Đáp án:

$a,\ C_2H_6;\ C_3H_8\\ b,\\ \%m_{C_2H_6}=40,54\%\\ \%m_{C_3h_8}=59,46\%$

Giải thích các bước giải:

 Gọi công thức chung của hai ankan là: $C_nH_{2n+2}$

Gọi hỗn hợp $CO_2;\ H_2O$ là $X$

$a,\\ n_{X}=\dfrac{2,688}{22,4}=0,12(mol)\\ M_X=\dfrac{173}{12}.2=\dfrac{173}{6}(đvc)\\ Đặt\ \begin{cases}n_{CO_2}=x(mol)\\n_{H_2O}=y(mol)\end{cases}\\ \Rightarrow \begin{cases}x+y=0,12\\44x+18y=\dfrac{173}{6}.0,12=3,46\end{cases}\\ \Rightarrow \begin{cases}x=0,05\\y=0,07\end{cases}\\ C_nH_{2n+2}+\dfrac{3n+1}{2}O_2\xrightarrow{t^0} nCO_2+(n+1)H_2O\\ \xrightarrow{\text{Từ PTHH}}\ \dfrac{n_{CO_2}}{n}=\dfrac{n_{H_2O}}{n+1}\\ \Rightarrow \dfrac{0,05}{n}=\dfrac{0,07}{n+1}\\ \Rightarrow n=2,5\\ \to Ankan \ gồm\ \begin{cases}C_2H_6\\C_3H_8\end{cases}$ $b,\\ Đặt\ \begin{cases}n_{C_2H_6}=a(mol)\\n_{C_3H_8}=b(mol)\end{cases}\\ \xrightarrow{\text{BTNT C}}\ n_{CO_2}=2.n_{C_2H_6}+3.n_{C_3H_8}=2a+3b(mol)\\ \Rightarrow 2a+3b=0,05\ (1)\\ \xrightarrow{\text{BTNT H}}\ n_{H_2O}=\dfrac{6.n_{C_2H_6}+8.n_{C_3H_8}}{2}=3a+4b(mol)\\ \Rightarrow 3a+4b=0,07\ (2)\\ Từ \ (1);(2)\ \Rightarrow \begin{cases}a=0,01\\b=0,01\end{cases}\\ m_{C_2H_6}=0,01.30=0,3(g)\\ m_{C_3H_8}=0,01.44=0,44(g)\\ m_{\text{ankan}}=0,44+0,3=0,74(g)\\ \%m_{C_2H_6}=\dfrac{0,3}{0,74}.100\%=40,54\%\\ \%m_{C_3H_8}=\dfrac{0,44}{0,74}.100\%=59,46\%$

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