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

Em tham khảo nha :

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