Ta có:
\(\left\{{}\begin{matrix}n_{CO2}=\frac{11,2}{22,4}=0,5\left(mol\right)\\n_{H2O}=\frac{5,4}{18}=0,3\left(mol\right)\end{matrix}\right.\)
\(n_{hh}=n_{CO2}-n_{H2O}=0,5-0,3=0,2\left(mol\right)\)
\(\overline{C}=\frac{n_{CO2}}{n_{hh}}=\frac{0,5}{0,2}=2,5\)
Vậy CTPT là C2H2 và C3H4
Gọi \(\left\{{}\begin{matrix}n_{C2H2}:a\left(mol\right)\\n_{C3H4}:b\left(mol\right)\end{matrix}\right.\)
Giải hệ PT:
\(\left\{{}\begin{matrix}2a+3b=0,5\\a+2b=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{C2H2}=\frac{0,1.26}{0,1.26+0,1.40}.100\%=39,39\%\\\%m_{C3H4}=100\%-39,39\%=60,61\%\end{matrix}\right.\)
