Em tham khảo nha:

\(\begin{array}{l}
2)\\
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
{C_2}{H_4} + 3{O_2} \xrightarrow{t^0} 2C{O_2} + 2{H_2}O\\
{C_3}{H_8} + 5{O_2} \xrightarrow{t^0} 3C{O_2} + 4{H_2}O\\
2{C_4}{H_{10}} + 13{O_2} \xrightarrow{t^0} 8C{O_2} + 10{H_2}O\\
{C_5}{H_{12}} + 8{O_2} \xrightarrow{t^0} 5C{O_2} + 6{H_2}O\\
3)\\
ancol\,metylic:{C_2}{H_5}OH\\
andehitfomic:HCHO\\
benzen:{C_6}{H_6}\\
axetilen:CH \equiv CH\\
axitaxetic:C{H_3}COOH\\
etilen:C{H_2} = C{H_2}\\
4)\\
2C{H_4} \xrightarrow{t^0,xt} {C_2}{H_2} + 3{H_2}\\
{C_2}{H_2} + {H_2} \xrightarrow{t^0,xt} {C_2}{H_4}\\
{C_2}{H_4} + {H_2}O \xrightarrow{xt} {C_2}{H_5}OH\\
{C_2}{H_5}OH + CuO \xrightarrow{t^0} C{H_3}CHO + Cu + {H_2}O\\
5)\\
a)\\
{n_{C{O_2}}} = \dfrac{{1,568}}{{22,4}} = 0,07\,mol\\
{n_{{H_2}O}} = \dfrac{{0,9}}{{18}} = 0,05\,mol\\
{n_{C{O_2}}} > {n_{{H_2}O}} \Rightarrow hh\,ankin\\
{n_{hh}} = 0,07 - 0,05 = 0,02\,mol\\
\overline C  = \dfrac{{0,07}}{{0,02}} = 3,5 \Rightarrow CTPT:{C_3}{H_4},{C_4}{H_6}\\
{C_3}{H_4}:\\
CH \equiv C - C{H_3}:propin\\
{C_4}{H_6}:\\
CH \equiv C - C{H_2} - C{H_3}:but - 1 - in\\
C{H_3} - C \equiv C - C{H_3}:but - 2 - in\\
b)\\
m = {m_C} + {m_H} = 0,07 \times 12 + 0,05 \times 2 \times 1 = 0,94g\\
{n_{{O_2}}} = \dfrac{{0,07 \times 2 + 0,05}}{2} = 0,095\,mol\\
V = 0,095 \times 22,4 = 2,128l
\end{array}\)