Đáp án:

$c\big)H\approx 71,02\%$ 

Giải thích các bước giải:

$c\big)$

$n_{CH_3COOH(lt)}=\frac{12}{60}=0,2(mol)$

$n_{C_2H_5OH(lt)}=\frac{7,36}{46}=0,16(mol)$

$n_{CH_3COOC_2H_5(tt)}=\frac{10}{88}=\frac{5}{44}(mol)$

$CH_3COOH+C_2H_5OH\buildrel{{t^o,H_2SO_{4(đ)}}}\over\rightleftharpoons CH_3COOC_2H_5+H_2O$

Vì $n_{CH_3COOH}>n_{C_2H_5OH}\to H$ tính theo $C_2H_5OH$

Theo PT: $n_{C_2H_5OH(tt)}=n_{CH_3COOC_2H_5(tt)}=\frac{5}{44}(mol)$

$\to H=\dfrac{\frac{5}{44}}{0,16}.100\%≈71,02\%$

Giải thích các bước giải:

$b)$

$CH_3COOH+C_2H_5OH \buildrel{{H_2SO_{4\ (đ)},\ t^o}}\over\rightleftharpoons CH_3COOC_2H_5+H_2O$

$n_{CH_3COOH}=\dfrac{12}{60}=0,2(mol)$

$n_{CH_3COOC_2H_5}=n_{CH_3COOH}=0,2(mol)$

$m_{CH_3COOC_2H_5}=88×0,2=17,6(g)$

$c)$

$n_{C_2H_5OH}=\dfrac{7,36}{46}=0,16(mol)$

$n_{C_2H_5OH}<n_{CH_3COOH}\to H$ tính theo $C_2H_5OH$

$n_{C_2H_5OH\ (tt)}=n_{CH_3COOC_2H_5\ (tt)}=\dfrac{10}{88}=\dfrac{5}{44}(mol)$

$H=\dfrac{\dfrac{5}{44}}{0,16}×100\%=71\%$