CHEM QUESTION!!?

As K is large, the reaction is almost to the right.

4NO + 3O₂ ⇌ 2N₂O₅ …… K = 7.25 × 10⁶
Mole ratio NO : O₂ = 4 : 3, and thus more NO is needed than O₂.
But now, (Initial NO concentration) < (Initial O₂ concentration)
Hence, NO almost completely reacts.
[NO] reacted ≈ 2.5 M
Hence, [O]₂ reacted = (2.5 M) × (3/4) = 1.875 M
and [N₂O₅] produced = (2.5 M) × (2/4) = 1.25 M

_____________ 4NO ____ + ____ 3O₂ ____ ⇌ ____ 2N₂O₅ _____ K = 7.25 × 10⁶
Initial: _______ 2.5 M _________ 3.5 M __________ 4.5 M
Change: ____ ≈ -2.5 M ______ - 1.875 M ________ +1.25 M
Eqm: _________ y M ___ _____ 1.625 M _________ 5.75 M

At eqm: K = [N₂O₅]² / ([NO]⁴ [O₂]³)
K = [N₂O₅]² / ([NO]⁴ [O₂]³)
7.25 × 10⁶ = 5.75² / {y⁴ × 1.625³}
y⁴ = 5.75² / [1.625³ × (7.25 × 10⁶)]
y = 0.0321

At eqm:
[NO] = 0.0321 M
[O₂] = 1.63 M
[N₂O₅] = 5.75 M