C

C

C

(1) \(2CH_4\rightarrow C_2H_2+2H_2\)

(2) \(C_2H_2+H_2\rightarrow C_2H_4\)

(3) \(C_2H_4+H_2O\rightarrow C_2H_5OH\)

(4) \(C_2H_4+H_2\rightarrow C_2H_6\)

(5) \(C_2H_6+Cl_2\rightarrow HCl+C_2H_5Cl\)

(6) \(3C_2H_2\rightarrow C_6H_6\)

(7) \(C_6H_6+Br_2\rightarrow HBr+C_6H_5Br\)

(8) \(2C_6H_6+15O_2\rightarrow12CO_2+6H_2O\)

câu A

\(\left(1\right)C_4H_{10}\rightarrow CH_4+C_3H_6\)  (PƯ cracking)

\(\left(2\right)CH_4\rightarrow C_2H_2+H_2\)  (điều kiện: làm lạnh nhanh)

\(\left(3\right)CH\equiv CH+H_2\rightarrow CH_2=CH_2\)   (điều kiện: Pd/PbCO₃, t⁰)

\(\left(4\right)CH_2=CH_2+KMnO_4+H_2O\)\(\rightarrow C_2H_6O_2+MnO_2+KOH\)

\(\left(5\right)nCH_2=CH_2\rightarrow\left(-CH_2-CH_2-\right)_n\) (điều kiện: nhiệt độ, xúc tác)

\(\left(6\right)3CH\equiv CH\rightarrow C_6H_6\)    (điều kiện: xúc tác carbon, nhiệt độ)

\(\left(7\right)C_6H_6+Br_2\rightarrow C_6H_5Br+HBr\)

\(\left(8\right)C_6H_6+C_2H_4\rightarrow C_6H_5-CH_2-CH_3\)

\(\left(9\right)C_6H_5-CH_2-CH_3\rightarrow C_8H_8+H_2\)

: a) CaCO3, CaO, CaC2, C2H2, C2H4, C2H6, CO2, CO

=> CaCO3-->CaO+CO2

CaO+C-->CaC2+CO

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

C2H2+H2-->C2H4

C2H4+H2-->C2H6

C2H6+O2-->CO2+H2O

CO2+C-->CO

C)Al+Cl2-->AlCl3

AlCl3+NaoH-->Al(OH)3+NaCl

Al(OH)3-->Al2O3+H2O

Al2o3+C-->Al4C3+CO

Al4C3+H2O-->Al(OH)3+CH4

CH4+H2-->C2H2

a) CaCO3--to-->CaO+CO2

CaO+2C--->CaC+CO

2CaC+2H2O--->Ca(OH)2+C2H2

C2H2+H2--to->C2H4

C2H4+H2--->C2H6

C2H6+7/2O2-->2CO2+3H2O

CO2+C---to->2CO

b) C2H6+Cl2-->C2H5Cl+HCl

C2H5Cl-->C2H4+HCl

C2H4+Br2--->C2H4Br2

C2H4Br+2NaOH-->C2H4(OH)2+2NaBr

c) 2Al+6HCl-->AlCl3+3H2

AlCl3+3NaOH-->Al(OH)3+3NaCl

2Al(OH)3-->Al2O3+3H2O

2Al2O3+9C--->Al4C3+6CO

Al4C3+12H2O--->3CH4+4Al(OH)3

2CH4-->C2H2+3H2

3C2H2-->C6H6

C6H6+CH3Cl--->C6H5CH3+HCl

C6H5CH3+2KMnO4--->H2O+KOH+2MnO2+C6H5COOK

: a) CaCO3, CaO, CaC2, C2H2, C2H4, C2H6, CO2, CO

=> CaCO3-->CaO+CO2

CaO+C-->CaC2+CO

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

C2H2+H2-->C2H4

C2H4+H2-->C2H6

C2H6+O2-->CO2+H2O

CO2+C-->CO

C)Al+Cl2-->AlCl3

AlCl3+NaoH-->Al(OH)3+NaCl

Al(OH)3-->Al2O3+H2O

Al2o3+C-->Al4C3+CO

Al4C3+H2O-->Al(OH)3+CH4

CH4+H2-->C2H2

Chọn B

B

B

I)

1) 

\(2CH_4\xrightarrow[lln]{1500^oC}C_2H_2+3H_2\\ 2CH\equiv CH\xrightarrow[]{t^o,p,xt}CH\equiv C-CH=CH_2\\ CH\equiv C-CH=CH_2+H_2\xrightarrow[]{Pd/PbCO_3,t^o}CH_2=CH-CH=CH_2\\ CH_2=CH-CH=CH_2+2H_2\xrightarrow[]{Ni,t^o}CH_3-CH_2-CH_2-CH_3\\ CH_3-CH_2-CH_2-CH_3\xrightarrow[]{cracking}CH_2=CH-CH_3+CH_4\)

2)

\(CH_3COONa+NaOH\xrightarrow[]{CaO,t^o}CH_4+Na_2CO_3\\ 2CH_4\xrightarrow[lln]{1500^oC}C_2H_2+3H_2\\ CH\equiv CH+H_2\xrightarrow[]{Pd/PbCO_3,t^o}CH_2=CH_2\\ CH_2=CH_2+Br_2\rightarrow CH_2Br-CH_2Br\)

