2CH4 -> (t°, làm lạnh nhanh) C2H2 + 3H2

C2H2 + H2 ->  (t°, xt, Pd) C2H4

C2H4 + HCl -> (askt) C2H5Cl

C2H5Cl + NaOH -> C2H5OH + NaCl

\(2CH_4\underrightarrow{t^o}C_2H_2+3H_2\)

\(C_2H_2\underrightarrow{t^o,xt,p}C_4H_4\)

\(C_4H_4+HCl\rightarrow C_2H_5Cl+C_2\)

\(C_2H_5Cl+NaOH\rightarrow C_2H_5OH+NaCl\)

\(BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)

\(x----x------x\)

\(n_{BaCl_2}=0,2\left(mol\right);n_{H_2SO_4}=0,125.0,4=0,05\left(mol\right)\)

Vậy H2SO4 phản ứng hết

\(\Rightarrow m_{BaSO_4}=0,05.233=11,65\left(g\right)\)

13Al + 48HNO3  → 3N2 + 3NO + 13Al(NO3)3 + 24H2O 

Câu 1.

\(CH_3COOH+NaOH\rightarrow CH_4\uparrow+Na_2CO_3\)

\(2CH_4\xrightarrow[LLN]{1500^oC}C_2H_2+2H_2\uparrow\)

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

a)\(C_2H_4+HBr\rightarrow C_2H_5Br\)

b)\(C_2H_4+HCl\underrightarrow{as}C_2H_5Cl\)

pt 1 và 2 bạn viết sai rồi :))

lỗi

???

Câu 5 : C etilen ; axetilen ; buta-1,3-đien ; propin

Câu 6 : B

Oxi hóa ancol A thu được andehit.

\(\Rightarrow A\) là ancol bậc \(I\).

Theo đáp án thì đáp án C thỏa mãn. Vì:

Đáp án A là ancol bậc ll.

Đáp án B là ancol bậc lll.

Đáp án D là ancol bậc ll.

\(\Rightarrow\)Chọn C là đáp án đúng.

Câu 1

a) \(CH_2=CH-CH=CH_2+H_2\underrightarrow{t^o,Ni}CH_3-CH_2-CH=CH_2\)

b) \(CH_2=CH-CH=CH_2+Br_2\rightarrow CH_2Br-CHBr-CH=CH_2\)

c) \(nCH_2=C\left(CH_3\right)-CH=CH_2\underrightarrow{t^o,p,xt}\left(-CH_2-C\left(CH_3\right)=CH-CH_2-\right)_n\)

d) \(CH\equiv C-CH_3+2H_2\underrightarrow{t^o,Ni}CH_3-CH_2-CH_3\)

e) \(CH\equiv C-CH_3+H_2\underrightarrow{t^o,Pd/PbCO_3}CH_2=CH-CH_3\)

f) \(CH\equiv CH+2HCl\underrightarrow{t^o,xt}CH_3-CHCl_2\)

g) \(CH\equiv C-CH_3+HOH\underrightarrow{Hg^{2+},H^+}CH_3-CO-CH_3\)

h) \(CH\equiv C-CH_3+AgNO_3+NH_3\underrightarrow{H_2O}CAg\equiv C-CH_3\downarrow+NH_4NO_3\)

Câu 2:

a) 

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

\(2CH_4\underrightarrow{1500^oC,làm.lạnh.nhanh}CH\equiv CH+3H_2\)

\(CH\equiv CH+HCl\underrightarrow{t^o,xt}CH_2=CHCl\)

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

b)

CaCO₃ --t^o--> CaO + CO₂

CaO + 3C --t^o--> CaC₂ + CO

CaC₂ + 2H₂O --> \(CH\equiv CH+Ca\left(OH\right)_2\)

\(2CH\equiv CH\underrightarrow{đime.hóa}CH\equiv C-CH=CH_2\)

\(CH\equiv C-CH=CH_2+H_2\underrightarrow{t^o,Pd/PbCO_3}CH_2=CH-CH=CH_2\)

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

c) X: CH₄, Y: C₂H₄

(1) \(CH_3-CH_2-CH_3\underrightarrow{t^o,xt}CH_4+CH_2=CH_2\)

(2) \(2CH_4\underrightarrow{1500^oC,làm.lạnh.nhanh}CH\equiv CH+3H_2\)

(3) \(CH\equiv CH+2H_2\underrightarrow{t^o,Ni}CH_3-CH_3\)

(4) \(CH\equiv CH+H_2\underrightarrow{t^o,Pd/PbCO_3}CH_2=CH_2\)

(5) \(CH_2=CH_2+H_2O\underrightarrow{H^+,t^o}CH_3-CH_2OH\)

(6) \(CH_3-CH_2OH\underrightarrow{H_2SO_4,170^oC}CH_2=CH_2\)