Em tham khảo nha :
\(\begin{array}{l}
a)\\
{n_{{O_2}}} = \dfrac{{7,84}}{{22,4}} = 0,35mol\\
{n_{{H_2}O}} = \dfrac{{5,4}}{{18}} = 0,3mol\\
BT\,Oxi:\\
2{n_{{O_2}}} = 2{n_{C{O_2}}} + {n_{{H_2}O}}\\
\Rightarrow {n_{C{O_2}}} = \dfrac{{2 \times 0,35 - 0,3}}{2} = 0,2mol\\
{n_{C{O_2}}} < {n_{{H_2}O}} \Rightarrow \text{ X thuộc dãy đồng đẳng của Ankan}\\
{n_X} = {n_{{H_2}O}} - {n_{C{O_2}}} = 0,3 - 0,2 = 0,1mol\\
CTTQ:{C_n}{H_{2n + 2}}(n \ge 1)\\
n = \dfrac{{{n_{C{O_2}}}}}{{{n_X}}} = \frac{{0,2}}{{0,1}} = 2\\
\Rightarrow CTPT:{C_2}{H_6}(etan)\\
b)\\
C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O\\
2{C_2}{H_6} + 7{O_2} \to 4C{O_2} + 6{H_2}O\\
{n_{hh}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
{n_{C{O_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4mol\\
hh:C{H_4}(a\,mol),{C_2}{H_6}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,3\\
a + 2b = 0,4
\end{array} \right.\\
\Rightarrow a = 0,2;b = 0,1\\
\% {V_{C{H_4}}} = \dfrac{{0,2 \times 22,4}}{{6,72}} \times 100\% = 66,7\% \\
\% {V_{{C_2}{H_6}}} = 100 - 66,7 = 33,3\%
\end{array}\)
