Chapter3Lecturepart2

Chapter3Lecturepart2 - 6 Chapter 3: Balancing simple...

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Unformatted text preview: 6 Chapter 3: Balancing simple equations Chapter 3: Balancing simple equations C + O 2 CO 2 reactants products 1 mole + 1 mole 1 mole This one is balanced. There are the same number of atoms of each type on each side, just rearranged —————————————————————————— Combustion (one type of oxidation) CH 4 + O 2 CO 2 + H 2 O This is not balanced. There are too many H on the left and too many O on the right. We need CH 4 + 2 O 2 CO 2 + 2 H 2 O In the balanced equation, the moles are 1 mole 2 moles 1 mole 2 moles It is just coincidence that the total on each side is three. That does NOT have to be Copyright: 2010 Prof. Magde 6 Chapter 3: Balancing simple equations Chapter 3: Balancing simple equations We previously had some data about mass for the combustion of a large organic molecule. Let us balance it: C 3 H 4 O 3 + O 2 CO 2 + H 2 O Take a few minutes and try it. —————————————————————————— A possible multiple choice question on an exam could be: How many moles of dioxygen do you need in the balanced equation, assuming you use all integers? 0 or 1 or 2 or 3 or 4 or 5 or none of the above? —————————————————————————— Many of you may know we should indicate the physical state of each reactant and product. C (s) + O 2 (g) CO 2 (g) Be patient; soon enough we will have to do that. Copyright: 2010 Prof. Magde 6 Chapter 3: Balancing simple equations Chapter 3: Balancing simple equations So we want to balance: C 3 H 4 O 3 + O 2 CO 2 + H 2 O The answer is: 2 C 3 H 4 O 3 + 5 O 2 6 CO 2 + 4 H 2 O 2 moles 5 moles...
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This note was uploaded on 01/12/2011 for the course CHEM 6A taught by Professor Pomeroy during the Fall '08 term at UCSD.

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Chapter3Lecturepart2 - 6 Chapter 3: Balancing simple...

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