January 12, 2025, 10:27:18 AM
Forum Rules: Read This Before Posting


Topic: Combustion Analysis  (Read 2508 times)

0 Members and 1 Guest are viewing this topic.

Offline qathetenkou

  • Very New Member
  • *
  • Posts: 2
  • Mole Snacks: +0/-0
Combustion Analysis
« on: October 21, 2010, 08:10:35 PM »
Maybe I'm just having a huge blonde moment or something, but currently working on a homework and I'm a little stuck on a question. My biggest fear is the fact this teacher likes wording questions in a trick manner and you can't miss a single step.

Part one of the question states "How many grams of hydrogen are needed to produce 10.00 grams of water when it is burned in excess oxygen?"

Consdering the wording of "excess", I keep thinking he wants to see the work written in the manner we use to get theoretical yields, etc. Only it's missing too much information.

I've tried using percent mass (Hydrogen is 11.21% of the mass of a water molecule, meaning I'd need 1.120 grams of hydrogen in this problem) but I can't help thinking he wants to see the work shown another way. Am I completely missing it?

The second half asks "How many grams of C7H11O3 are needed to produce 10.00 grams of water when it is burned in excess oxygen". The third part is just another variation of this.

I will happily take examples instead of actual answers, I'm just not sure how I'm expected to work it out when it's missing the majority of information we've always been provided before.

Offline opti384

  • Full Member
  • ****
  • Posts: 434
  • Mole Snacks: +33/-25
  • Gender: Male
    • In the Search for the Laws of Nature
Re: Combustion Analysis
« Reply #1 on: October 21, 2010, 09:10:46 PM »
I suggest that you write a balanced equation for all the reactions.

Offline qathetenkou

  • Very New Member
  • *
  • Posts: 2
  • Mole Snacks: +0/-0
Re: Combustion Analysis
« Reply #2 on: October 21, 2010, 09:44:12 PM »
Probably the most obvious answer in the world and now I feel like an idiot for posting, haha. Thanks! Reminded me of something else and I think I've got it. It really helped, promise!

Online Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27896
  • Mole Snacks: +1816/-412
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Combustion Analysis
« Reply #3 on: October 22, 2010, 02:49:59 AM »
Now that you got it from the most obvious side - there is another, perfectly valid and faster approach.

Think, how number of moles of water produced depends on number of moles of hydrogen present in the given reactant. Do you have to balance full reaction to find out stoichiometric ratio of water to this reactant?

Note, that it will work here only as you are known there is an excess of oxygen, so you can be sure hydrogen is a limiting reagent.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Sponsored Links