3)

\(2CH_4\xrightarrow[]{1500^oC}C_2H_2+3H_2\\ 2CH\equiv CH\xrightarrow[]{t^o,p,xt}CH\equiv C-CH=CH_2\\ CH\equiv C-CH=CH_2+H_2\xrightarrow[]{Pd/PbCO_3,t^o}CH_2=CH-CH=CH_2\\ nCH_2=CH-CH=CH_2\xrightarrow[]{t^o,p,xt}\left(-CH_2-CH=CH-CH_2-\right)_n\)

4) 

\(C_4H_8+H_2\xrightarrow[]{Ni,t^o}C_4H_{10}\\ C_4H_{10}\xrightarrow[]{cracking}CH_4+C_3H_6\\ 2CH_4\xrightarrow[lln]{1500^oC}C_2H_2+3H_2\\ CH\equiv CH\xrightarrow[]{t^o,p,xt}CH\equiv C-CH=CH_2\)

II)

1)

\(CH\equiv C-CH_2-CH_3+2Br_2\rightarrow CHBr_2-CBr_2-CH_2-CH_3\\ CH_3-C\equiv C-CH_3+2Br_2\rightarrow CH_3-CBr_2-CBr_2-CH_3\\ CH\equiv C-CH_2-CH_3+AgNO_3+NH_3\rightarrow CAg\equiv C-CH_2-CH_3\downarrow+NH_4NO_3\)

2)

\(CH\equiv CH+2AgNO_3+2NH_3\rightarrow CAg\equiv CAg\downarrow+2NH_4NO_3\\ CH_2=CH_2+Br_2\rightarrow CH_2Br-CH_2Br\)

III)

1) \(n_C=n_{CO_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)

\(\Rightarrow n_H=\dfrac{0,88-0,06.12}{1}=0,16\left(mol\right)\Rightarrow n_{H_2O}=\dfrac{1}{2}n_{H_2O}=0,08\left(mol\right)\\ \Rightarrow n_{ankan}=n_{H_2O}-n_{CO_2}=0,02\left(mol\right)\\ \Rightarrow M_{ankan}=\dfrac{0,88}{0,02}=44\left(g/mol\right)\)

Đặt CTHH của ankan là C_nH_2n+2

=> 14n + 2 = 44 => n = 3

Vậy X là C₃H₈ \(\left(CTCT:CH_3-CH_2-CH_3:propan\right)\)

2) \(n_{H_2O}=\dfrac{1,62}{18}=0,09\left(mol\right)\Rightarrow n_H=2n_{H_2O}=0,18\left(mol\right)\)

\(\Rightarrow n_{CO_2}=n_C=\dfrac{1,62-0,18}{12}=0,12\left(mol\right)\\ \Rightarrow n_{ankin}=n_{CO_2}-n_{H_2O}=0,03\left(mol\right)\\ \Rightarrow nM_{ankin}=\dfrac{1,62}{0,03}=54\left(g/mol\right)\)

Đặt CTHH của ankin là C_nH_2n-2

=> 14n - 2 = 54 => n = 4

Vậy X là C₄H₆

CTCT: 

\(CH\equiv C-CH_2-CH_3:\) but-1-in

\(CH_3-C\equiv C-CH_3:\) but-2-in

3)

Sửa đề: 1,17 -> 11,7

\(\left\{{}\begin{matrix}n_{H_2O}=\dfrac{11,7}{18}=0,65\left(mol\right)\\n_{CO_2}=\dfrac{17,6}{44}=0,4\left(mol\right)\end{matrix}\right.\Rightarrow n_{H_2O}>n_{CO_2}\Rightarrow\) hh thuốc dãy đồng đẳng ankan

Ta có: \(n_{hh}=n_{H_2O}-n_{CO_2}=0,25\left(mol\right)\)

Theo BTNT: \(\left\{{}\begin{matrix}n_C=n_{CO_2}=0,4\left(mol\right)\\n_H=2n_{H_2O}=1,3\left(mol\right)\end{matrix}\right.\Rightarrow m_{hh}=0,4.12+1,3=6,1\left(g\right)\)

\(\Rightarrow M_{hh}=\dfrac{6,1}{0,25}=24,4\left(g/mol\right)\)

Đặt CT chung của hh là C_nH_2n+2

=> 14n + 2 = 24,4 => n = 1,6 

=> Hai hiđrocacbon là CH₄ và C₂H₆

\(CH_3COONa+NaOH\underrightarrow{^{t^o,CaO}}CH_4+Na_2CO_3\)

\(2CH_4\underrightarrow{^{1500^oC,lln}}C_2H_2+3H_2\)

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

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

\(CH_4+Cl_2\underrightarrow{as}CH_3Cl+HCl\)

\(C_2H_2+2H_2\underrightarrow{t^o,Ni}C_2H_6\)

\(nCH_2=CH_2\underrightarrow{t^o,p,xt}\left(-CH_2-CH_2-\right)_n\)

Câu 1: Dãy các chất nào sau đây đều là hiđrocacbon ? A. C2H6, C4H10, C2H4. B. CH4, C2H2, C3H7Cl. C. C2H4, CH4, C2H5Cl. D. C2H6O, C3H8, C2H2.