Partial pressure and moles problem?

In summary, the problem involves a combustion reaction between carbon monoxide and oxygen at 300 K. The balanced equation is 2CO + O2 → 2CO2. Before the reaction begins, there are 1 mole of O2 and 2 moles of CO present. When the reaction goes to completion, 2 moles of CO2 remain. The partial pressure of O2 will be 0 as it is the limiting reagent, while the partial pressure of CO2 will be 1 atm. There is not enough information provided to determine the partial pressure of CO.
  • #1
gh_pluvilias
8
0

Homework Statement


Here is the problem: Carbon monoxide and oxygen are kept at exactly 300 K in separate chambers of the apparatus shown below. When the stopcock is opened, the combustion reaction begins and is allowed to proceed to completion, while being maintained at exactly 300 K.


Homework Equations


PV=nRT
I believe the balanced equation is: 2CO + O2 → 2CO2

The Attempt at a Solution


I have 3 questions to answer regarding it. I have typed up what I think are the answer or what I have so far. If you can help me answer these it would be greatly appreciated.

1. How many moles of each reactant are present before the reaction begins?
2 moles of CO2 react with 1 mole of O2 in this reaction.

2. How many moles of each gaseous reactant and product are present when the reaction goes to completion? Assume the connecting tube has negligible volume.
After the reaction is complete, 2 moles of CO2 remain (equal amount of starting CO).

3. What is the partial pressure of each gas when the reaction goes to completion? Assume the connecting tube has negligible volume.
CO2:O2 ratio = 2:1 For NO, 0.5*2/(0.082*300) = 0.041 mol of NO
For O2, 1*1/(0.082*300) = 0.041 mol of O2, 0.02 corrected for ratio.
O2 is the limiting reagent (there won't be any left), so pressure due to O2=0, mass of O2=0.
Not sure what to do from here. =/
 
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  • #2
gh_pluvilias said:
1. How many moles of each reactant are present before the reaction begins?
2 moles of CO2 react with 1 mole of O2 in this reaction.

The answer is not related to the question. And the data you have listed so far is not enough to answer the question, could be there was additional data on the image.

For NO, 0.5*2/(0.082*300) = 0.041 mol of NO

Where did you got nitrogen from? You started with CO and O2 only.
 
  • #3
Oh sorry... it was late. Must've have just typed NO instead of CO. lol Guess I get my oxides mixed up.

There is an image attached, but I thought I had included it in my attempts. Sorry. There's 1L of O2 at 1atm and 2L of CO at 0.5atm. And that is all the information I'm given.
 

FAQ: Partial pressure and moles problem?

What is partial pressure?

Partial pressure is the pressure exerted by an individual gas in a mixture of gases. It is the pressure that the gas would exert if it occupied the same volume as the entire mixture at the same temperature.

How do you calculate partial pressure?

Partial pressure can be calculated by multiplying the total pressure of the gas mixture by the mole fraction of the gas in question. The mole fraction is the ratio of the number of moles of the gas to the total number of moles in the mixture.

3. What is the ideal gas law?

The ideal gas law is a fundamental equation in chemistry that relates the properties of gases (pressure, volume, temperature, and number of moles) to each other. It is expressed as PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.

4. How do you convert between moles and pressure?

To convert between moles and pressure, you can use the ideal gas law. Rearrange the equation to solve for pressure (P = nRT/V) and plug in the given values for the number of moles, gas constant, and volume. This will give you the pressure in units of Pascals (Pa). You can then convert to other units, such as atmospheres or millimeters of mercury, if needed.

5. What is the relationship between partial pressure and mole fraction?

The relationship between partial pressure and mole fraction is directly proportional. This means that as the mole fraction of a gas increases, its partial pressure also increases. This relationship is described by Dalton's Law of Partial Pressures, which states that the total pressure of a gas mixture is equal to the sum of the partial pressures of each individual gas in the mixture.

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