1.
\(n_{anken}=\frac{168:1000}{22,4}=0,0075\left(mol\right)\)
\(\Rightarrow\overline{M_{anken}}=\frac{0,345}{0,0075}=46=14\overline{n}\)
\(\Rightarrow\overline{n}=3,2\) [C3H6(a mol); C4H8 (b mol) ]
2.
Ta có:
\(\left\{{}\begin{matrix}a+b=0,0075\\42a+56b=0,345\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\frac{3}{560}\\b=\frac{3}{1400}\end{matrix}\right.\)
\(\%V_{C3H6}=\frac{\frac{3}{560}.100}{0,0075}=71,43\%\)
\(\%V_{C4H8}=100\%-71,43\%=28,57\%\)
3.\(\%m_{C3H6}=\frac{\frac{3}{560}.42.100}{0,345}=65,22\%\)
\(\%m_{C4H8}=100\%-65,22\%=34,78\%\)