[satha13]

The problem is:

NO(g) and O2(g) are separated in two separated containers, when the stopcock between two containers is open the gas reaction is as the following:
2NO (g) + O2(g) ------> 2NO2 (g)
If the temperature is kept stable at 25 C. Identify the remained gas in the containers and caculate pressure of them.

The solution is  Pressure of O2 = 0.167 atm  and Pressure of NO2 is 0.333 atm

But I think that

1. we don't know Volume and pressure in each container before opening stoppock
2. So we can't calculate mol of NO and O2 exactly
3. Then, we cannot use mole ratio in equation to calculate limiting reagent and product amount

In my opinion, this question is incomplete, but why they can calculate  Pressure of O2 and NO2 ??

I try to make assumption by give V = 1 dm3 and P = 1 atm  but the result is not the same as answer.... Can any expert in Gas calculation help me about this?

Satha

[Borek]

Not enough data to solve the question as given.

Have you copied the entire question?

[satha13]

Nitric Oxide (NO) reacts with molecular oxygen (O2) as follows :

2NO (Gas) + O2 (Gas) --> 2NO2 (gas)

Initially NO and O2 are seperated, when the valve is opened, the reaction quickely goes to completion. Determine what gases remain at the end and calculate their partial pressures. Assume that the temp. remains constant at 25 Celcius.
The answer is PO2 = 0.167 and PNO2 = 0.333

This is a original Question....

But I cannot calculate the answer to be the same as above? 
Can anyone give me a good explanation?

Satha

[sdekivit]

the only assumption you can make is that every gas at 25 degrees C has a volume of 24,5 L / mol

[satha13]

IS this methode collect?

1. Assume P = 1 atm and volume for each container = V
2. We get mol O2 and mol NO = V / RT
from 2NO  +  O2  --------> 2NO2  mole ration of NO:O2 = 2:1
So the limiting reagent is NO and after completeion
Mole NO2 = V / RT
Mole O2 = V / 2RT

We get NO2 = 1 mole and O2 = 0.5 mole
The total mole is 1.5 mole

From the Partial presuure  P_O2 = X_O2 x P_total
we get mole fraction of O2 = 0.5 / 1.5  = 0.333
but for Ptotal , when the stopcock is openesd, V double and total pressure is 1/2

So PO2 = 0.333 x 1/2 = 0.167 atm
and PNO2 = 0.666 x 1/2 = 0.300 atm

This will give the same answer in the book, but I am not sure that We can use the relationship in Boyle' law between P and V or not? because mole is not constant, If yes, I assume that answer in the book (PO2 = 0.167, PNO2 = 0.333) is absolutely wrong....

Can Any expert in chemistry tell me about above method that it is correct or not?

[Borek]

You have made several assumptions that were not given in the question - identical pressures&volumes of both containers, 1 atm pressure and reaction goes to completion. While that seems reasonable, that's not in the question.

Sorry, no more comments nor results checking from me ATM, I am leaving Warsaw tomorrow and I have packed schedule