Stoichiometry Calculations.....NOW WITH MOLARITY AND VOLUME!

OK, so last blog, you learned about doing some simple stoichiometry calculations that involved converting the mass of one substance into the mass of another substance.


Now, let's pull out the mole map


We're going to try some harder conversions: by including Molarity conversions and Molar Volume.

So, as we all remember, molarity is the molar concentration. The formula is M = mol/V



Let's try an example:
We have a chemical equation of
NaCl + CaO ---> Na2O + CaCl2

and we have 120 mL of a 0.300 M solution of NaCl. How many grams of CaCl2 are produced?

First, we balance:
2 NaCl + CaO ---> Na2O + CaCl2

Then, we figure out how many moles of NaCl there are, based on the information given:

M = mol/V   ---> mol = MV
mol NaCl = (0.120)*(0.300) = 0.0360 mol NaCl

0.0360 mol NaCl X 1 mol CaCl2 X 111.1g           =  1.9998 (2 SF = 2.0g CaCl2)
                                 2 mol NaCl     1 mol CaCl2

This answer tells us that 2.0 g of CaCl2 is produced when 120 mL of 0.300 M solution of NaCl reacts with sufficient CaO.

Now, we can also add on Molar Volume calculations.
Recall that there are 22.4L per mole in STP conditions (Standard temperature and pressure)


Let's try another example:
We have the chemical reaction:  Li2O + MgCl2 ---> LiCl + MgO
How many grams of Li2O will produce 9.0 L of MgO at STP?

First, we balance:
Li2O + MgCl2 ---> 2 LiCl + MgO

Now, we convert:
9.0L MgO X 1 mol MgO X 1 mol Li2O X     29.8g      =  11.97 (2 SF = 12 g)
                         22.4 L          1 mol MgO    1 mol Li2O

So this means that 12 g of Li2O reacts with sufficient MgCl2 in order to produce 9.0 L of MgO

Now, here are some problems, as well as a video: