Em tham khảo nha:

\(\begin{array}{l}
3)\\
{n_{C{O_2}}} = \dfrac{{6,6}}{{44}} = 0,15\,mol \Rightarrow {n_C} = {n_{C{O_2}}} = 0,15\,mol\\
{n_{{H_2}O}} = \dfrac{{3,6}}{{18}} = 0,2\,mol \Rightarrow {n_H} = 2{n_{{H_2}O}} = 0,4\,mol\\
{m_O} = 3 - 0,15 \times 12 - 0,4 = 0,8g\\
{n_O} = \dfrac{{0,8}}{{16}} = 0,05\,mol\\
{n_C}:{n_H}:{n_O} = 0,15:0,4:0,05 = 3:8:1 \Rightarrow CTPT:{C_3}{H_8}O\\
4)\\
2{C_2}{H_2} + 5{O_2} \xrightarrow{t^0} 4C{O_2} + 2{H_2}O\\
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
hh:C{H_4}(a\,ml),{C_2}{H_2}(b\,ml)\\
\left\{ \begin{array}{l}
2a + 2,5b = 33,6\\
a + b = 14
\end{array} \right.\\
 \Rightarrow a = 2,8;b = 11,2\\
\% {V_{C{H_4}}} = \dfrac{{2,8}}{{14}} \times 100\%  = 20\% \\
\% {V_{{C_2}{H_2}}} = 100 - 20 = 80\% 
\end{array}\)