Find Mass in Equilibrium mixture?

N₂O₄(g) ⇄ 2 NO₂(g)

So the equilibrium concentrations (in mol/L) are given by:

Kc = [NO₂]² / [N₂O₄]

with Kc = 0.133

You could solve the problem step by step.In first step calculate concentrations when equilibrium has established after 50g N₂O₄ have been introduced to the vessel. Then shift NO₂ concentration due to addition of the 5g, and compute the concentrations when equilibrium has reestablishes. Actually you get the same result if you introduce 50g N₂O₄ and 5g NO₂ at the same time to the vessel.So i would treat the problem in that way, because i takes less mathematical effort to solve.

Initially you got pure dinitrogen tetroxide at a concentration of

[NO₂]₀ = m(NO₂) / (M(NO₂)·V)

= 5g / (46g/mol · 2L)

= 5.435×10⁻² mol/L

[N₂O₄]₀ = m(N₂O₄) / (M(N₂O₄)·V)

= 50g / (92g/mol · 2L)

= 2.7174×10⁻¹ mol/L

Set up ICE table:

............... [N₂O₄]............... [NO₂]

I........... 2.7174×10⁻¹........ 5.435×10⁻²

C.............. - x................... + 2·x

E....... 2.7174×10⁻¹ -x..... 5.435×10⁻² + 2·x

Substitute the expressions to equilibrium equation

Kc = [NO₂]² / [N₂O₄]

<=>

0.133 = (5.435×10⁻² + 2·x)² / (2.7174×10⁻¹ - x)

<=>

0.133·(2.7174×10⁻¹ - x) = (5.435×10⁻²)² + 4·5.435×10⁻²·x + 4·x²

you can rewrite this as simple quadratic equation of the form

a·x² + b·x + c = 0

with

a = 4

b = 4·5.435×10⁻² + 0.133 = 0.35039

c = (5.435×10⁻²)² - 0.133·2.7174×10⁻¹ = -0.033188

Solutions are

x = (-b ± √(b² - 4·a·c) ) / (2·a)

=>

x = 0.05727

or

x = -0.1449

Second solution is infeasible, because it would to a negative equilibrium concentration of NO₂.

The final concentration od dinitrogen tetroxide is

[N₂O₄] = 0.27174 - 0.05727 = 0.21447 mol/L

So the mass of N₂O₄ in final equilibrium mixture is:

m(N₂O₄) = [N₂O₄] · M(N₂O₄) · V

= 0.21447 mol/L · 92g/mol · 2L

= 39.46g