\(\left(-C_6H_{10}O_5-\right)_n+nH_2O\rightarrow nC_6H_{12}O_6\\ C_6H_{12}O_6\underrightarrow{\text{men rượu}}2C_2H_5OH+2CO_2\\ C_2H_5OH+O_2\underrightarrow{\text{men giấm}}CH_3COOH+H_2O\\ CH_3COOH+C_2H_5OH\xrightarrow[t^o]{H_2SO_{4\left(đ\right)}}CH_3COOC_2H_5+H_2O\)

1)nH2O+(C6H10O5)n→nC6H12O6
(2)C6H12O6lm→2C2H5OH+2CO2
(3)C2H5OH+O2to,xt→CH3COOH+H2O
(4)C2H5OH+HCOOH→H2O+HCOOC2H5
 

Chuỗi 1:

\(\left(1\right)CaCO_3\rightarrow\left(t^o\right)CaO+CO_2\\ \left(2\right)CaO+3C\rightarrow\left(2000^oC,lò.điện\right)CaC_2+CO\uparrow\\ \left(3\right)CaC_2+2H_2O\rightarrow C_2H_2+Ca\left(OH\right)_2\\ \left(4\right)C_2H_2+H_2\rightarrow\left(Ni,t^o\right)C_2H_4\\ \left(5\right)C_2H_4+H_2O\rightarrow\left(t^o,H^+\right)C_2H_5OH\\ \left(6\right)C_2H_5OH+2NaOH+CH_3COOH\rightarrow CH_3COONa+C_2H_5ONa+2H_2O\)

a) CaC2 + 2H2O --> Ca(OH)2 + C2H2

C2H2 + H2 -Ni-> C2H4

C2H4 + H2O --> C2H5OH

C2H5OH + O2 -mg-> CH3COOH + H2O

CH3COOH + C2H5OH <-H2SO4đ,to-> CH3COOC2H5 + H2O

b) (-C6H10O5-)n + nH2O -axit-> nC6H12O6

C6H12O6 -mr-> 2C2H5OH + 2CO2

C2H5OH + O2 -mg-> CH3COOH + H2O

CH3COOH + C2H5OH <-H2SO4đ,to-> CH3COOC2H5 + H2O

CH3COOC2H5 + NaOH -to-> CH3COONa + C2H5OH

a/ CaC2 + 2H2O => Ca(OH)2 + C2H2

C2H2 + H2 => (Pd,t^o) C2H4

C2H4 + H2O => (140^oC,H2SO4đ) C2H5OH

C2H5OH + O2 => (men giấm) CH3COOH + H2O

CH3COOH + C2H5OH => (H2SO4đ,t^o,pứ hai chiều) CH3COOC2H5 + H2O: pứ este hóa

\(a,-\left(-C_6H_{10}O_5-\right)-_n+nH_2O\rightarrow nC_6H_{12}O_6\\ C_6H_{12}O_6\underrightarrow{\text{men rượu}}2C_2H_5OH+2CO_2\uparrow\\ C_2H_5OH+O_2\underrightarrow{\text{men giấm}}CH_3COOH+H_2O\\ CH_3COOH+C_2H_5OH\xrightarrow[H_2SO_{4\left(đ\right)}]{t^o}CH_3COOC_2H_5+H_2O\)

\(b,C_{12}H_{22}O_{11}+H_2O\xrightarrow[t^o]{H^+}C_6H_{12}O_6+C_6H_{12}O_6\)

Mấy pthh sau giống ở trên á bạn

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

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

\(C_2H_4+H_2O\underrightarrow{t^o,xt}C_2H_5OH\)

\(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)

\(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (xt: H₂SO₄ đặc, t^o)

\(CH_3COOC_2H_5+NaOH\underrightarrow{t^o}CH_3COONa+C_2H_5OH\)

Câu 2:

\(1,C_{12}H_{22}O_{11}+H_2O\underrightarrow{t^o}C_6H_{12}O_6+C_6H_{12}O_6\\ C_6H_{12}O_6\xrightarrow[\text{men rượu}]{H^+,t^o}2C_2H_5OH+2CO_2\uparrow\\ C_2H_5OH+O_2\underrightarrow{\text{men giấm}}CH_3COOH+H_2O\\ 2CH_3COOH+ZnO\rightarrow\left(CH_3COO\right)_2Zn+H_2O\)

2, tất cả đều giống 1 trừ PTHH cuối:

\(2CH_3COOH+MgO\rightarrow\left(CH_3COO\right)_2Mg+H_2O\)

