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STOICHIOMETRY

Stoichiometry problems generally work this way:

Quantity of A Mols of A Mols of B Quantity of B

We will use this pattern for most of this tutorial.

Convert Quantity of A to Mols of A

Set up your conversion factors that contains mass & mol units. Molar mass contains these units. Ex.: you have 25.0g CaCO3 & need to find out the quantity of mols that are present. Begin by calculating CaCO3's molar mass:

Now, convert the mass of CaCO3 into mols.

Convert mols to grams

Convert this type of problem in the same manner as the previous example. You have 0.750 mol CaCO3. Convert this to grams. The problem is worked out below.


Mol of A to Mol of B Conversions

When you know the mols of a substance, you can make a valid comparison. From mols, we can convert to grams, to number of molecules with the help of Avogadro's number, or to mols of a different substance. First we'll convert mols of substance A to mols of substance B.

Suppose you have 0.100 mol Na2CO3 (sodium carbonate). We want to find out the mols of Na+ ions. One mole of Na2CO3 contains 2 mols of Na, 1 mol of C and 3 mols of O. We will compare the mols of Na2CO3 and mols of sodium ions.

In the above example, we performed a comparison within one compound. Sometimes, a reaction involves a comparing two compounds. Let's work through a typical problem:

In this problem, 2 mols of KClO3 (potassium chlorate) decomposes to 2 mols KCl (potassium chloride) and 3 mols O2 (oxygen). If you have 0.400 mol KClO3, and want to calculate the mols O2 formed, use the coefficients to formulate a ratio where the mols of KClO3 cancel and the mols of O2 remain:

Once you have mols you can also use Avogadro's number to find particles. Avogadro's number is 6.02 x 1023 and tells us the number of particles found in 1 mol. We can use it to convert mols to particles or particles to mols. Let's use the above example to calculate the number of oxygen molecules that were formed:


The Entire Pattern

Now that we can convert grams mols, and mols of A mols of B, and mols grams, you can now use the entire pattern, (Quantity of A Mols of A Mols of B Quantity of B) to work a stoichiometry problem where you calculate theoretical yield. Theoretical yield is the maximum quantity of a certain product that is produced from the reactants. Let's look at the example problem below.

If NH3, ammonia, burns in air, this reaction will take place:

If you start with 51.0 g NH3 (ammonia), how many grams of H2O will be produced?

The gameplan is to convert the g of NH3 to mols of NH3 (Quantity of A to mols of A), then convert mols of NH3 to mols H2O (mols of A to mols of B), and then convert the mols of H2O to g of H2O (mols of B to Quantity of B):

Quantity of A Mols of A Mols of B Quantity of B

The 81.0 grams H2O that was produced (theoretical yield) is the maximum quantity of H2O produced.

In this example, we made the assumption that there was enough O2 to totally use up the ammonia. We could make this assumption because no information was given about how much O2 was in the reaction. If this were a limiting reagent problem, we would have been given the starting amounts of both reactants and you don't know which reactant is going to be consumed first. For a review of limiting reactant problems, see the tutorial that covers this topic.

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