Chương 5. Hiđrocacbon no

a)

C₂H₂ + 2AgNO₃ + 2NH₃ --> C₂Ag₂ + 2NH₄NO₃

C₃H₆ + Br₂ --> C₃H₆Br₂ 

b) 

\(n_{C_2Ag_2}=\dfrac{3,6}{240}=0,015\left(mol\right)\)

=> n_C2H2 = 0,015 (mol)

\(m_{tăng}=m_{C_3H_6}=2,1\left(g\right)\)

=> \(n_{C_3H_6}=\dfrac{2,1}{42}=0,05\left(mol\right)\)

\(n_{C_2H_6}=\dfrac{0,784}{22,4}=0,035\left(mol\right)\)

\(\left\{{}\begin{matrix}\%V_{C_2H_6}=\dfrac{0,035}{0,035+0,05+0,015}.100\%=35\%\\\%V_{C_3H_6}=\dfrac{0,05}{0,035+0,05+0,015}.100\%=50\%\\\%V_{C_2H_2}=\dfrac{0,015}{0,035+0,05+0,015}.100\%=15\%\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{C_2H_6}=\dfrac{0,035.30}{0,035.30+0,05.42+0,015.26}.100\%=29,661\%\\\%m_{C_3H_6}=\dfrac{0,05.42}{0,035.30+0,05.42+0,015.26}.100\%=59,322\%\\\%m_{C_2H_2}=\dfrac{0,015.26}{0,035.30+0,05.42+0,015.26}.100\%=11,017\%\end{matrix}\right.\)

c)

\(n_{CaCO_3}=\dfrac{5}{100}=0,05\left(mol\right)\)

Bảo toàn C: \(n_{CaCO_3}+2.n_{Ca\left(HCO_3\right)_2}=0,25\left(mol\right)\)

=> \(n_{Ca\left(HCO_3\right)_2}=0,1\left(mol\right)\)

Bảo toàn Ca: \(n_{Ca\left(OH\right)_2}=0,15\left(mol\right)\)

=> \(V_{dd.Ca\left(OH\right)_2}=\dfrac{0,15}{1}=0,15\left(l\right)=150\left(ml\right)\)

a) \(n_{C_2Ag_2}=\dfrac{48}{240}=0,2\left(mol\right)\)

=> n_C2H2 = 0,2 (mol)

\(n_{Br_2}=\dfrac{24}{160}=0,15\left(mol\right)\)

PTHH: C₂H₄ + Br₂ --> C₂H₄Br₂

            0,15<--0,15

\(n_{C_3H_8}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(\left\{{}\begin{matrix}\%V_{C_3H_8}=\dfrac{0,15}{0,15+0,15+0,2}.100\%=30\%\\\%V_{C_2H_4}=\dfrac{0,15}{0,15+0,15+0,2}.100\%=30\%\\\%V_{C_2H_2}=\dfrac{0,2}{0,15+0,15+0,2}.100\%=40\%\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{C_3H_8}=\dfrac{0,15.44}{0,15.44+0,15.28+0,2.26}.100\%=41,25\%\\\%m_{C_2H_4}=\dfrac{0,15.28}{0,15.44+0,15.28+0,2.26}.100\%=26,25\%\\\%m_{C_2H_2}=\dfrac{0,2.26}{0,15.44+0,15.28+0,2.26}.100\%=32,5\%\end{matrix}\right.\)

b) Bảo toàn C; n_CaCO3 = 1,15 (mol)

=> m_CaCO3 = 1,15.100 = 115 (g)

c) \(n_{C_2H_2\left(bđ\right)}=\dfrac{0,2}{2}=0,1\left(mol\right)\)

=> n_{C2Ag2(lý thuyết)} = 0,1 (mol)

\(H\%=\dfrac{19,2}{0,1.240}.100\%=80\%\)

Ankan có CTHH là \(C_nH_{2n+2}\)

BTKL: \(m_X=m_Y\)

Mà \(n_X=3n_Y\Rightarrow\dfrac{m_X}{M_X}=3\cdot\dfrac{m_Y}{M_Y}\Rightarrow M_X=3M_Y\)

\(M_Y=12\cdot2=24\Rightarrow M_X=24\cdot3=72đvC\)

\(\Rightarrow14n+2=72\Rightarrow n=5\)

\(\Rightarrow C_5H_{12}\)

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

V (khí thoát ra) = VC3H8 = 2,24 (l)

%VCH4 = 2,24/3,36 = 66,66%

%VC2H2 = 100% - 66,66% = 33,34%

V khí thoát =V _C3H8=2,24l

C2H4+Br2-->C2H4Br2

%C3H8=\(\dfrac{2,24}{3,36}\).100=66,67%

=>%C2H4=100-66,67=33,33%

\(CH_2\)

n(CH2=CH2)-to\Xt->-(-CH2-CH2-)-_n

C2H4+Br2-to>C2H4Br2

0.6----0,6

n Br2=\(\dfrac{96}{160}\)=0,6 mol

m CH4=24,8-0,6.28=8g

=>n CH4=\(\dfrac{8}{16}\)=0,5 mol

=>%CH4=\(\dfrac{0,5.22,4}{0,5.22,4+0,6.22,4}\).100=45,45%

=>%C2H4=54,55%

a)

\(n_{BaCO_3}=\dfrac{78,8}{197}=0,4\left(mol\right)\)

=> \(n_{CO_2}=0,4\left(mol\right)\)

=> \(n_{H_2O}=\dfrac{22,1-0,4.44}{18}=0,25\left(mol\right)\)

Do \(n_{CO_2}>n_{H_2O}\)

=> 2 hidrocacbon thuộc dãy đồng đẳng ankin hoặc ankadien

Bảo toàn C: n_C = 0,4 (mol)

Do \(\overline{C}=\dfrac{0,4}{0,15}=2,67< 3\)

=> 2 hidrocacbon thuộc dãy đồng đẳng ankin

b)

Do \(\overline{C}=2,67\)

=> Trong X có C₂H₂ (etin)

Giả sử A là C₂H₂

Gọi CTPT của B là C_nH_2n-2

TH1: \(\left\{{}\begin{matrix}n_A+n_B=0,15\\n_A:n_B=1:2\end{matrix}\right.\)

=> n_A = 0,05 (mol); n_B = 0,1 (mol)

Bảo toàn C: 2.0,05 + 0,1.n = 0,4

=> n = 3

=> B là C₃H₄ (propin)

TH2: \(\left\{{}\begin{matrix}n_A+n_B=0,15\\n_B:n_A=1:2\end{matrix}\right.\)

=> n_A = 0,1 (mol); n_B = 0,05 (mol)

Bảo toàn C: 0,1.2 + 0,05.n = 0,4

=> n = 4

=> B là C₄H₆

C₄H₆ có 2 đồng phân ankin

+ \(CH\equiv C-CH_2-CH_3\) (but-1-in)

+ \(CH_3-C\equiv C-CH_3\) (but-2-in)