March 11, 2011

Why hello, hello, hello. Today I am just going to get straight to it because first of all it is Monday today which just says it all and I am tired. But don't worry, I still gaurentee you that I will not dissapoint any of you fellow bloggers and viewers.

Excess and Limiting Reactants
* Excess Quantity
- a balanced equation that describes what should happen in a chemical reaction.
- one reactant  is the excess quantity and some of it will be left over. The second reactant is used up completely, and is called the limiting reactant.

- Calculating amount of products formed when one of the reactants is in excess.
ex 1. How many grams of OCl2 will be formed when 44.0grams of O2 rect with 97.0grams of Cl2
= convert both reactants to the desired product and the smaller amount of product will actaully be produced
= 1O2 + 2Cl2 (arrow) 2OCl2
= convert Cl2 to OCl2 = 97.0g Cl2 X 1 mol Cl2  X 2 mol OCl2 X 87.0g OCl2
                                                      71.0g Cl2     2 mol Cl2      1 mol OCl2
= 118.86grams OCl2
= convert O2 to OCl2 = 44.0gO2 X 1mol O2 X 2 mol OCl2 X 87.0gOCl2
                                                    32.0g O2   1 mol O2       1 mol OCl2
= 239.25g OCl2
= Cl2 is the limiting quantity and O2 is the excess quantity

- Finding limiting quantity, excess quality, and amount excess.
ex 2. When 41.0g of O2 is reacted with 164g of Cl2, how many grams of which rectant will be left over (in excess)
= 1O2 + 2Cl2 (arrow) 2OCl2
= calculate how many grams of Cl2 would be required to react with 41.0g of oxygen gas
= 41.0g O2 X 1 mol O2 X 2 mol Cl2 X 71.0g Cl2
                    32.0g O2    1 mol O2     1 mol Cl2
= 181.93g Cl2
= since there are only 164g of Cl2, not all 41.0g of O2 can react therefore, chlorine is the limiting and oxygen is the excess.
= however, to calculate how much of the excess quantity is left over, we need to find out how much of O2 would react with the limiting reactant (164g Cl2)
= convert Cl2 to O2 = 164g Cl2 X 1 mol Cl2 X 1 mol O2 X 32.0gO2
                                                  71.0g Cl2    2 mol Cl2   1 mol O2
= 36.958g O2
= 37.0g O2 would react with 164g Cl2
= 41.0g (have) MINUS 37.0g O2 (reacts)
= 4.0g O2 would be excess when 164g Cl2 reacts with 41.0g O2.

Whewww man that was alot of calculations!! But dont stress out if you dont get this on the first try; it is perfectly normal and trust me, practice makes perfect!!!
Here are some more excercises that you can & should check out:
1) http://www.science.uwaterloo.ca/~cchieh/cact/c120/limitn.html
2) http://www.science.uwaterloo.ca/~cchieh/cact/c120/stoichio.html
3) http://www.ausetute.com.au/exceslim.html
4) http://wiki.answers.com/Q/How_do_you_solve_limiting_reagent_problems
5) http://members.shaw.ca/mrkwan/chemistry11/Excess%20Reactant%20Notes.pdf

Here are some videos if you are more of a visual learner:
1) http://www.youtube.com/watch?v=JOQz2rIFpVM
2) http://www.youtube.com/watch?v=rESzyhPOJ7I&feature=fvwrel
3) http://www.youtube.com/watch?v=4OeKFsVRmfQ&feature=related
4) http://www.youtube.com/watch?v=dAkbcRgecBo&feature=related
5) http://www.youtube.com/watch?v=SjQG3rKSZUQ&feature=relmfu
6) http://www1.teachertube.com/viewVideo.php?ref=kvanderveen&title=Limiting_Reactant&video_id=75396

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And of course there is always enough time to have some laughs!