Bài 1 :

\(n_{H_2O}>n_{CO_2}\Rightarrow X:ankan\)

\(Đặt:CTHH:C_nH_{2n+2}\)

\(\dfrac{n}{2n+2}=\dfrac{0.1}{0.3}\Rightarrow n=2\)

\(Vậy:Xlà:C_2H_6\left(etan\right)\)

Bài 1 

\(n_{CO_2} < n_{H_2O} \to\) X là ankan (C_nH_2n+2)

\(n_X = n_{H_2O} - n_{CO_2} = 0,15 - 0,1 = 0,05(mol)\)

Suy ra:  \(n = \dfrac{n_{CO_2}}{n_X} = \dfrac{0,1}{0,05} = 2\)

Vậy X là C₂H₆(etan)

Bài 2 : 

 Hỗn hợp có dạng C_nH_2n+2

\(n_{hỗn\ hợp} = \dfrac{4,48}{22,4} = 0,2(mol)\\ n_{H_2O} = \dfrac{18}{18} = 1(mol)\\ \Rightarrow n + 2 = \dfrac{2n_{H_2O}}{n_{hh}} = 5\\ Suy\ ra\ n = 3\)

\(\Rightarrow n_{CO_2} = 3n_{hh} = 0,2.3 = 0,6(mol)\\ \Rightarrow V = 0,6.22,4 = 13,44(lít)\)

Bài 2 :

\(n_{H_2O}=\dfrac{18}{18}=1\left(mol\right)\)

\(n_{ankan}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)

\(n_{CO_2}=n_{H_2O}-n_{ankan}=1-0.2=0.8\left(mol\right)\)

\(V_{CO_2}=0.8\cdot22.4=17.92\left(l\right)\)

CTTQ của X : \(C_nH_{2n-2}\)

\(C_nH_{2n-2} + \dfrac{3n-1}{2}O_2 \xrightarrow{t^o} nCO_2 + (n-1)H_2O\\ n_{O_2} = 0,1.\dfrac{3n-1}{2} = \dfrac{12,992}{22,4} =0,58\\ \Rightarrow n = 4,2\)

Vậy X gồm : \(C_4H_6(x\ mol) ;C_5H_8(y\ mol)\)

Ta có :

\(n_X = x + y = 0,1\\ n_{O_2} = \dfrac{3.4-1}{2}x + \dfrac{3.5-1}{2} = 0,58\\ \Rightarrow x = 0,08 ; y = 0,02\\ \Rightarrow \%n_{C_4H_6} = \dfrac{0,08}{0,1}.100\% = 80\%\)

\(CT:C_{\overline{n}}H_{2\overline{n}-2}\)

\(C_{\overline{n}}H_{2\overline{n}-2}+\dfrac{3\overline{n}-1}{2}O_2\underrightarrow{t^0}\overline{n}CO_2+\left(\overline{n}-1\right)H_2O\)

\(1.................\dfrac{3\overline{n}-1}{2}\)

\(0.1.................0.58\)

\(\Rightarrow3\overline{n}-1=\left(\dfrac{0.58}{0.1}\right)\cdot2=11.6\)

\(\Rightarrow\overline{n}=4.2\)

\(CT:C_4H_6,C_5H_8\)

\(n_{C_4H_6}=a\left(mol\right),n_{C_5H_8}=b\left(mol\right)\)

\(n_X=a+b=0.1\left(1\right)\)

\(C_4H_6+\dfrac{11}{2}O_2\underrightarrow{t^0}4CO_2+3H_2O\)

\(C_5H_8+7O_2\underrightarrow{t^0}5CO_2+4H_2O\)

\(n_{O_2}=\dfrac{11}{2}a+7b=0.58\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.08,b=0.02\)

\(\%C_4H_6=80\%,\%C_5H_8=20\%\)

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