C2H6 + \(\frac{7}{2}\)O2 \(\underrightarrow{to}\) 2CO2 + 3H2O (1)
CnH2n + \(\frac{3n}{2}\)O2 \(\underrightarrow{to}\) nCO2 + nH2O (2)
\(n_{CO_2}=\frac{13,44}{22,4}=0,6\left(mol\right)\)
\(n_{H_2O}=\frac{12,6}{18}=0,7\left(mol\right)\)
Gọi x,y lần lượt là số mol của C2H6 và CnH2n
Theo Pt1: \(n_{CO_2}=2n_{C_2H_6}=2x\left(mol\right)\)
Theo Pt2: \(n_{CO_2}=nn_{C_nH_{2n}}=ny\left(mol\right)\)
Ta có: \(23+ny=0,6\) (*)
Theo Pt1: \(n_{H_2O}=3n_{C_2H_6}=3x\left(mol\right)\)
Theo PT2: \(n_{H_2O}=nn_{C_nH_{2n}}=ny\left(mol\right)\)
Ta có: \(3x+ny=0,7\) (**)
Từ (*)(**) ta có: \(\left\{{}\begin{matrix}2x+ny=0,6\\3x+ny=0,7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\ny=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=\frac{0,4}{n}\end{matrix}\right.\)
Vậy \(n_{C_2H_6}=0,1\left(mol\right)\Rightarrow m_{C_2H_6}=0,1\times30=3\left(g\right)\)
\(n_{C_nH_{2n}}=\frac{0,4}{n}\left(mol\right)\Rightarrow m_{C_nH_{2n}}=14n\times\frac{0,4}{n}=\frac{5,6n}{n}\left(g\right)\)
Ta có: \(M_{hh}=\frac{m_{hh}}{n_{hh}}=\frac{8,6}{x+y}=\frac{8,6}{0,1+\frac{0,4}{n}}=\frac{30+14n}{2}\)
\(\Rightarrow17,2=8,6+1,4n+\frac{12}{n}\)
\(\Leftrightarrow1,4n+\frac{12}{n}-8,6=0\)
\(\Leftrightarrow1,4n^2-8,6n+12=0\)
\(\Leftrightarrow\left[{}\begin{matrix}n=4\\n=\frac{3}{1,4}\left(loại\right)\end{matrix}\right.\)
Vậy CTHH là C4H8