Em tham khảo nha :
\(\begin{array}{l}
a)\\
{C_n}{H_{2n + 2}} + \dfrac{{3n + 1}}{2}{O_2} \to nC{O_2} + (n + 1){H_2}O\\
{n_{C{O_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25mol\\
{n_{{H_2}O}} = \dfrac{{6,3}}{{18}} = 0,35mol\\
{n_{C{O_2}}} = \dfrac{{n{n_{{H_2}O}}}}{{n + 1}}\\
\Rightarrow 0,25 = \dfrac{n}{{n + 1}} \times 0,35\\
\Rightarrow n = 2,5\\
\Rightarrow CTPT:{C_2}{H_6},{C_3}{H_8}\\
b)\\
{C_3}{H_8} + 5{O_2} \to 3C{O_2} + 4{H_2}O\\
2{C_2}{H_6} + 7{O_2} \to 4C{O_2} + 6{H_2}O\\
hh:{C_3}{H_8}(a\,mol),{C_2}{H_6}(b\,mol)\\
3a + 2b = 0,25(1)\\
4a + 3b = 0,35(2)\\
\text{Từ (1) và (2)}\Rightarrow a = 0,05;b = 0,05\\
\% {V_{{C_3}{H_8}}} = \dfrac{{0,05}}{{0,05 + 0,05}} \times 100\% = 50\% \\
\% {V_{{C_2}H{ 6}}} = 100 - 50 = 50\%
\end{array}\)
