Reread my post. You have not figured the moles of propane, the first thing I suggested.
moles propane=masspropane/molmassPropane
=desity*volume/44
I am still not quite grasping how to complete this problem:
Posted by K on Sunday, October 14, 2007 at 6:43pm.
Many home barbeques are fueled with propane gas C3H8.
What mass of carbon dioxide (in kg) is produced upon the complete combustion of 18.9L of propane (approximate contents of one 5-gallon tank)? Assume that the density of the liquid propane in the tank is 0.621g/ml.
How do I start & work this question?
I wrote a balanced equation:
C3H8 + 5O2 -> 4H2O + 3CO2
Now, where do I go from here?
For Further Reading
CHEM - bobpursley, Sunday, October 14, 2007 at 7:08pm
The propane is liquid. Determine the mass from the volume and liquid. Convert that mass to moles.
For each mole of propane, you get (from the balanced equation) three times as much CO2. Convert moles of carbon dioxide to kg
---from my balanced equation I have:
C3H8 = 44g/mol
5O2 = 160g/mol
4H2O = 72g/mol
3CO2 = 130g/mol
The density of the liquid propane = 0.621g/mL.
The original amount of propane is 18.9L.
The problem looks simple enough, but I still cannot get it correct. Please help.
4 answers
does this sound correct?
8.00X10^-4 kg?
8.00X10^-4 kg?
No, I don't think so. Follow the instructions Bob Pursley gave and show your work.
1.55 x 10^9