Bài 46. Mối liên hệ giữa etilen, rượu etylic và axit axetic

Đặt : \(n_{C2H4}=a\left(mol\right),n_{C2H2}=b\left(mol\right)\rightarrow a+b=\dfrac{13,44}{22,4}\left(1\right)\)

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

Ta có : \(n_{Br2}=0,8.1=0,8\left(mol\right)\rightarrow a+2b=0,8\left(2\right)\)

\(\left(1\right),\left(2\right)\rightarrow a=0,4,b=0,2\)

\(\rightarrow\left\{{}\begin{matrix}\%V_{C2H4}=\dfrac{0,4.22,4}{13,44}.100\%=66,67\%\\\%V_{C2H2}=100\%-66,67\%=33,33\%\end{matrix}\right.\)

\(n_{CO2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(C_2H_5OH+3O_2\xrightarrow[]{t^o}2CO_2+3H_2O\)

  0,1             0,3         0,2

b) \(m_{C2H5OH}=0,1.46=4,6\left(g\right)\)

c) \(V_{O2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)

 Chúc bạn học tốt

a)

- Xét TN2:

n_NaOH = 0,2.0,2 = 0,04 (mol)

PTHH: CH₃COOH + NaOH --> CH₃COONa + H₂O

                 0,04<-----0,04

- Xét TN1:

\(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

PTHH: 2CH₃COOH + 2Na --> 2CH₃COONa + H₂

                 0,04------------------------------>0,02

            2C₂H₅OH + 2Na --> 2C₂H₅ONa + H₂

                 0,02<--------------------------0,01

=> m = 0,04.60 + 0,02.46 = 3,32 (g)

b) \(\left\{{}\begin{matrix}\%m_{CH_3COOH}=\dfrac{0,04.60}{3,32}.100\%=72,29\%\\\%m_{C_2H_5OH}=\dfrac{0,02.46}{3,32}.100\%=27,71\%\end{matrix}\right.\)

a, PTHH: CH₃COOH + C₂H₅OH --t^o, H₂SO_4(đ)--> CH₃COOC₂H₅ + H₂O

b, \(\left\{{}\begin{matrix}n_{CH_3COOH}=\dfrac{30}{60}=0,5\left(mol\right)\\n_{C_2H_5OH}=\dfrac{80}{46}=\dfrac{40}{23}\left(mol\right)\end{matrix}\right.\)

LTL: \(0,5< \dfrac{40}{23}\) => C₂H₅OH dư

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

=> \(m_{este}=0,5.80\%.88=35,2\left(g\right)\)

\(CH_3COOH+C_2H_5OH\underrightarrow{H_2SO_4đ,t^o}CH_3COOC_2H_5+H_2O\)

\(nCH_3COOH=\dfrac{30}{60}=0,5\left(mol\right)\)

\(nC_2H_5OH=\dfrac{80}{46}=1,74\left(mol\right)\) 

=> CH3COOH đủ  , nC2H5OH dư

=> \(CH_3COOC_2H_5=0,5\left(mol\right)\)

=> \(mCH_3COOC_2H_5=0,5.88=44\left(g\right)\)

=> \(mCH_3COOC_2H_{5\left(thựcte\right)}=\dfrac{44.80}{100}=35,2\left(g\right)\)

CH4+2O2-to>CO2+2H2O

x---------2x-------x---------2x

C2H4+3O2-to>2CO2+2H2O

y-------------3y------2y-------2y

=>\(\left\{{}\begin{matrix}16x+28y=1,36\\2x+3y=0,16\end{matrix}\right.\)

=>x=0,05 mol , y=0,02 mol

=>%m CH4=\(\dfrac{0,05.16}{1,36}100\)=58,82%

=>%m C2H4=41,17%

=>mH2O=(0,1+0,04).18=2,52g

a, \(n_{O_2}=\dfrac{3,584}{22,4}=0,16\left(mol\right)\)

