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

\(C_2H_5OH+O_2\underrightarrow{men.giấm}CH_3COOH+H_2O\)

\(CH_3COOH+C_2H_5OH\underrightarrow{H_2SO_{4\left(đ\right)},t^o}CH_3COOC_2H_5+H_2O\)

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

 C₂H₄ → C₂H₅OH → CH₃COOH → CH₃COOC₂H₅ → C₂H₅OH

(1)   C₂H₄ + H₂O \(\underrightarrow{axit}\) C₂H₅OH

(2)  C₂H₅OH  + O₂  \(\xrightarrow[25^0-30^0C]{mengiam}\) CH₃COOH + H₂O

(3)   CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O

(4) CH₃COOC₂H₅ + NaOH \(\underrightarrow{t^0}\) CH₃COONa + C₂H₅OH

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

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

\(\left(-C_6H_{10}O_5-\right)n+nH_2O\underrightarrow{t^o}nC_6H_{12}O_6\)

\(C_6H_{12}O_6\underrightarrow{men.rượu}2C_2H_5OH+2CO_2\uparrow\)

\(C_2H_5OH+O_2\underrightarrow{men.giấm}CH_3COOH+H_2O\)

\(CH_3COOH+C_2H_5OH\underrightarrow{t^o}CH_3COOC_2H_5+H_2O\)

1)

a)

$C_2H_4 + H_2O \xrightarrow{t^o,xt} C_2H_5OH$
$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$

$CH_3COOH + NaOH \to CH_3COONa + H_2O$

b)

$CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$

2)

a) $n_{CO_2} = \dfrac{16,8}{22,4} = 0,75(mol)$
$C_6H_{12}O_6 \xrightarrow{men\ rượu} 2CO_2 + 2C_2H_5OH$
$n_{glucozo} = \dfrac{1}{2}n_{CO_2} = 0,375(mol)$
$m_{glucozo} = 0,375.180 = 67,5(gam)$

b) $n_{C_2H_5OH} = n_{CO_2} = 0,75(mol)$
$m_{C_2H_5OH} = 0,75.46 = 34,5(gam)$

$V_{C_2H_5OH} = \dfrac{34,5}{0,8}=  43,125(ml)$

Câu 1:

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

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

\(CH_3COOH+Na\rightarrow CH_3COOH+\dfrac{1}{2}H_2\)

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

Câu 2:

a, \(n_{CO_2}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\)

\(C_6H_{12}O_6\underrightarrow{t^o,xt}2C_2H_5OH+2CO_2\)

Theo PT: \(n_{C_6H_{12}O_6}=\dfrac{1}{2}n_{CO_2}=0,375\left(mol\right)\)

\(\Rightarrow m_{C_6H_{12}O_6}=0,375.180=67,5\left(g\right)\)

b, \(n_{C_2H_5OH}=n_{CO_2}=0,75\left(mol\right)\Rightarrow m_{C_2H_5OH}=0,75.46=34,5\left(g\right)\)

\(\Rightarrow V_{C_2H_5OH}=\dfrac{34,5}{0,8}=43,125\left(ml\right)\)

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

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

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

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

\(C_2H_2 + H_2 \xrightarrow{t^o,xt} C_2H_4\\ C_2H_4 + H_2O \xrightarrow{t^o,xt} C_2H_5OH\\ C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O\\ CH_3COOH + NaOH \to CH_3COONa + H_2O\)

\(C_2H_2+H_2\underrightarrow{^{Pd,t^0}}C_2H_4\)

\(C_2H_4+H_2O\underrightarrow{^{H_2SO_4,170^0C}}C_2H_5OH\)

\(C_2H_5OH+O_2\underrightarrow{^{\text{men giấm}}}CH_3COOH+H_2O\)

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

CH3COONa + NaOH---CaO,\(t^o\)---> CH4 + Na2CO3

2CH4 --t\(^o=1500^oC,\)làm lạnh nhanh--> C2H2 + 3H2

C2H2 + H2 --\(t^o,xt,Pd\)--> C2H4

C2H4 + H2O --\(t^o,H2SO4\)--> C2H5OH

3C2H5OH + 3C6H6O -> 4C6H6 + 6H2O

C6H6 + HNO3 --\(t^o,H2SO4\)--> C6H5NO2 + H2O

\(C_2H_5OH+O_2\text{ }\underrightarrow{^{\text{men giấm}}}CH_3COOH+H_2O\)

\(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\left(Dk:H_2SO_{4\left(đ\right)},t^0\right)\)

\(CH_3COOC_2H_5+NaOH\underrightarrow{^{t^0}}CH_3COONa+C_2H_5OH\)

Câu 1:

\(a.\left(1\right)CaSO_3+2HCl\rightarrow CaCl_2+SO_2+H_2O\\ \left(2\right)SO_2+\dfrac{1}{2}O_2⇌\left(t^o,xt\right)SO_3\\ \left(3\right)SO_3+H_2O\rightarrow H_2SO_4\\ \left(4\right)Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

Câu 2:

a. Hiện tượng: Kẽm tan, tạo thành dung dịch khoomg màu, có sủi bọt khí.

b. Hiện tượng: Bột đồng (II) oxit có màu đen chuyển sang màu đỏ, đồng thời có hơi nước bám ở thành ống nghiệm.

c. Nếu là dung dịch H₂SO₄ loãng thì không có hiện tượng gì xảy ra. Nhưng nếu là dd H₂SO₄  đặc, nóng thì bột Ag tan đồng thời có xuất hiện chất khí mùi hắc nhé!

\(PTHH:\\ a.Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\\ b.CuO\left(đen\right)+H_2\underrightarrow{t^o}Cu\left(đỏ\right)+H_2O\\ c.2Ag+2H_2SO_4\left(đặc\right)\underrightarrow{t^o}Ag_2SO_4+SO_2\uparrow\left(mùi.hắc\right)+2H_2O\)

\(Mg+Cl_2\underrightarrow{^{^{t^0}}}MgCl_2\)

\(Na_2O+H_2SO_4\rightarrow Na_2SO_4+H_2O\)

\(Fe+CuCl_2\rightarrow FeCl_2+Cu\)

\(2KOH+FeSO_4\rightarrow K_2SO_4+Fe\left(OH\right)_2\)