\(C_2H_2+AgNO_3+NH_3->C_2Ag_2+NH_4NO_3\\ CH_2=CH-CH_3+Br_2->CH_2Br-CHBr-CH_3\\ n_{C_2Ag_2}=0,325mol=n_{C_2H_2}\\ m_{C_2H_2}=8,45g\\ n_{Br_2}=n_{C_3H_6}=\dfrac{32}{160}=0,2mol\\ m_{C_3H_6}=8,4g\\ m_{CH_4}=20-8,45-8,4=3,15g\\ n_{CH_4}=0,196875mol\\ n_{hh}=n_{CH_4}+n_{C_2H_2}+n_{C_3H_6}=0,721875mol\\ \%V_{C_3H_6}=\dfrac{0,2}{n_{hh}}.100\%=27,71\%\\ \%V_{C_2H_2}=\dfrac{0,325}{n_{hh}}.100\%=45,02\%\\ \%V_{CH_{_34}}=27,27\%\)

Ta có: \(n_X=\dfrac{3,584}{22,4}=0,16\left(mol\right)\)

- Cho hh khí quá bình 1:

PT: \(C_3H_4+AgNO_3+NH_3\rightarrow AgC_3H_3+NH_4NO_3\)

Ta có: \(n_{AgC_3H_3}=\dfrac{5,88}{147}=0,04\left(mol\right)\)

Theo PT: \(n_{C_3H_4}=n_{AgC_3H_3}=0,04\left(mol\right)\)

- Cho tiếp hh khí còn lại qua bình 2, thấy dd Br₂ nhạt màu.

→ Br₂ dư, C₂H₄ pư hết. Khí thoát ra là C₃H₈.

PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

Ta có: \(n_{C_3H_8}=\dfrac{1,792}{22,4}=0,08\left(mol\right)\)

a, \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,16-0,08-0,04}{0,16}.100\%=25\%\\\%V_{C_3H_4}=\dfrac{0,04}{0,16}.100\%=25\%\\\%V_{C_3H_8}=50\%\end{matrix}\right.\)

b, Ta có: \(n_X=\dfrac{1,792}{22,4}=0,08\left(mol\right)\)

\(\Rightarrow X\left\{{}\begin{matrix}n_{C_2H_4}=0,08.25\%=0,02\left(mol\right)\\n_{C_3H_4}=0,08.25\%=0,02\left(mol\right)\\n_{C_3H_8}=0,04\left(mol\right)\end{matrix}\right.\) \(\Rightarrow n_{\pi}=n_{C_2H_4}+2n_{C_3H_4}=0,06\left(mol\right)\)

\(n_{Br_2}=0,08.1=0,08\left(mol\right)\)

⇒ n_π < n_Br2

→ Br₂ dư. Hiện tượng: Dd Br₂ nhạt màu dần.

Đáp án : A

Hỗn hợp X có : 0,2 mol C₂H₂ ; 0,1 mol C₃H₆ ; 0,3 mol H₂

Bảo toàn khối lượng : m_X = m_Y = 10g

M_Y : M_X = 12 : 7 => n_X : n_Y = 12 : 7

=> n_Y = 0,35 mol => n_{H2 pứ} = 0,25 mol

Có n_{pi (X)} = 2n_C2H2 + n_C3H6 = 0,5 mol => n_Pi(Y) = n_{pi (X)} – n_{H2 pứ} = 0,25 mol

Y phản ứng được với AgNO₃/NH₃ => n_Ag2C2 = n_C2H2 = 0,05 mol

=> n_{pi (Z)} = n_pi(Y) – 2n_{C2H2 (Y)} = 0,15 mol

=> n_Br2 = n_{pi (Z)} = 0,15 mol => m = 24g

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)

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

\(n_{C_3H_3Ag}=\dfrac{7,35}{147}=0,05\left(mol\right)\)

=> \(n_{C_3H_4}=0,05\left(mol\right)\)

\(n_{Br_2}=\dfrac{6,4}{160}=0,04\left(mol\right)\)

=> \(n_{C_2H_4}=0,04\left(mol\right)\)

=> \(\left\{{}\begin{matrix}\%C_3H_4=\dfrac{0,05.40}{5,52}.100\%=36,23\%\\\%C_2H_4=\dfrac{0,04.28}{5,52}.100\%=20,29\%\\\%C_2H_6=100\%-36,23\%-20,29\%=43,48\%\end{matrix}\right.\)

=> A

Đáp án D