n_CO2 = 0,4 mol

C_nH_2n + \(\dfrac{3n}{2}\)O₂ \(\underrightarrow{t^o}\) nCO₂ + nH₂O 

\(\dfrac{0,4}{n}\)                        .0,4

ta có : MCnH2n = \(\dfrac{5,6n}{0,4}\) = 14n

↔ n=1(loại)

do n≥ 2 

\(n_C=n_{CO_2}=0,5\left(mol\right);n_H=2.n_{H_2O}=2.\dfrac{10,8}{18}=1,2\left(mol\right)\\ m_C+m_H=0,5.12+1,2.1=7,7,2\\ \Rightarrow X.ko.có.oxi\left(O\right)\\ \Rightarrow CTTQ:C_xH_y\left(x,y:nguyên,dương\right)\\ x:y=n_C:n_H=0,5:1,2=5:12\\a, \Rightarrow x=5;y=12\Rightarrow CTPT.X:C_5H_{12}\\ b,CH_3-CH_2-CH_2-CH_2-CH_3\\ CH_3-CH\left(CH_3\right)-CH_2-CH_3\\ CH_3-C\left(CH_3\right)_2-CH_3\)

a) \(n_{CO_2}=\dfrac{6,6}{44}=0,15\left(mol\right)\Rightarrow n_C=0,15\left(mol\right)\)

\(n_{H_2O}=\dfrac{3,6}{18}=0,2\left(mol\right)\Rightarrow n_H=0,4\left(mol\right)\)

=> \(n_O=\dfrac{3-0,15.12-0,4.1}{16}=0,05\left(mol\right)\)

Xét n_C : n_H : n_O = 0,15 : 0,4 : 0,05 = 3 : 8 : 1

=> CTPT: (C₃H₈O)_n

Mà M_A = 60 g/mol

=> n= 1

=> CTPT: C₃H₈O

b) 

(1) \(CH_3-CH_2-CH_2OH\)

(2) \(CH_3-CH\left(OH\right)-CH_3\)

c)

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

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

nCO2 = 8.96/22.4 = 0.4 (mol) 

nH2O = 9/18 = 0.5 (mol) 

mO = mA - mC - mH = 5.8 - 0.4*12 - 0.5*2 = 0 

n_A = 5.8/58 = 0.1 (mol) 

Số nguyên tử C : 0.4/0.1 = 4 

Số nguyên tử H : 0.5*2/0.1 = 10 

CT : C₄H₁₀

CTCT:

CH₃ - CH₂ - CH₂ - CH₃

CH₃ - CH(CH₃) - CH₃

n_CO2 = 0,4 mol => n_C = 0,4 mol

n_H2O = 0,5 mol => n_H = 0,5 . 2 = 1 mol

C : H = 0,4 : 1 = 4 : 10

          X là C₄H₁₀