Ta có:
\(\left\{{}\begin{matrix}n_{CO2}=0,4\left(mol\right)\\n_{H2O}=0,56\left(mol\right)\end{matrix}\right.\Rightarrow n_{ancol}=n_{H2O}-n_{CO2}=0,56-0,4=0,16\left(mol\right)\)
\(\Rightarrow C=\frac{0,4}{0,56-0,4}=2,5\)
Vậy CT của 2 ancol là C2H5OH và C3H7OH
Gọi \(\left\{{}\begin{matrix}n_{C2H5OH}:x\left(mol\right)\\n_{C3H7OH}:y\left(mol\right)\end{matrix}\right.\)
Giải hệ PT:
\(\left\{{}\begin{matrix}2x+3y=0,4\\x+y=0,16\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,08\\y=0,08\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{C2H5OH}=\frac{0,08.46}{0,08.\left(46+60\right)}.100\%=43,40\%\\\%m_{C3H7OH}=100\%-43,40\%=56,60\%\end{matrix}\right.\)
