Bài 25: Ankan

Chọn D (Vì từ C₁₈H₃₈ trở đi ankan có tồn tại trạng thái rắn)

a, Gọi CTPT chung 2 ankan là \(C_{\overline{n}}H_{2\overline{n}+2}\)

\(n_{H_2O}=\dfrac{28,08}{18}=1,56\left(mol\right)\Rightarrow n_H=1,56.2=3,12\left(mol\right)\)

\(\Rightarrow n_{CO_2}=n_C=\dfrac{16,44-m_H}{12}=1,11\left(mol\right)\)

⇒ \(n_{C_{\overline{n}}H_{2\overline{n}+2}}=n_{H_2O}-n_{CO_2}=0,45\left(mol\right)\)

\(\Rightarrow\overline{n}=\dfrac{1,11}{0,45}=2,467\)

Mà: 2 ankan liên tiếp.

→ CTPT: C₂H₆ và C₃H₈.

CTCT: CH₃-CH₃ và CH₃-CH₂-CH₃.

b, Có: \(30n_{C_2H_6}+44n_{C_3H_8}=16,44\left(1\right)\)

BTNT H, có: \(6n_{C_2H_6}+8n_{C_3H_8}=1,56.2\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{C_2H_6}=0,24\left(mol\right)\\n_{C_3H_8}=0,21\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%n_{C_2H_6}=\dfrac{0,24}{0,24+0,21}.100\%\approx53,33\%\\\%n_{C_3H_8}\approx46,67\%\end{matrix}\right.\)

Yêu cầu đề là gì em?

1) Đặt CTPT của ankan là \(C_nH_{2n+2}\left(n\in N;n\ge1\right)\)

\(\Rightarrow\dfrac{12n}{14n+2}.100\%=84\%\Leftrightarrow n=7\left(t/m\right)\)

Vậy ankan là C₇H₁₆

2) Đặt CTPT của X là \(C_nH_{2n+2}\left(n\in N;n\ge1\right)\)

\(\Rightarrow\dfrac{2n+2}{14n+2}.100\%=17,24\%\Leftrightarrow n=4\left(t/m\right)\)

Vậy X là C₄H₁₀

3) Đặt CTPT của X là \(C_nH_{2n+2}\left(n\in N;n\ge1\right)\)

\(\Rightarrow\dfrac{12n}{2n+2}=\dfrac{36}{7}\Leftrightarrow n=6\left(t/m\right)\)

Vậy X là C₆H₁₄

4) Đặt CTPT của X là \(\left(CH_3\right)_n\left(n\in N;n\ge1\right)\)

\(\Rightarrow3n=2n+2\Leftrightarrow n=2\left(t/m\right)\)

Vẫy X là C₂H₆

5) \(n_{ankan}=\dfrac{P.V}{R.T}=\dfrac{2.2,464}{0,082.\left(27,3+273\right)}=0,2\left(mol\right)\)

\(\Rightarrow M_{ankan}=\dfrac{14,4}{0,2}=72\left(g/mol\right)\)

Đặt CTPT của X là \(C_nH_{2n+2}\left(n\in N,n\ge1\right)\)

\(\Rightarrow14n+2=72\Leftrightarrow n=5\left(t/m\right)\)

Vậy X là C₅H₁₂

6) \(n_A=n_{O_2}=\dfrac{0,64}{32}=0,02\left(mol\right)\)

\(\Rightarrow M_A=\dfrac{0,88}{0,02}=44\left(g/mol\right)\)

Đặt CTPT của A là \(C_nH_{2n+2}\left(n\in N;n\ge1\right)\)

\(\Rightarrow14n+2=44\Leftrightarrow n=3\left(t/m\right)\)

Vậy A là C₃H₈

a)

\(4Al+3C\xrightarrow[]{1500-1700^oC}Al_4C_3\\ Al_4C_3+12H_2O\rightarrow4Al\left(OH\right)_3\downarrow+3CH_4\uparrow\\ CH_4+Cl_2\xrightarrow[]{a/s}CH_3Cl+HCl\\ CH_3Cl+Cl_2\xrightarrow[]{a/s}CH_2Cl_2+HCl\\ CH_2Cl_2+Cl_2\xrightarrow[]{a/s}CHCl_3\\ CHCl_3+Cl_2\xrightarrow[]{a/s}CCl_4+HCl\)

b)

\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\\ CH_3COONa+NaOH\xrightarrow[]{CaO,t^o}CH_4\uparrow+Na_2CO_3\)

\(CH_4+Cl_2\xrightarrow[]{a/s}CH_3Cl+HCl\\ 2CH_3Cl+2Na\xrightarrow[]{t^o}2NaCl+C_2H_6\\ C_2H_6\xrightarrow[]{t^o,xt}C_2H_4\)

c)

\(C_4H_{10}\xrightarrow[]{crackinh}C_2H_6+C_2H_4\\ C_2H_6+Cl_2\xrightarrow[]{a/s}C_2H_5Cl+HCl\\ 2C_2H_5Cl+2Na\xrightarrow[]{t^o,xt}C_4H_{10}+2NaCl\\ C_4H_{10}\xrightarrow[]{crackinh}CH_4+C_3H_6\\ C_3H_6+H_2\xrightarrow[]{Ni,t^o}C_3H_8\)

a) \(CH_3-CH_2-C\left(CH_3\right)_2-CH\left(C_2H_5\right)-CH_2-CH_3\)

b) \(CH_2Br-CHCl-CH\left(CH_3\right)-CH_2-CH_3\)

c) \(CH_2Cl-CCl\left(CH_3\right)-CH_2-CH_2-CH_2-CH_3\)

d) \(CH_3-CH\left(CH_3\right)-CH_2-CH_3\)

1) 2-metylbutan

2) Từ \(CH_3-CH_2-CH\left(CH_3\right)-CH\left(C_2H_5\right)-CH_3\) ta bẻ mạch thành

\(CH_3-CH_2-CH\left(CH_3\right)-CH\left(CH_3\right)-CH_2-CH_3\)

2,3-đimetylhexan

3) 2,2-đimetylbutan

4) Từ \(CH_3-CHBr-CH_2-CH\left(C_2H_5\right)-CH_3\) bẻ mạch ta được:

\(CH_3-CHBr-CH_2-CH\left(CH_3\right)-CH_2-CH_3\)

2-brom-4-metylhexan