N2(g) + 2O2(g) --> N2O4(g)?

(A)
Change in [N₂] = -(5.9 × 10⁻³ mol) / (500/1000 L) = -1.18 × 10⁻² mol/L
(The negative sign indicates that N₂ is consumed.)


(B)
N₂(g) + 2O₂(g) ⇌ N₂O₄(g)
Mole ratio N₂ : O₂ : N₂O₄ = 1 : 2 : 1

Change in [O₂] = -(1.18 × 10⁻² M) × (2/1) = -2.36 × 10⁻² mol/L
Change in [N₂O] = +(1.18 × 10⁻² M) × (1/1) = +1.18 × 10⁻² mol/L


(C)
Initial [N₂] = (2.0 × 10⁻² mol) / (500/1000 L) = 4.0 × 10⁻² mol/L
Initial [O₂] = (2.0 × 10⁻² mol) / (500/1000 L) = 4.0 × 10⁻² mol/L
Initial [N₂O₄] = 0 mol/L

[N₂] at equilibrium = [(4.0 × 10⁻²) - (1.18 × 10⁻²)] mol/L = 2.92 × 10⁻² mol/L
[O₂] at equilibrium = [(4.0 × 10⁻²) - (2.36 × 10⁻²)] mol/L = 1.84 × 10⁻² mol/L
[N₂O] at equilibrium = [0 + (1.18 × 10⁻²)] mol/L = 1.18 × 10⁻² mol/L

(A)
You're told the change in concentration of N2, so supposing the question meant to ask about the concentration of N2 after the equilibrium is established:
(2.0 x 10^-2 mol of N2 initially) - (5.9 x 10^-3 mol N2 reacted) = 0.0141 mol N2 remains

(B)
(5.9 x 10^-3 mol N2 reacted) x (2 mol O2 / 1 mol N2) = 0.0118 mol O2 reacted (so maybe the answer should be -0.0118 mol)

(C)
(5.9 x 10^-3 mol N2 reacted) x (1 mol N2O4 / 1 mol N2) = 0.0059 mol N2O produced
(2.0 x 10^-2 mol O2 initially) - (0.0118 mol O2 reacted) = 0.0082 mol O2 remains
0.0141 mol N2 remains

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