CTTQankan:CnH2n+2
CTTQanken:CmH2m
\(n_{ankan}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
\(\Rightarrow M_{ankan}=\frac{3}{0,1}=30\left(\frac{g}{mol}\right)\)
\(14n+2=30\Rightarrow n=2\)
Vậy ankan là C2H6
\(n_{hh}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
\(\Rightarrow n_{anken}=0,25-0,1=0,15\left(mol\right)\)
\(M_{anken}=\frac{6,3}{0,15}=42\left(\frac{g}{mol}\right)\)
\(14m=42\Rightarrow m=3\)
Vậy anken là C3H6