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

Wednesday, March 9, 2011

Stoichiometry with Molarity + Gases (STP)

Stoichiometry's basic concept is simple. The thing that makes it challenging sometimes is organizing your information. Sometimes there are calculations that need to be done to find the information you need to actually do the stoichiometry part. For example, there are questions that won't tell you how many moles or grams of a reactant you have. Instead, it may only tell you the volume and molarity of it. In which case, you would have to use the molarity formula to determine the moles of the substance in question. 

Molarity = mol

Remember that this formula can be rearranged to solve for any of the 3 pieces of information-- as long as you have the other two pieces. 

There are also gas stoichiometry questions that will require knowledge of STP as well. Recall that...

                        One mole of any gas at STP is 22.4L gas            1 mole
                                                                     ---------------   OR ------------
                                                                      1 mole                 22.4 L gas

Use that STP rule to convert from moles to volume or volume to moles, depending on what the question requires you to do.

Molarity and STP are both concepts that you have learned earlier this year. Now it is time to apply these concepts to slightly more complicated situations... stoichiometry! 

We trust that you should be able to handle questions like these with no problem,  but here is an example anyways:

CaCO3 -> CO2 + CaO

How many grams of Calcium Carbonate are needed to produce 9.0L of Carbon Dioxide measured at STP?

9.0L CO2    1mole CO2       1mole CaCO3       100.1 g CaCO3
               x -----------------  x   ---------------------- x  ------------------------ = 4.0 x 10^1 g CaCO3 required
                    22.4 L CO2      1 mole CO2          1 mole CaCO3

(The bolded part of the work shown above is where STP comes into play.)

Friday, March 4, 2011

Stoichiometry (Pronounced: stoi'kē-ŏm'ĭ-trē)

The main idea of stoichiometry is to measure the amounts of elements and compounds that are involved in a reaction. (Ex/ the relationship between the amounts of reactants and products in a reaction)

Mole Ratio is something that will be used in pretty much ever stoichiometry related question that you will need to solve this year. It can be found by looking at the coefficients of a balanced equation.

4NH3 + 5O2 -> 6H2O + 4NO

Mole ration of this equation:
4 : 5 : 6 : 4

If a question asks you to use the information provided in the equation to find the grams of NO produced if 3.0 moles of O2 were combined with excess 4NH3...

3.0 mol O2          4 mol NO       30.0 g NO
                       X -------------- X -------------- = 72 g NO should be produced
                            5 mol O2       1 mol NO

**The bolded part of the work shown above is where mole ratios come in.**

  • We start with 3.0 mol O2 because that's only solid piece of information we are given. 
  • We use mole ratio to determine the amount of O2 and NO needed for a reaction to occur successfully. We do this by looking at the balanced equation. We see that the coefficient of O2 is 5 and the coefficient of NO is 4. That is the ratio of O2 : NO. 
  • Add that to your work as the next step. Doing this has allowed us to figure out the moles for NO required for 3.0 moles of O2 to react successfully in this particular situation. We now need to convert moles of NO into grams, because that what the question asked us for. 
  • Using the periodic table, we see that one mole of NO is 30.0g. Add this to the work, and calculate. 
  • We have determined the grams of NO that should be produced if 3.0 moles of O2 were to be reacted. 
If you still don't understand how to do stoichiometry, have a look at these images below which outline the key points.