\(C_xH_y+\left(x+\dfrac{y}{4}\right)O_2\rightarrow xCO_2+\dfrac{y}{2}H_2O\)
0,1----------------------------0,1x---------\(0,1.\dfrac{y}{2}\)
\(n_{hidrocacbon}=\dfrac{2,24}{22,4}=0,1mol\); \(n_{CO_2}=0,3mol;n_{H_2O}=0,3mol\)
\(\Rightarrow0,1x=0,3\rightarrow x=3\)
\(0,1\dfrac{y}{2}=0,3\rightarrow y=6\)
Vậy chất hữu cơ cần tìm là \(C_3H_6\)