a)

Do đốt cháy A thu được sản phẩm chứa các nguyên tố C, H, O

=> A chứa C, H và có thể có O

 \(n_{CO_2}=\dfrac{10,8}{24}=0,45\left(mol\right)\)

=> n_C = 0,45 (mol)

\(n_{H_2O}=\dfrac{10,8}{18}=0,6\left(mol\right)\)

=> n_H = 1,2 (mol)

Xét m_C + m_H = 0,45.12 + 1,2.1 = 6,6 (g) < 9 (g)

=> A chứa C, H, O

=> A là dẫn xuất

b) \(n_O=\dfrac{9-6,6}{16}=0,15\left(mol\right)\)

Xét n_C : n_H : n_O = 0,45 : 1,2 : 0,15 = 3 : 8 : 1

=> CTPT: (C₃H₈O)_n

Mà PTK_A = 2,143.28 = 60 (đvC)

=> n = 1

=> CTPT: C₃H₈O

c) 

CTCT:

(1) \(CH_3-CH_2-CH_2OH\)

(2) \(CH_3-CH\left(OH\right)-CH_3\)

d) 

\(2CH_3-CH_2-CH_2OH+2Na\rightarrow2CH_3-CH_2-CH_2ONa+H_2\)

\(2CH_3-CH\left(OH\right)-CH_3+2Na\rightarrow2CH_3-CH\left(ONa\right)-CH_3+H_2\)

\(n_{H_2O} = \dfrac{9}{18} = 0,5(mol)\\ n_{CO_2} = \dfrac{8,96}{22,4} = 0,4(mol)\\ \Rightarrow n_X = n_{H_2O} - n_{CO_2}= 0,5 - 0,4 = 0,1(mol)\\ X\ :\ C_nH_{2n+2}(n \geq 1)\\ n = \dfrac{n_{CO_2}}{n_X} = \dfrac{0,4}{0,1} = 4\\ \Rightarrow CTPT\ X\ : C_4H_{10}\)

CTCT :

\(CH_3-CH_2-CH_2-CH_3 : butan\\ CH_3-CH(CH_3)-CH_3 : 2-metylpropan\)

\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

=> n_C = 0,3 (mol)

=> \(n_H=\dfrac{4-0,3.12}{1}=0,4\left(mol\right)\)

Xét n_C : n_H = 0,3 : 0,4 = 3 : 4

=> CTPT: (C₃H₄)_n

Mà M_X = 1,25.32 = 40 (g/mol)

=> n = 1

=> CTPT: C₃H₄

\(M_A=15.M_{H_2}=15.2=30\left(\dfrac{g}{mol}\right)\\ Đặt:C_aH_b\left(a,b:nguyên,dương\right)\\ n_C=n_{CO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ \Rightarrow m_C=0,5.12=6\left(g\right)\\ m_H=7,5-6=1,5\left(g\right)\\ \Rightarrow n_H=\dfrac{1,5}{1}=1,5\left(mol\right)\\ \Rightarrow a:b=0,5:1,5=1:3\\ \Rightarrow CTTQ:\left(CH_3\right)_t\left(t:nguyên,dương\right)\\ \Leftrightarrow15t=30\\ \Leftrightarrow t=2\\ \Rightarrow CTPT:C_2H_6\\ CTCT:CH_3-CH_3\)

a) CTPT: C_nH_2n-2

\(n_{C_nH_{2n-2}}=\dfrac{10,8}{14n-2}\left(mol\right)\)

=> \(n_{CO_2}=\dfrac{10,8}{14n-2}.n=\dfrac{17,92}{22,4}=0,8\)

=> n = 4

=> CTPT: C₄H₆

b) 

CTCT:
(1) \(CH\equiv C-CH_2-CH_3\) (but-1-in)

(2) \(CH_3-C\equiv C-CH_3\) (but-2-in)

\(CT:C_nH_{2n}\)

\(n_{CO_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

\(BảotoànC:\)

\(n\cdot n_X=n_{CO_2}\)

\(\Rightarrow n=\dfrac{0.3}{0.15}=2\)

\(CTPT:C_2H_4\)

\(CTXT:CH_2=CH_2\left(etilen\right)\)

a) $n_{CO_2} = \dfrac{26,4}{44} = 0,6(mol)$

Gọi CTHH của ankan là $C_nH_{2n+2}$

Bảo toàn nguyên tố với C : $n_{ankan} = \dfrac{1}{n}n_{CO_2} = \dfrac{0,6}{n}(mol)$

$\Rightarrow \dfrac{0,6}{n}(14n+2} = 8,7$

$\Rightarrow n = 4$

Vậy CTHH cần tìm : $C_4H_{10}$

CTCT : 

$CH_3-CH_2-CH_2-CH_3$

$CH_3-CH(CH_3)-CH_3$

b) $CH_3-CH_2-CH_2-CH_3 + Cl_2 \to CH_2Cl-CH_2-CH_2-CH_3 + HCl$

c) $n_{C_4H_{10}} = \dfrac{5,8}{58} = 0,1(mol)$

$C_4H_{10} + \dfrac{13}{2}O_2 \xrightarrow{t^o} 4CO_2 + 5H_2O$

$n_{O_2} = \dfrac{13}{2}n_{O_2} = 0,65(mol)$
$V_{O_2} = 0,65.22,4 = 14,56(l)$

$V_{kk} = 14,56 : 21\% = 69,33(l)$

\(n_X=\dfrac{0.112}{22.4}=0.005\left(mol\right)\)

\(CT:C_nH_{2n+2}\)

\(m_{tăng}=m_{H_2O}=0.36g\)

\(\Rightarrow n_{_{ }H_2O}=\dfrac{0.36}{18}=0.02\left(mol\right)\)

\(BTNTH:\)

\(\left(2n+2\right)\cdot n_X=2\cdot n_{H_2O}\)

\(\Leftrightarrow\left(2n+2\right)\cdot0.005=2\cdot0.02\)

\(\Leftrightarrow n=3\)

\(CTPT:C_3H_6\)

\(CH_3-CH_2-CH_3\)