Câu 3:

Cho mẩu Na tác dụng với từng chất:

- Na tan dần, có sủi bọt khí không màu, mùi: C₂H₅OH, CH₃COOH (*)

\(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\uparrow\\ CH_3COOH+Na\rightarrow CH_3COONa+\dfrac{1}{2}H_2\uparrow\)

- Không hiện tượng: C₆H₁₂O₆

Cho QT vào các chất (*):

- Hoá hồng: CH₃COOH

- Không hiện tượng: C₂H₅OH

Câu 4:

a) Bảo toàn C: \(n_C=n_{CO_2}=\dfrac{44}{44}=1\left(mol\right)\)

Bảo toàn H: \(n_H=2n_{H_2O}=2.\dfrac{27}{18}=3\left(mol\right)\)

Xét m_H + m_C = 3 + 12 = 15 (g)

=> A chỉ chứa C và H

b) M_A = 2.15 = 30 (g/mol)

CTPT: C_xH_y

=> x : y = 1 : 3

=> (CH₃)_n = 30 

=> n = 2

CTCT: CH₃-CH₃

Câu 5:

a) Bảo toàn C: \(n_C=n_{CO_2}=\dfrac{44}{44}=1\left(mol\right)\)

Bảo toàn H: \(n_H=2n_{H_2O}=2.\dfrac{18}{18}=2\left(mol\right)\)

Xét m_H + m_C = 2 + 12 = 14 (g)

=> A chỉ chứa C và H

b) M_A = 2.14 = 28 (g/mol)

CTPT: C_xH_y

=> x : y = 1 : 2

=> (CH₂)_n = 28

=> n = 2

CTPT: C₂H₄

CTCT: CH₂=CH₂

Câu 6:

\(a,n_{CH_3COOH}=\dfrac{30}{60}=0,5\left(mol\right)\\ n_{C_2H_5OH}=\dfrac{46}{46}=1\left(mol\right)\)

PTHH: CH₃COOH + C₂H₅OH -H₂SO_{4 (đặc)}, t^o-> CH₃COOC₂H₅ + H₂O

LTL: 0,5 < 1 => C₂H₅OH dư

Theo pthh: n_CH3COOC2H5 = n_CH3COOH = 0,5 (mol)

=> m_CH3COOC2H5 = 0,5.60%.88 = 26,4 (g)

Câu 7:

\(a,n_{CH_3COOH}=\dfrac{60}{60}=1\left(mol\right)\\ n_{C_2H_5OH}=\dfrac{23}{46}=0,5\left(mol\right)\)

PTHH: 

CH₃COOH + C₂H₅OH -H₂SO_{4 (đặc)}, t^o-> CH₃COOC₂H₅ + H₂O

LTL: 1 > 0,5 => CH₃COOH dư

Theo pthh: n_CH3COOC2H5 = n_C2H5OH = 0,5 (mol)

=> m_CH3COOC2H5 = 0,5.70%.88 = 30,8 (g)

thank bạn nhoa

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

$CH_3COOC_2H_5 + KOH \to CH_3COOK + C_2H_5OH$

\(C_2H_4+H_2O\rightarrow C_2H_5OH\)

\(C_2H_5OH+\left(CH_3CO\right)_2O\rightarrow CH_3COOH+CH_3COOC_2H_5\)

\(CH_3COOH+C_2H_5ONa\rightarrow NaOH+CH_3COOC_2H_5\)

\(KOH+CH_3COOC_2H_5\rightarrow C_2H_5OH+CH_3COOK\)

\(2C_4H_{10}+5O_2\underrightarrow{t^o}4CH_3COOH+2H_2O\)

\(CH_3COOH+C_2H_5OH\xrightarrow[H_2SO_4đặc]{t^o}CH_3COOC_2H_5+H_2O\)

CH₃COOC₂H₅ + NaOH \(\rightarrow\) CH₃COONa + C₂H₅OH

\(C_2H_5OH+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)

CaC₂ + 2H₂O\(\rightarrow\) C₂H₂ + Ca(OH)₂

\(C_2H_2+H_2\underrightarrow{t^o}C_2H_4\)

\(C_2H_4+H_2O\underrightarrow{axit}C_2H_5OH\)

\(C_2H_5OH+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)

CO₂ + 2NaOH \(\rightarrow\) Na₂CO₃ + H₂O

2CH₃COOH + Na₂CO₃ \(\rightarrow\) 2CH₃COONa + CO₂ + H₂O