Friday, March 11, 2011

Excess and Limiting Reagents WOO WOO EXCITMENT

What is UP my friends? Today we are going to learn something reallyyyy special. It's called STOICHIOMETRY OF EXCESS QUANTITIES.
So whats going to happen is I want you to clear your mind of eeeeverything you've learned so far about stoichiometry. (okay well not actually still have to use 99% of the information but clear out that 1%)

So far, the stoichiometry calculations in the previous section just ASSUME that a given reactant is completely used up during a reaction. PSHT that is soooo lame. It's like when you see a fat woman and ASSUME she is pregnant. If you're wrong, run.....just run

In reality, most reactions are carried out in such a way that one or more of the reactants are present in EXCESS amounts. Some reasons this happens is:
1. When you add and excess of a reactant to make sure the second reactant is completely used up, because the second reactant may be too expensive to waste or too harmful to the environment if left unreacted
2. When there is a limiting amount of one reactant so having an excess of another reactant is unavoidable

One reactant is the EXCESS QUANTITY and some of it will be left over. The second reactant is used up completely, and is the LIMITING REACTANT

Now let's have an example equation shall we? : )

If 234 grams of CS2 is reacted with 619 grams O2 what is the limiting reactant? How many grams of the excess reactant remain after the reaction is complete?
                                        CS2 (l) + 3 O2 (g) ---> CO2 (g) + 2 SO2 (g)

Step 1: Randomly select one product to use. In this case, I'll randomly select CO2

Step 2: Convert both CS2 and O2 into grams of the same product, In this case, grams of CO2
234g CS2 x 1 mol CS2  x 1 mol CO2  x 44.0g CO2 = 135.12g CO2
                     76.2g CS2      1 mol CS2    1 mol CO2
619g O2 x 1 mol O2   1 mol CO2   x 44.0g CO  = 283.71g CO2
                  32.0g O2      3 mol O2           1 mol CO2

Step 3: Now you can see that CS2 is the limiting reactant since it is the smaller amount. Now you must find out the grams of the excess reactant remaining. You do this by taking the ORIGINAL grams of the limiting reactant (CS2) from the question, and change that number into grams of the excess reactant (O2)
234g CS2 x 1 mol CS2  x 3 mol O2   x 32.0g O2 = 294.80g O2
                     76.2g CS2     1 mol CS2     1 mol O2

Step 4: After, you take the grams of the limiting reactant from the ORIGINAL equation, and subtract the grams you found in the equation above
619g O2 - 294.80g O2 = 324.2g O2 excess

Now for a joke because all the seriousness is starting to make me nauseous

If you need any further help, you can watch this video

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