April 25, 2024, 12:05:27 PM
Forum Rules: Read This Before Posting


Topic: Using Heats of Combustion  (Read 5146 times)

0 Members and 1 Guest are viewing this topic.

Offline hockey101

  • Regular Member
  • ***
  • Posts: 32
  • Mole Snacks: +0/-3
Using Heats of Combustion
« on: February 26, 2013, 05:30:47 PM »
hey everyone, I tried searching for this topic as well as hydrocarbons, but none of them helped me answer my question.  I need to know if I did this correctly or if Im on the right track.


The Question states:  How much energy, in kilojoules, is released by completely burning 25.0 moles of hexane?


My Work:  86 - 25 = 61 g/mol x 4825 KJ/g = 2940.2 kJ/mol

Is that anywhere close?

Offline UG

  • Full Member
  • ****
  • Posts: 822
  • Mole Snacks: +134/-15
  • Gender: Male
Re: Using Heats of Combustion
« Reply #1 on: February 26, 2013, 05:34:57 PM »
Whats the reason for 86 - 25?

Offline hockey101

  • Regular Member
  • ***
  • Posts: 32
  • Mole Snacks: +0/-3
Re: Using Heats of Combustion
« Reply #2 on: February 26, 2013, 05:37:03 PM »
I thought I needed to subtract that from the original molar mass.  Guess not

Offline UG

  • Full Member
  • ****
  • Posts: 822
  • Mole Snacks: +134/-15
  • Gender: Male
Re: Using Heats of Combustion
« Reply #3 on: February 26, 2013, 05:43:36 PM »
The molar mass shouldn't change. The question asks for the energy in kilojoules whereas you have the answer in kJ/mol.
The first question I would ask is, how many grams of hexane do you have if you have 25.0 moles of it?

Offline hockey101

  • Regular Member
  • ***
  • Posts: 32
  • Mole Snacks: +0/-3
Re: Using Heats of Combustion
« Reply #4 on: February 26, 2013, 06:00:09 PM »
so for that I would use the proportion grams of the substance/moles of the substance = molar mass of the substance/1 mol.

So: 
grams of the substance = x
moles of the substance = 25
molar mass of the substance = 86 (C6H14
1 mol = 1 mol

Solution:   1 mol = 2150 g/mol

Offline UG

  • Full Member
  • ****
  • Posts: 822
  • Mole Snacks: +134/-15
  • Gender: Male
Re: Using Heats of Combustion
« Reply #5 on: February 26, 2013, 06:08:40 PM »
Ok, very good you have the correct numerical value but the units are no longer g/mol. It is simply grams.
Next step is easy, how many kilojoules of energy are released upon combustion of this mass of hexane if the heat of combustion is 4825 kJ per gram.

Offline hockey101

  • Regular Member
  • ***
  • Posts: 32
  • Mole Snacks: +0/-3
Re: Using Heats of Combustion
« Reply #6 on: February 26, 2013, 06:13:57 PM »
4825 kJ/g     divided by       2150 g = 2.2 kJ?

Offline UG

  • Full Member
  • ****
  • Posts: 822
  • Mole Snacks: +134/-15
  • Gender: Male
Re: Using Heats of Combustion
« Reply #7 on: February 26, 2013, 06:16:29 PM »
Not quite like that. You have 2150 grams of hexane, you know each gram combusted releases 4825 kJ. How much energy in total is released?

Offline hockey101

  • Regular Member
  • ***
  • Posts: 32
  • Mole Snacks: +0/-3
Re: Using Heats of Combustion
« Reply #8 on: February 26, 2013, 06:17:11 PM »
wait a minute...thats supposed to be 48.2 kJ...so the answer should be 44.6 kJ

Offline hockey101

  • Regular Member
  • ***
  • Posts: 32
  • Mole Snacks: +0/-3
Re: Using Heats of Combustion
« Reply #9 on: February 26, 2013, 06:20:51 PM »
Wait, am I using the Combustion formula of C6H14 +O2  :rarrow: CO2 + H2O?  And then balance it?

Offline UG

  • Full Member
  • ****
  • Posts: 822
  • Mole Snacks: +134/-15
  • Gender: Male
Re: Using Heats of Combustion
« Reply #10 on: February 26, 2013, 06:25:01 PM »
Wait, am I using the Combustion formula of C6H14 +O2  :rarrow: CO2 + H2O?  And then balance it?
Oh no, you are over thinking this. It is a simple matter of multiplication with the numbers you already have..

Offline hockey101

  • Regular Member
  • ***
  • Posts: 32
  • Mole Snacks: +0/-3
Re: Using Heats of Combustion
« Reply #11 on: February 26, 2013, 06:43:05 PM »
im totally confused

Offline UG

  • Full Member
  • ****
  • Posts: 822
  • Mole Snacks: +134/-15
  • Gender: Male
Re: Using Heats of Combustion
« Reply #12 on: February 26, 2013, 06:51:40 PM »
Well thats not too good.
You are combusting 25.0 moles of hexane, which we worked out to be equal to 2150 grams of hexane. You are given information that the heat of combustion is 48.2 kJ per gram of hexane. Hence if you had one gram of hexane, complete combustion would release 48.2 kJ of energy. So if you had 2150 grams of hexane you would have 48.2 kJ/g x 2150 g = 104000 kJ of energy released upon combustion. Still confused?

Offline hockey101

  • Regular Member
  • ***
  • Posts: 32
  • Mole Snacks: +0/-3
Re: Using Heats of Combustion
« Reply #13 on: February 26, 2013, 06:55:46 PM »
Wow it was that easy??  Thank you that helped a lot

Offline UG

  • Full Member
  • ****
  • Posts: 822
  • Mole Snacks: +134/-15
  • Gender: Male
Re: Using Heats of Combustion
« Reply #14 on: February 26, 2013, 07:01:34 PM »

Sponsored Links