Đáp án+Giải thích các bước giải:
$\rm n_{Ankan}=\dfrac{11,2}{22,4}=0,5(mol)\\\rightarrow M_{Ankan}=\dfrac{24,8}{0,5}=49,6g/mol\\CTPT:C_nH_{2n+2}\\\rightarrow 14n+2=49,6\\\rightarrow n={3,4}\\\rightarrow \text{2 Ankan} :C_3H_8,C_4H_{10}\\n_{C_3H_8}=a(mol),n_{C_4H_{10}}=b(mol)\\\rightarrow 44a+58b=24,8(1)\\∑n_{Ankan}=a+b=0,5(2)\\(1),(2)\rightarrow a=0,3,b=0,2\\\rightarrow m_{C_3H_8}=0,3.44=13,2g\\\%m_{C_3H_8}=\dfrac{13,2}{24,8}.100\%=53,22\%\\\%m_{C_4H_{10}}=100\%-53,22\%=46,78\%$