Đáp án + Giải thích các bước giải:

Bài 6:

$n_{C_2H_5COOH}=\frac{7,4}{74}=0,1(mol)$

$n_{CH_3OH}=\frac{4,8}{32}=0,15(mol)$

$n_{CH_3COOC_2H_5(tt)}=\frac{5,72}{88}=0,065(mol)$

$C_2H_5COOH+CH_3OH\buildrel{{t^o,H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5+H_2O$

Vì $0,1<0,15\to H$ tính theo $C_2H_5COOH$

Theo PT: $n_{C_2H_5COOH(tt)}=n_{CH_3COOC_2H_5(tt)}=0,065(mol)$

$\to H=\frac{0,065}{0,1}.100\%=65\%$

Bài 7:

Đổi $200ml=0,2l$

$\to n_{CH_3COOH}=0,2.1,5=0,3(mol)$

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

Theo PT: $n_{H_2}=\frac{1}{2}.0,3=0,15(mol)$

$\to V_{H_2(đktc)}=0,15.22,4=3,36(l)$

Bài 8:

$n_{(C_{15}H_{31}COO)_3C_3H_5}=\frac{10}{806}=\frac{5}{403}(kmol)$

$(C_{15}H_{31}COO)_3C_3H_5+3NaOH\xrightarrow{t^o}3C_{15}H_{31}COONa+C_3H_5(OH)_3$

Theo PT: $n_{C_{15}H_{31}COONa}=3.\frac{5}{403}=\frac{15}{403}(kmol)$

$\to m_{C_{15}H_{31}COONa}=\frac{15}{403}.278=\frac{4170}{403}(kg)$

$\to m_{xà\,phòng}=\dfrac{\frac{4170}{403}}{72\%}\approx 14,37(kg)$

Đáp án + Giải thích các bước giải:

 Bài 6:

$n_{C_2H_5COOH} = \dfrac{m}{M} = \dfrac{7,4}{74} = 0,1\text{ mol}$

$n_{CH_3OH} = \dfrac{m}{M} = \dfrac{4,8}{32} = 0,15\text{ mol}$
PTHH:
$C_2H_5COOH + CH_3OH \longrightarrow C_2H_5COOCH_3 + H_2O$

Ta có:

$n_{CH_3OH} > n_{C_2H_5COOH}(0,15 > 0,1)$
$\Rightarrow CH_3OH$ dư, tính theo $C_2H_5COOH$
Theo PTHH:
$n_{C_2H_5COOCH_3(lt)} = n_{C_2H_5COOH} = 0,1\text{ mol}$

$\Rightarrow m_{C_2H_5COOCH_3(lt)} = n . M = 0,1 . 88 = 8,8g$

$H\% = \dfrac{m_{tt}}{m_{lt}} . 100\% = \dfrac{5,72}{8,8} . 100\% = 65\%$

Bài 7:

$200ml = 0,2l$

$n_{CH_3COOH} = C_M . V = 1,5 . 0,2 = 0,3\text{ mol}$

PTHH:
$2CH_3COOH + Mg \longrightarrow (CH_3COO)_2Mg + H_2 \uparrow$
Theo PTHH:
$n_{H_2} = \dfrac{1}{2}n_{CH_3COOH} = \dfrac{1}{2} . 0,3 = 0,15\text{ mol}$

$\Rightarrow V_{H_2} = n . 22,4 = 0,15 . 22,4 = 3,36l$

Bài 8:

$10kg = 10000g$

$n_{(C_{15}H_{31}COO)_3C_3H_5} = \dfrac{m}{M} = \dfrac{10000}{806} \approx 12,41\text{ mol}$

PTHH:
$(C_{15}H_{31}COO)_3C_3H_5 + 3NaOH \xrightarrow{H_2O, t^o} 3C_{15}H_{31}COONa + C_3H_5(OH)_3$

Theo PTHH:
$n_{C_{15}H_{31}COONa} = 3n_{(C_{15}H_{31}COO)_3C_3H_5} = 3 . 12,41 = 37,23 \text{ mol}$

$\Rightarrow m_{C_{15}H_{31}COONa} = n . M = 37,23 . 278 = 10349,94g$

$m_{xp} = 10349,94 : 72\% \approx 14374,92g$