46 = 23g$

$\Rightarrow V_{C_2H_5OH} = \dfrac{m}{D} = \dfrac{23}{0,8} = 28,75ml$

$\Rightarrow Đ_r = \dfrac{V_{C_2H_5OH} . 100}{V_{dd}} = \dfrac{28,75 . 100}{45} \approx 64^o$

c) $n_{CH_3COOH} = \dfrac{m}{M} = \dfrac{30}{60} = 0,5\text{ mol}$

PTHH:
$CH_3COOH + C_2H_5OH \overset{H_2SO_4\text{ đặc}, t^o}{\rightleftharpoons} CH_3COOC_2H_5 + H_2O$

Ta có:
$n_{CH_3COOH} = n_{C_2H_5OH}(= 0,5)$

$\Rightarrow$ Phản ứng xảy ra vừa đủ

Theo PTHH:
$n_{CH_3COOC_2H_5(lt)} = n_{CH_3COOH} = 0,5\text{ mol}$
$\Rightarrow m_{CH_3COOC_2H_5(lt)} = n . M = 0,5 . 88 = 44g$

$\Rightarrow m_{CH_3COOC_2H_5(tt)} = 44 . 60\% = 26,4g$