Giải hộ ạ...để mình so với đáp án mình làm

Em tham khảo nha :

\(\begin{array}{l}
a)\\
2{C_x}{H_y} + \dfrac{{4x + y}}{2}{O_2} \to 2xC{O_2} + y{H_2}O\\
{n_{C{O_2}}} = \dfrac{{22}}{{44}} = 0,5mol\\
{n_{{H_2}O}} = \dfrac{{10,8}}{{18}} = 0,6mol\\
{n_A} = \dfrac{{7,2}}{{72}} = 0,1mol\\
{n_C} = {n_{C{O_2}}} = 0,5mol\\
{n_H} = 2{n_{{H_2}O}} = 1,2mol\\
x = \dfrac{{{n_C}}}{{{n_A}}} = \dfrac{{0,5}}{{0,1}} = 5\\
y = \dfrac{{{n_H}}}{{{n_A}}} = \dfrac{{1,2}}{{0,1}} = 12\\
 \Rightarrow CTPT\,A:{C_5}{H_{12}}\\
b)\\
C{H_3} - C{H_2} - C{H_2} - C{H_2} - C{H_3}\\
C{H_3} - CH(C{H_3}) - C{H_2} - C{H_3}\\
C{H_3} - C{(C{H_3})_2} - C{H_3}\\
c)\\
C{l_2} + {C_5}{H_{12}} \to {C_5}{H_{11}}Cl + HCl
\end{array}\)