Answer :
[tex]2NO_{(g)} + O_{2}_{(g)} -----\ \textgreater \ 2NO _{2}_{(g)} [/tex]
moles of NO₂ = [tex] \frac{Mass}{Mr} [/tex]
= [tex] \frac{46.0 g}{((14 * 1) (16 * 2))} [/tex]
= 1 mol
ratio of NO₂ : O₂ = 1 : 2
∴ since mole of NO₂ = 1 mol
the mole of O₂ = [tex] \frac{1}{2} [/tex] mol
= 0.5 mol
moles of NO₂ = [tex] \frac{Mass}{Mr} [/tex]
= [tex] \frac{46.0 g}{((14 * 1) (16 * 2))} [/tex]
= 1 mol
ratio of NO₂ : O₂ = 1 : 2
∴ since mole of NO₂ = 1 mol
the mole of O₂ = [tex] \frac{1}{2} [/tex] mol
= 0.5 mol