a,

$V_2-V_1=3,36(l)$

$\Leftrightarrow 22,4n_2-22,4n_1=3,36$

$\Leftrightarrow n_2-n_1=0,15(mol)$

Đặt $n_1=x (mol)\Rightarrow n_2=0,15+x (mol)$

TN1:

$m_{hh}=8,4g$

$\Rightarrow x.M_Y+16(0,15+x)=8,4$

$\Leftrightarrow x.M_Y+16x+2,4=8,4$ $(1)$

TN2:

$m_{hh}=12,6g$

$\Rightarrow (0,15+x)M_Y+16x=12,6$

$\Leftrightarrow 0,15M_Y+x.M_Y+16x=12,6$             $(2)$

$(2)-(1)$: $0,15M_Y-2,4=12,6-8,4$

$\Leftrightarrow M_Y=44$

$\Rightarrow M_X+16.2=44$

$\Leftrightarrow M_X=12(C)$

$\to CO_2$

b,

Giả sử trộn $2$ mol $CO_2$ với $5$ mol $CH_4$

$\overline{M}_A=\dfrac{2.44+5.16}{2+5}=24$

$\overline{M}_A<29\to A$ nhẹ hơn không khí.