Question

Answer 1

\(n_{H_2O}=\dfrac{5,4}{18}=0,3\left(mol\right)\\ n_{CO_2}=\dfrac{13,2}{44}=0,3\left(mol\right)\\ V\text{ì}:n_{H_2O}=n_{CO_2}\\ \Rightarrow A:C_nH_{2n}\left(n\ge2\right)\\ M_{C_nH_{2n}}=28\\ \Leftrightarrow14n=28\\ \Leftrightarrow n=2\\ \Rightarrow CTPT.A:C_2H_4\)