[calmwater3000]

Hi,
I have been trying to figure this problem out myself and I'm not getting the same answer as the one in my book.

The question:

A 2.50 mole quantity of NOCl was initially in a 1.50-L reaction chamber at 400 Celsius. After equilibrium was established, it was found that 28.0 percent of the NOCl had dissociated:

2NOCl(gas) <--> 2NO(gas) + Cl₂(gas)

The equation would be- K= [NO]^2[Cl]/[NOCl]^2

2.50 / 1.50 L= 1.67 m/L of initial concentration of NOCl

1.67 m/L * .28 = 0.467 , 1.67 - 0.467 = 1.2 m/L amount of concentration at equilibrium for NOCl,

My question is how do I calculate the equilibrium concentration for NO and Cl₂?

[Borek]

Stoichiometry.

[calmwater3000]

I don't get how stochiometry apply to this,
So 1.2 m/L of NOCl multiply by 2 = 2.4 m/L of NOCl, NO2 would also be 2.4 m/L and Cl₂ would be 1.2 m/L?

If I work this out, K= [2.4^2][1.2]/[2.4^2] then K would be 1.2 which wouldn't be the right answer, the text book have an answer of 3.53x10^-2

[Borek]

Take a look at the reaction equation. You know that 28% of 2.5 moles NOCl dissociated - how many moles of NO were produced?

[calmwater3000]

Ok, I know what you meant now. I got the same answer as the one in the textbook, thanks!