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.)

No comments:

Post a Comment