What is the Correct Volume of the Reaction Mixture at STP After Completion?

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SUMMARY

The discussion centers on calculating the volume of the reaction mixture at STP after the reaction of 400 mL of NO with 500 mL of O2. The initial calculation incorrectly summed the volumes to 900 mL without accounting for the stoichiometry of the reaction. The correct volume of the product NO2 is determined to be 700 mL, as the reaction consumes the reactants in a specific molar ratio. The key to solving this problem lies in applying stoichiometric principles to the gas volumes involved.

PREREQUISITES
  • Understanding of gas laws, specifically the Ideal Gas Law (PV=nRT)
  • Knowledge of stoichiometry in chemical reactions
  • Familiarity with Standard Temperature and Pressure (STP) conditions
  • Basic skills in unit conversion between mL and L
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  • Study stoichiometric calculations for gas reactions
  • Learn about the Ideal Gas Law and its applications in chemistry
  • Explore the concept of molar volume of gases at STP
  • Practice problems involving gas mixtures and reactions
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Chemistry students, educators, and anyone involved in gas phase reaction calculations will benefit from this discussion, particularly those focusing on stoichiometry and gas laws.

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Homework Statement


Consider the following gas phase reaction:
http://scholar.uh.edu/webct/RelativeResourceManager/Template/Imported_Resources/20053H_CHEM1331_08946_QIZ_200511041405291131475017966%20folder/QIZ_6442871_M/my_files/test_images/f5g1q49g1.gif
400. mL of NO at STP is reacted with 500. mL of O2 at STP. Calculate the volume of the reaction mixture at STP after the reaction goes to completion.

V of NO= 400.mL= .400L
V of O2= 500.mL = .500L
At STP:
P= 1atm
T= 273K
V= 22.4L/mol of gas
V of mixed gas NO2= ?

Homework Equations


n=PV/RT;
M=mRT/PV;


The Attempt at a Solution


1, I added two volumes of gases together to get the total volume.
.400L+.500L= .900L

2. I calculated the mol of NO2 at STP
n= PV/RT= (1atm*22.4L)/(0.0821L*atm/K*mol)
n= 0.9994 mol of NO2

3. Took that mol multiply with .9L to get the Volume of the mixed gas NO2, I got ~900mL
But the right answer is 700mL. Is there any idea on what I did wrong?
 
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The gases reacted. You end with two gases - but you need to calculate their amounts with simple stoichiometry.
 
Thank you! I'll try again!
 
Last edited:

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