- Xét TN1:
Gọi \(\left\{{}\begin{matrix}n_{CH_4}=a\left(mol\right)\\n_{C_2H_2}=b\left(mol\right)\\C_3H_6=c\left(mol\right)\end{matrix}\right.\)
=> 16a + 26b + 42c = 1,1 (1)
\(n_{CO_2}=\dfrac{3,52}{44}=0,08\left(mol\right)\)
Bảo toàn C: a + 2b + 3c = 0,08 (2)
- Xét TN2:
Gọi \(\left\{{}\begin{matrix}n_{CH_4}=ak\left(mol\right)\\n_{C_2H_2}=bk\left(mol\right)\\n_{C_3H_6}=ck\left(mol\right)\end{matrix}\right.\)
=> \(ak+bk+ck=\dfrac{0,448}{22,4}=0,02\left(mol\right)\) (*)
\(n_{Br_2}=\dfrac{4}{160}=0,025\left(mol\right)\)
Bảo toàn liên kết \(\pi\): 2bk + ck = 0,025 (**)
(*)(**) => \(\dfrac{a+b+c}{2b+c}=\dfrac{4}{5}\)
=> 5a - 3b + c = 0 (3)
(1)(2)(3) => a = 0,01 (mol); b = 0,02 (mol); c = 0,01 (mol)
\(\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,01.16}{1,1}.100\%=14,55\%\\\%m_{C_2H_2}=\dfrac{0,02.26}{1,1}.100\%=47,27\%\\\%m_{C_3H_6}=\dfrac{0,01.42}{1,1}.100\%=38,18\%\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,01}{0,01+0,02+0,01}.100\%=25\%\\\%V_{C_2H_2}=\dfrac{0,02}{0,01+0,02+0,01}.100\%=50\%\\\%V_{C_3H_6}=\dfrac{0,01}{0,01+0,02+0,01}.100\%=25\%\end{matrix}\right.\)
