1)

$n_{H_2O} = \dfrac{3,6}{18}= 0,2(mol)$
Bảo toàn nguyên tố C,H : 

$C_nH_{2n}O_2 \to nCO_2 + nH_2O$

$\Rightarrow n_{CO_2} = n_{H_2O} = 0,2(mol)$

$\Rightarrow n_C = 0,2(mol) ; n_H = 0,4(mol)$
$\Rightarrow n_O = \dfrac{4,4 - 0,2.12-0,4.1}{16} = 0,1(mol)$

$n_X = \dfrac{1}{2}n_O = 0,05(mol)$
$\Rightarrow M_X = 12n + 2n + 32 = \dfrac{4,4}{0,06} = 88$
$\Rightarrow n = 4$

Vậy CTPT của X là $C_4H_8O_2$

2)

$n_{Ba(OH)_2} = 0,15(mol) < n_{CO_2} = 0,2(mol)$

\(Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3+H_2O\)

0,15                 0,15            0,15                           (mol)

\(BaCO_3+CO_2+H_2O\rightarrow Ba\left(HCO_3\right)_2\)

0,05               0,05                                              (mol)

$\Rightarrow n_{BaCO_3} = 0,15 - 0,05 = 0,1(mol)$
$\Rightarrow m = 0,1.197 = 19,7(gam)$

3)

CTTQ của muối : $RCOONa$
$n_{muối} = n_X = 0,05(mol) \Rightarrow M_{muối} = R + 67 = \dfrac{4,1}{0,05} = 82$
$\Rightarrow R = 15(-CH_3)$

Vậy CTCT của X là $CH_3COOC_2H_5$

$C + O_2 \xrightarrow{t^o} CO_2$
$CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O$
$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
$CO_2 + C \to 2CO$

$C_2H_4 + H_2O \xrightarrow{t^o,H^+} C_2H_5OH$
$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$
$CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4,t^o}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$
$CH_3COOC_2H_5 + NaOH \to CH_3COONa + C_2H_5OH$

1)

$C_6H_{12}O_6 \xrightarrow{t^o,men\ rượu} 2CO_2 + 2C_2H_5OH$

$C_2H_5OH \xrightarrow{t^o,H_2SO_4} C_2H_4 + H_2O$
$C_2H_4 + H_2O \xrightarrow{t^o,H^+} C_2H_5OH$
$C_2H_5OH + HBr \to C_2H_5Br + H_2O$

$C_2H_5OH + CuO \xrightarrow{t^o} Cu  + CH_3-CHO + H_2O$
$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH  + H_2O$
$CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$

2)

$(C_6H_{10}O_5)_n + nH_2O \xrightarrow{t^o,H^+} nC_6H_{12}O_6$

$C_6H_{12}O_6 \xrightarrow{men\ rượu} 2CO_2 + 2C_2H_5OH$

Trích mẫu thử

Cho giấy quỳ tím vào các mẫu thử

- mẫu thử nào làm quỳ tím hóa đỏ là $CH_3COOH$

Đốt cháy hai mẫu thử còn lại rồi dẫn sản phẩm vào dung dịch nước vôi trong :

- mẫu thử nào tạo kết tủa trắng là $C_2H_5OH$
$C_2H_5OH + 3O_2 \xrightarrow{t^o} 2CO_2 + 3H_2O$
$CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O$
- mẫu thử không hiện tượng gì là nước cất

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

b) $n_{C_2H_5OH} = \dfrac{4,6}{46} = 0,1(mol)$

$n_{O_2} = 3n_{C_2H_5OH} = 0,3(mol)$
$V_{O_2} = 0,3.22,4 = 6,72(lít)$

c)

Theo PTHH : 

$n_{CO_2} = 2n_{C_2H_5OH} = 0,2(mol) \Rightarrow V_{CO_2} = 0,2.22,4 = 4,48(lít)$

$n_{H_2O} = 3n_{C_2H_5OH} = 0,3(mol) \Rightarrow m_{H_2O} = 0,3.18 = 5,4(gam)$

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

b) $n_{C_2H_5OH} = \dfrac{9,2}{46} = 0,2(mol)$

Theo PTHH : 

$n_{CO_2} = 2n_{C_2H_5OH} = 0,4(mol) \Rightarrow V_{CO_2} = 0,4.22,4 = 8,96(lít)$

$n_{H_2O} = 3n_{C_2H_5OH} = 0,6(mol) \Rightarrow m_{H_2O} = 0,6.18 = 10,8(gam)$

c) $CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O$
$n_{CaCO_3} = n_{CO_2} = 0,4(mol)$

$m_{CaCO_3} = 0,4.100 = 40(gam)$

a) $n_{(CH_3COO)_2Mg} = \dfrac{14,2}{142} = 0,1(mol)$

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

Theo PTHH : $n_{CH_3COOH} = 2n_{(CH_3COO)_2Mg} = 0,2(mol)$
$C_{M_{CH_3COOH}} = \dfrac{0,2}{0,4} = 0,5M$

b) $n_{H_2} = n_{(CH_3COO)_2Mg} = 0,1(mol)$
$V_{H_2} = 0,1.22,4 = 2,24(lít)$

c)

$CH_3COOH + NaOH \to CH_3COONa + H_2O$
$n_{NaOH} = n_{CH_3COOH} = 0,1(mol)$
$V_{dd\ NaOH} = \dfrac{0,1}{0,75} = 0,13(lít)$

a) $C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$
b) $C_2H_5OH + 3O_2 \xrightarrow{t^o} 2CO_2 + 3H_2O$
c) $C_2H_5OH + CH_3COOH \to CH_3COOC_2H_5 + H_2O$
d) $2C_2H_5OH + 2K \to 2C_2H_5OK + H_2O$
e) $CH_3COOH + KOH\to CH_3COOK + H_2O$
f) $CH_3COOC_2H_5 + H_2O \to CH_3COOH + C_2H_5OH$

g) $2CH_3COOH + Mg \to (CH_3COO)_2Mg + H_2O$

h) $2CH_3COOH + MgCO_3 \to (CH_3COO)_2Mg + CO_2 + H_2O$

$n_{CH_3COOH} = \dfrac{12}{60} = 0,2(mol)$
$n_{C_2H_5OH} = \dfrac{1,38}{46} = 0,03(mol)$
\(CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O\)

Ta thấy :

$n_{CH_3COOH} : 1 > n_{C_2H_5OH}:1$ nên hiệu suất tính theo số mol $C_2H_5OH$
$n_{CH_3COOC_2H_5} = n_{C_2H_5OH\ pư} = 0,03.75\% = 0,0375(mol)$

$m_{CH_3COOC_2H_5} = 0,0375.88 =3,3(gam)$

a)

1,5 lít = 1500 ml

$V_{rượu} = 1500.\dfrac{20}{100} = 300(ml)$

b)
$m_{rượu} = D.V = 0,8.300 = 240(gam)$

$n_{C_2H_5OH} = \dfrac{240}{46} = \dfrac{120}{23}(mol)$
$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$

$n_{CH_3COOH} = n_{C_2H_5OH} = \dfrac{120}{23}(mol)$
$m_{CH_3COOH} = \dfrac{120}{23}.60 = 313,04(gam)$