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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: hockey101 on February 26, 2013, 05:30:47 PM

Title: Using Heats of Combustion
Post by: hockey101 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?
Title: Re: Using Heats of Combustion
Post by: UG on February 26, 2013, 05:34:57 PM: Whats the reason for 86 - 25?
Title: Re: Using Heats of Combustion
Post by: hockey101 on February 26, 2013, 05:37:03 PM: I thought I needed to subtract that from the original molar mass. Guess not
Title: Re: Using Heats of Combustion
Post by: UG 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?
Title: Re: Using Heats of Combustion
Post by: hockey101 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
Title: Re: Using Heats of Combustion
Post by: UG 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.
Title: Re: Using Heats of Combustion
Post by: hockey101 on February 26, 2013, 06:13:57 PM: 4825 kJ/g divided by 2150 g = 2.2 kJ?
Title: Re: Using Heats of Combustion
Post by: UG 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?
Title: Re: Using Heats of Combustion
Post by: hockey101 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
Title: Re: Using Heats of Combustion
Post by: hockey101 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?
Title: Re: Using Heats of Combustion
Post by: UG on February 26, 2013, 06:25:01 PM: Quote from: hockey101 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?
Oh no, you are over thinking this. It is a simple matter of multiplication with the numbers you already have..
Title: Re: Using Heats of Combustion
Post by: hockey101 on February 26, 2013, 06:43:05 PM: im totally confused
Title: Re: Using Heats of Combustion
Post by: UG 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?
Title: Re: Using Heats of Combustion
Post by: hockey101 on February 26, 2013, 06:55:46 PM: Wow it was that easy?? Thank you that helped a lot
Title: Re: Using Heats of Combustion
Post by: UG on February 26, 2013, 07:01:34 PM: Quote from: hockey101 on February 26, 2013, 06:55:46 PM
Wow it was that easy??
Yea.... :P
Title: Re: Using Heats of Combustion
Post by: hockey101 on February 26, 2013, 07:09:33 PM: As you can see, I suck at Chemistry. ;D