Calculating Combustion Products: Propane, Oxygen, and Water Vapor

  • Thread starter Thread starter Paulsmith13
  • Start date Start date
  • Tags Tags
    Chemistry Exam
AI Thread Summary
To calculate the combustion products of propane (C3H8) with 60% pure oxygen, it is essential to set up a chart detailing the number of moles, molar mass, and mass. The equation n = m/M can be used to find the mass and number of moles, focusing on one mole of propane. It's important to consider the water vapor present since the propane is collected over water at SATP, which affects the calculations. The vapor pressure of water at STP must be accounted for to determine how much of the collected gas is actually propane versus water vapor. Accurate calculations will lead to a better understanding of the combustion process and its products.
Paulsmith13
Messages
1
Reaction score
0

Homework Statement


Incomplete combustion of C6H8, set up chart showing number of moles, molar mass and mass
the propane is burned with 60% pure oxygen, the propane is collected over water at SATP

Homework Equations



n=m/M

The Attempt at a Solution


How do I find the mass and number of moles if all I know is molar mass, and amount of propane and 60% pure oxygen?
 
Physics news on Phys.org
Prepare table for 1 mole of propane.
 
It might help with your mole calculations to denote propane as C3H8 rather than C6H8.

I assume the question is asking you to determine the combustion product(s) for a gas mixture that contains 40% propane (collected over water) and 60% oxygen? Do you think that it might be necessary to account for the water vapor in this experiment? Remember that the propane was collected over water at STP. What is the vapor pressure of water at STP and how much of the propane gas is actually water vapor?

Hope it is helpful...
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...

Similar threads

Back
Top