PT: \(C_4H_{10}+\dfrac{13}{2}O_2\underrightarrow{t^o}4CO_2+5H_2O\)
\(C_6H_{14}+\dfrac{19}{2}O_2\underrightarrow{t^o}6CO_2+7H_2O\)
Ta có: \(n_{C_4H_{10}}+n_{C_6H_{14}}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\left(1\right)\)
Theo PT: \(n_{CO_2}=4n_{C_4H_{10}}+6n_{C_6H_{14}}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{C_4H_{10}}=0,025\left(mol\right)\\n_{C_6H_{14}}=0,05\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{hh}=m_{C_4H_{10}}+m_{C_6H_{14}}=5,75\left(g\right)\)
\(\left\{{}\begin{matrix}\%n_{C_4H_{10}}=\dfrac{0,025}{0,025+0,05}.100\%\approx33,33\%\\\%n_{C_6H_{14}}\approx66,67\%\end{matrix}\right.\)
