Ta có: \(\%m_O=\dfrac{16y}{14x+16y}.100\%=69,6\%\)
=> \(\dfrac{x}{y}=\dfrac{1}{2}\)
=> NxOy có là NxO2x hay (NO2)x
Giả sử có 1 mol khí
=> \(\left\{{}\begin{matrix}n_{NO}=0,45\left(mol\right)\\n_{NO_2}=0,15\left(mol\right)\\n_{\left(NO_2\right)_x}=0,4\left(mol\right)\end{matrix}\right.\)
=> \(\%m_{NO}=\dfrac{0,45.30}{0,45.30+0,15.46+0,4.46x}.100\%=23,6\%\)
=> \(x=2\)
=> NxOy là N2O4