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

\(n_{H_2O}=\dfrac{7,2}{18}=0,4\left(mol\right)\)

\(n_X=n_{H_2O}-n_{CO_2}=0,4-0,3=0,1\left(mol\right)\)

=> \(M_X=\dfrac{4,4}{0,1}=44g/mol\)

<=> \(14n+2=44\)

=> n = 3

Vậy CTPT của X: \(C_3H_8\)

Chọn đáp án A.

Ta có: \(n_{CO_2}=\dfrac{4,4}{44}=0,1\left(mol\right)=n_C\)

\(\Rightarrow m_H=1,45-m_C=1,45-0,1.12=0,25\left(g\right)\)

\(\Rightarrow n_H=0,25\left(mol\right)\Rightarrow n_{H_2O}=\dfrac{0,25}{2}=0,125\left(mol\right)\)

⇒ n _ankan = 0,125 - 0,1 = 0,025 (mol)

Gọi CTPT của ankan cần tìm là C_nH_2n+2 ( n ≥ 1)

\(\Rightarrow n=\dfrac{0,1}{0,025}=4\left(tm\right)\)

Vậy: Ankan đó là C₄H₁₀.

Bạn tham khảo nhé!

\(Ankan : C_nH_{2n+2}(n\geq1)\\ n_C = n_{CO_2} = \dfrac{4,4}{44} = 0,1(mol)\\ m_{ankan} = m_C + m_H \Rightarrow m_H = 1,45 - 0,2.12 = -0,95<0 \)

(Sai đề)

Ankan X : \(C_nH_{2n+2}\)

\(n_{CO_2} = \dfrac{2,24}{22,4} = 0,1(mol)\)

Bảo toàn nguyên tố với C:

\(n_X = \dfrac{n_{CO_2}}{n} = \dfrac{0,1}{n}mol\\ \Rightarrow m_X = \dfrac{0,1}{n}.(14n+2) = 1,6\ gam\\ \Rightarrow n = 1\)

Vậy X : CH₄(metan)

nCO2 = 2.24/22.4 = 0.1 (mol) 

BT cacbon : 

nCnH2n+2 = nCO2/n = 0.1/n (mol) 

M = 1.6/0.1/n = 16n

=> 14n + 2 = 16n

=> 2n = 2 

=> n = 1 

CT : CH4 ( metan) 

\(n_C=n_{CO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ \Rightarrow n_H=\dfrac{0,72-0,05.12}{1}=0,12\left(mol\right)\\ \Rightarrow n_{H_2O}=\dfrac{1}{2}n_H=0,06\left(mol\right)\\ \Rightarrow n_{ankan}=n_{H_2O}-n_{CO_2}=0,01\left(mol\right)\\ \Rightarrow M_{C_nH_{2n+2}}=\dfrac{0,72}{0,01}=72\left(g/mol\right)\\ \Rightarrow14n+2=72\\ \Leftrightarrow n=5\)

Vậy X là C₅H₁₂

X : C_nH_2n+2

\(n_{CO_2} = \dfrac{5,6}{22,4} = 0,25(mol)\\ \Rightarrow n_X = \dfrac{0,25}{n}(mol)\\ \Rightarrow M_X = 14n + 2 = \dfrac{3,6}{\dfrac{0,25}{n}} = 14,4n\\ \Rightarrow n = 5\)

Vậy CTPT của X: C₅H₁₂

CTCT :

\(CH_3-CH_2-CH_2-CH_2-CH_3\ : pentan\\ CH_3-CH(CH_3)-CH_2-CH_3 : 2-metylbutan\\ CH_3-C(CH_3)_2-CH_3 : 2,2-đimetylpropan\\ \)

\(n_{CO_2}=\dfrac{17.6}{44}=0.4\left(mol\right)\)

\(\Rightarrow n_X=\dfrac{0.4}{n}\left(mol\right)\)

\(M_X=\dfrac{6}{\dfrac{0.4}{n}}=15n\)

\(\Rightarrow14n+2=15n\)

\(\Rightarrow n=2\)

\(CT:C_2H_6\)

\(\)

nCO2=8,96/22,4=0,4 mol

=> nC=nCO2=0,4 mol

mC=0,4.12=4,8g

=> mH=5,8-4,8=1g

nH=1 mol -> nH2O=0,5

nH=0,5 mol

Pt: CnH2n+2 + (3n+1/2)O2-> nCO2 + (n+1)H2O

                                                0,4          0,5 mol

=> n/0,4=n+1/0,5 -> n=4

Vậy cthh cần tìm là C4H10

\(n_{H_2O}=\dfrac{12.6}{18}=0.7\left(mol\right)\)

\(n_{CO_2}=\dfrac{13.44}{22.4}=0.6\left(mol\right)\)

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

\(\dfrac{n}{2n+2}=\dfrac{0.6}{0.7\cdot2}\Rightarrow n=6\)

\(CT:C_6H_{14}\)