$n_X=0,1875 mol\Rightarrow n_{2 hc}=0,1875-\frac{4,95}{32}=0,0328125 mol$$\Rightarrow \overline{M}_{2 hc}=39,6\Rightarrow X:\begin{cases}O_2:0,1875 mol \\C_2H_4:x mol\\C_4H_6:y mol \end{cases} $
$C_2H_4+3O_2\rightarrow 2CO_2+2H_2O$
$x 3x 2x 2x$
$C_4H_6+5,5O_2\rightarrow 4CO_2+3H_2O$
$y 5,5y 4y 3y$
$\Rightarrow \begin{cases}x+y=0,0328125 \\2x+4 y=\frac{4,18}{44} \end{cases}\Rightarrow \begin{cases}x=0,018125 mol \\ y=0,0146875 mol \end{cases}$
$\Rightarrow n_{Sau}=n_X+0,5y=0,19484375 mol$
$\frac{P_o.V_o}{T_o}=\frac{P_1.V_1}{T_1}$$\Rightarrow P_1=\frac{1.22,4.0,19484375.(273+136,5)}{273.8,4}=0,78 atm$
$\Rightarrow Chon:D$