Đáp án C

\(C_2H_6 + Cl_2 \xrightarrow{as} C_2H_5Cl + HCl\)

Đáp án A :

\(CH_3-CH=CH-CH_3 : \text{But-2-en}\)

Ankin : C_nH_2n-2

\(n_{CO_2} = \dfrac{0,448}{22,4}=0,02(mol)\\ c_nH_{2n-2} + \dfrac{3n-1}{2}O_2 \xrightarrow{t^o} nCO_2 + (n-1)H_2O\\ n_{ankin} = \dfrac{n_{CO_2}}{n}=\dfrac{0,02}{n}(mol)\\ \Rightarrow \dfrac{0,02}{n}(14n-2) = 1,08\\ \Rightarrow n = -0,05<0\)

(Sai đề(

a, Ta có: \(n_{Ag_2C_2}=\dfrac{43,2}{240}=0,18\left(mol\right)\)

BTNT C, có: \(n_{C_2H_2}=n_{Ag_2C_2}=0,18\left(mol\right)\)

Mà: \(n_{hhk}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(\Rightarrow n_{C_3H_8}=0,07\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_2}=\dfrac{0,18}{0,25}.100\%=72\%\\\%V_{C_3H_8}=28\%\end{matrix}\right.\)

và \(\left\{{}\begin{matrix}\%m_{C_2H_2}=\dfrac{0,18.26}{0,18.26+0,07.44}.100\%\approx60,3\%\\\%m_{C_3H_8}=39,7\%\end{matrix}\right.\)

b, BTNT C, có: \(n_{CO_2}=3n_{C_3H_8}=0,21\left(mol\right)\)

\(\Rightarrow m_{CO_2}=0,21.44=9,24\left(g\right)\)

Bạn tham khảo nhé!

M_Y = 2 * 11 = 22 (g/mol) 

m_X = 0.1*26 + 0.2*28 + 0.3*2 = 8.8 (g) 

BTKL : 

m_X = m_Y = 8.8 (g) 

=> n_Y = 8.8 / 22 = 0.4 (mol) 

n_H2(pư) = n_X - n_Y = 0.1 + 0.2 + 0.3 - 0.4 = 0.2 (mol) 

Bảo toàn liên kết pi : 

2n_C2H2 + n_C2H4 = n_H2(pư) + n_Br2 

=> n_Br2 = 2 * 0.1 + 0.2 - 0.2 = 0.2 (mol) 

nC3H3Ag = 17.64/147 = 0.12 (mol) 

=> nC3H4 = 0.12 (mol) 

C3H4 + 2H2 -Ni,to-> C3H8 

0.12........0.24 

C2H4 + H2 -Ni,to-> C2H6 

0.1........0.34-0.24 

nX = 0.12 + 0.1 = 0.22 (mol) 

n_ankan = n_H2O - n_CO2 = 0.4 - 0.35 = 0.05 (mol) 

=> n_anken = 0.2 - 0.05 = 0.15 (mol) 

%anken = 0.15 / 0.2 * 100% = 75% 

Bảo toàn O :

nO2 = nCO2 + 1/2 * nH2O 

=> nO2 = 2.4 + 2.4 * 1/2 = 3.6 (mol) 

VO2 = 3.6*22.4 = 80.64(l)

\(X : C_nH_{2n}\\ m_X = m_{tăng} = 7,7(gam)\\ \Rightarrow M_X = 14n = \dfrac{7,7}{\dfrac{3,36}{22,4}}\\ \Rightarrow n = 3,6\)

Vậy hai anken là \(C_3H_6(a\ mol) ; C_4H_8(b\ mol)\)

Ta có: 

\(a + b = 0,15 \\ 42a + 56b = 7,7\\ \Rightarrow a = 0,05 ; b = 0,1\\ \%V_{C_3H_6} = \dfrac{0,05}{0,15} = 33,33\%\\ \%V_{C_4H_8} = 100\% -33,33\% = 66,67\%\)

Em làm tương tự như câu này nhé !!