Em tham khảo nha :
\(\begin{array}{l}
a)\\
{n_{C{O_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
CTPT:{C_n}{H_{2n - 2}}\\
{n_X} = \dfrac{{{n_{C{O_2}}}}}{n} = \dfrac{{0,2}}{n}\\
{M_X} = 2,7:\dfrac{{0,2}}{n} = 13,5n(g/mol)\\
\Rightarrow 14n - 2 = 13,5n\\
\Rightarrow n = 4\\
\Rightarrow CTPT:{C_4}{H_6}\\
b)\\
{C_4}{H_6} + \dfrac{{11}}{2}{O_2} \to 4C{O_2} + 3{H_2}O\\
{n_{{O_2}}} = \dfrac{{11}}{8}{n_{C{O_2}}} = 0,275mol\\
{V_{{O_2}}} = 0,275 \times 22,4 = 6,16l
\end{array}\)