Đặt \(n_{H_2O}=n_{CO_2}=a\left(mol\right)\)
\(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\Rightarrow m_{O_2}=0,4.32=12,8\left(g\right)\)
Theo ĐLBTKL: \(m_{CO_2}+m_{H_2O}=5,8+12,8=18,6\left(g\right)\)
=> 44a + 18a = 18,6
=> a = 0,3 (mol)
Theo ĐLBTNT: \(\left\{{}\begin{matrix}n_C=n_{CO_2}=0,3\left(mol\right)\\n_H=2n_{H_2O}=0,6\left(mol\right)\\n_{O\left(X\right)}=2n_{CO_2}+n_{H_2O}-2n_{O_2}=0,1\left(mol\right)\end{matrix}\right.\)
Ta có: \(n_C:n_H:n_O=0,3:0,6:0,1=3:6:1\)
=> CTĐGN của X là `C_3H_6O`