When 62.7 g nitrogen and excess oxygen react they generate nitrogen dioxide. if the no2 is collected at 625 k and 0.724 atm, what volume will it occupy?

Okay, so this problem has a few layers to it, but I will do my best to walk you through it.

Firstly, the equation we will want to use for this problem is our trusty ‘ol PV = nRT. In its original form, however, it,s not going to be as useful for us. As you may know, we can manipulate this equation for the different types of problems we are trying to solve. In this particular problem, we are trying to find the final volume the NO2 will occupy. This requires a very simple manipulation. The only thing we need to do is divide each side of the equation by dividing each side by P. By doing this, we will get:

V = nRT/P

Our next step is to start by looking at all of our values we have to work with:
- 62.7 g N2 reacts with excess O2
- The NO2 is collected at a temperature of 625K
- The NO2 is also collected at pressure of 0.724 atm
- Our gas constant of 0.0806 L*atm/K*mol

The only thing we need to do is convert our first value into moles of gas so that we can use it for our n value:

(62.7 g N2)/(28.01344 g N2/mol)* (2 mol NO2)/(1 mol N2) = 4.4764 mol NO2

Now, we can use all of our values and simply plug them into our equation:

V = [(4.4764 mol) * (0.08206 Latm/Kmol) * (625K) / (0.724K)
V = 317 L NO2

The volume occupied by the NO2 gas is 317 L.

I’m hoping this makes sense to you! There are usually several levels to these types of problems, so it’s best to keep your work organized and always double check your math and units. Good luck!

257 grams h2o i think