$\%V_{CO_2}=100-30-30=40\%$
Giả sử có $1$ mol hỗn hợp $X$
$\to n_{N_xO}=1.30\%=0,3(mol); n_{SO_2}=1.30\%=0,3(mol); n_{CO_2}=1.40\%=0,4(mol)$
$\to m_X=0,3.(14x+16)+0,3.64+0,4.44=4,2x+41,6(g)$
$\%m_{N_xO}=19,651\%$
$\to\dfrac{0,3.(14x+16).100}{4,2x+41,6}=19,651$
$\to 4,2x+41,6=1,527(14x+16)$
$\to x=1$
Vậy CTHH khí $N_xO$ là $NO$
$m_X=4,2+41,6=45,8g$
$\to \overline{M}_X=\dfrac{45,8}{1}=45,8$
$\to d_{X/H_2}=\dfrac{45,8}{2}=22,9$