mtăng = manken = 11,2 (g)
Gọi \(\left\{{}\begin{matrix}n_{C_3H_8}=a\left(mol\right)\\n_{C_nH_{2n}}=b\left(mol\right)\end{matrix}\right.\)
=> \(a+b=\dfrac{11,2}{22,4}=0,5\) (1)
\(n_{CO_2}=\dfrac{38,08}{22,4}=1,7\left(mol\right)\)
Bảo toàn C: 3a + bn = 1,7 (2)
(1)(2) => bn - 3b = 0,2
=> \(b=\dfrac{0,2}{n-3}\left(n\ne3\right)\)
Có \(m_{anken}=14n.b=14n.\dfrac{0,2}{n-3}=11,2\left(g\right)\)
=> n = 4 (Tm)
=> CTPT: C4H8
b) \(\left\{{}\begin{matrix}n_{C_3H_8}=0,3\left(mol\right)\\n_{C_4H_8}=0,2\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\%m_{C_3H_8}=\dfrac{0,3.44}{0,3.44+0,2.56}.100\%=54,1\%\\\%m_{C_4H_8}=\dfrac{0,2.56}{0,3.44+0,2.56}.100\%=45,9\%\end{matrix}\right.\)