PTHH: CH₄ + 2O₂ ---t^o→ CO₂ + 2H₂O

Mol:       x        2x                           2x

PTHH: C₂H₄ + 3O₂ ---t^o→ 2CO₂ + 2H₂O

Mol:        y         3y                               2y

b,Ta có hệ pt: \(\left\{{}\begin{matrix}16x+28y=1,36\\2x+3y=0,16\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,05\left(mol\right)\\y=0,02\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{CH_4}=0,05.16=0,8\left(g\right)\Rightarrow\%m_{CH_4}=\dfrac{0,8.100\%}{1,36}=58,82\%\)

\(\Rightarrow\%m_{C_2H_4}=100\%-58,82\%=41,18\%\)

c, \(m_{H_2O}=\left(2.0,05+2.0,02\right).18=2,52\left(g\right)\)

CH4+2O2-to>CO2+2H2O

x---------2x-------x

C2H4+3O2-to>2CO2+2H2O

y-------------3y------2y

=>Ta có :

\(\left\{{}\begin{matrix}x+y=0,04\\2x+3y=0,11\end{matrix}\right.\)

=>x=0,01, y=0,03 mol

=>%VCH4=\(\dfrac{0,01.22,4}{0,896}100\)=25%

=>%VC2H4=75%

=>m CO2=(0,01+0,03.2).44=3,08g

a, ta có PTHH : 

CH₄ + 2O₂ \(\Rightarrow\) CO₂ + 2H₂O (*)

C₂H₄ + 3O₂ \(\Rightarrow\) 2CO₂ + 2H₂O(+)

b, Gọi x, y lần lượ là số mol của CH₄ , C₂H₄ ( x, y > 0 )

Ta có : n_{h =} \(\dfrac{0,896}{22,4}=0,04\left(mol\right)\)

\(\Rightarrow x+y=0,04\left(mol\right)\) (1)

lại có n_{o2 pứ} = \(\dfrac{3,52}{32}=0,11\left(mol\right)\)

\(\Rightarrow2x+3y=0,11\left(mol\right)\) (2)

Từ (1) và (2) ta có hệ pt:\(\left\{{}\begin{matrix}x+y=0,04\\2x+3y=0,11\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,01\left(mol\right)\\y=0,03\left(mol\right)\end{matrix}\right.\)

%V_CH4 = \(\dfrac{0,01}{0,04}\times100\%=25\%\)

%V_C2H4 = \(\dfrac{0,03}{0,04}\times100\%=75\%\)

c, Theo pt n_CO2(*) = n_CH4 = 0,01 (mol)

n_CO2(+) = 2n_C2H4 = 0,6 (mol)

\(\Rightarrow m_{co_2}=\left(0,01+0,6\right)\times44=26,84\left(g\right)\)

a)\(2CH_3COOH+Mg\rightarrow\left(CH_3COO\right)_2Mg+H_2\uparrow\)

b)\(n_{H_2}=\dfrac{1}{2}n_{CH_3COOH}\Rightarrow V_{H_2}=\dfrac{1}{2}V_{CH_3COOh}=\dfrac{1}{2}\cdot100=50ml\)

c)\(CH_3COOH+C_2H_5OH\xrightarrow[xtH_2SO_4đặc]{140^oC}CH_3COOC_2H_5+H_2O\)

\(n_{este}=n_{CH_3COOC_2H_5}=\dfrac{130,68}{88}=1,485mol\)

\(\Rightarrow n_{CH_3COOH}=1,485mol\Rightarrow m_{líthuyết}=1,485\cdot60=89,1g\)

\(m_{thựctế}=V_{CH_3COOH}\cdot D_{CH_3COOH}=100\cdot1,05=105g\)

Hiệu suất phản ứng:

\(H=\dfrac{m_{líthuyết}}{m_{thựctế}}\cdot100\%=\dfrac{89,1}{105}\cdot100\%=84,86\%\)

A tác dụng với Br₂ theo tỉ lệ mol 1 : 1

=> A có dạng C_nH_2n

Với lp 9 thì chắc là C₂H₄ đấy :)

vì 0,4 mol A td với 0,4 mol Br2 

=> trong nguyên tử sẽ có 1 nối đôi 

=>thuộc loại Anken

NGU VL

B. C2H4, CO2, CaCO3. 

B

B