Stoichiometry
From the Greek stoikheion "element" and metriā "measure."
A good overview of stoichiometry problems with worked out examples can be found here. It's another Chemistry Corner! By Vicki Klawinski from Commerce High School in Commerce, Texas. Click on "Lecture Notes" and then choose "Stoichiometry" from the drop-down menu.
Here is another good site introducing stoichiometry, with practice problems, from John L. Park's ChemTeam site.
You might also want to look at the Wikipedia article about stoichiometry here.
Stoichiometry calculations are about calculating the amounts of substances that react and form in a chemical reaction. For example, based on the balanced chemical equation, we can calculate the amount of a product substance that will form if we begin with a specific amount of one or more reactants. Or, you may have a target amount of product to prepare. How much starting compounds are needed to prepare this amount? These are practical calculations that are done frequently by chemists.
The Method
For practically all stoichiometry problems, we want to use the. . . .
Fabulous Four Steps 
Step 1: Write the balanced chemical equation for the reaction.
Step 2: Calculate the moles of "given" substance. If more than one reactant amount is given, calculate the moles of each to determine which is the limiting reactant.
Step 3: Calculate the moles of "desired" substance from your answer in Step 2 using the coefficients from the balanced chemical equation. If more than one reactant was given originally, you can calculate the moles of product twice, based on the moles of each reactant. The reactant that gives the smaller moles of product is the limiting reactant. Keep this answer for Step 4.
Step 4: Convert your answer in Step 3 to the units the problem asks for. Usually this is grams, but it could be volume (for gases or liquid solutions) or concentration (such as molarity, for solutions).
Again in brief:
1. Balanced reaction
2. Moles of "given" substance(s)
3. Moles of "desired" substance such as a product
4. Convert Step 3 answer to the units asked for
Examples
Example 1: How many grams of ferric oxide, Fe2O3, are formed from the reaction of 5.00 g of iron metal with excess oxygen gas?
Step 1: Balanced reaction.
4 Fe (s) + 3 O2 (g) ––�?gt; 2 Fe2O3 (s)
Step 2: Moles of "given" substance, the iron metal, since its amount is given.
moles of Fe = grams of Fe / molar mass of Fe = 5.00 g / 55.845 g/mol = 0.08953353 mol (not rounding yet)
Step 3: Moles of "desired" substance, ferric oxide.
Here's where the balanced reaction comes in. The coefficients in the balanced reaction represent the moles of the substances that react and form. As such, balanced reactions are "in" moles by default. That is why we always have to convert amounts to moles when working these kinds of problems.
moles of Fe2O3 = 0.08953353 mol Fe X 2 mol Fe2O3 = 0.044766765 mol Fe2O3
1 4 mol Fe
Step 4: Grams of Fe2O3
Grams of Fe2O3 = moles of Fe2O3 X molar mass of Fe2O3
= 0.044766765 mol X 159.6882 g/mol = 7.15 g rounded to 3 significant figures.
And that's it! We can also do steps 2�? like a conversion problem. Once you are familiar with the steps, you can work these problems more quickly this way.
5.00 g Fe X 1 mol Fe X 2 mol Fe2O3 X 159.6882 g Fe2O3 = 7.15 g of Fe2O3
1 55.845 g Fe 4 mol Fe mol Fe2O3
Remember to 1) write the units on all numbers and check that the units cancel properly to give the correct unit for the answer, and
2) avoid rounding numbers too much during the calculation, or you will have roundoff error in your answer.
Example 2: If 10.0 g of iron metal is reacted with 15.0 g of Cl2 gas, how many grams of ferric chloride, FeCl3, will form?
In this problem, the amounts of both reactants are given, so we will have to determine which is the limiting reactant (the one that "limits" the amount of product that is formed). The other reactant is in excess amount. We'll use the Fab Four Steps just as before.
Step 1: Balanced reaction.
2 Fe (s) + 3 Cl2 (g) ––�?gt; 2 FeCl3 (s)
Step 2: Moles of given substances.
moles of Fe = 10.0 g / 55.845 g/mol = 0.17906706 mol
moles of Cl2 = 15.0 g / 70.906 g/mol = 0.211547682 mol
Step 3: Moles of desired substance, FeCl3.
Since we have two given amounts, a straightforward approach to this step is to calculate the moles of FeCl3 twice, first based on the moles of Fe and second based on the moles of Cl2. Keep the smaller answer. The reactant that gives this smaller answer is the limiting reactant. The other reactant is in excess amount.
moles of FeCl3 = 0.17906706 mol Fe X 2 mol FeCl3 = 0.17906706 mol of FeCl3
based on Fe 1 2 mol Fe
moles of FeCl3 = 0.211547682 mol Cl2 X 2 mol FeCl3 = 0.141031788 mol of FeCl3 <–�?Keep this answer!
based on Cl2 1 3 mol Cl2
Since the moles of FeCl3 based on moles of Cl2 is the smaller answer, Cl2 is the limiting reactant. Iron metal is therefore in excess amount, so there will be some Fe left over unreacted.
Note that we might have reasonably assumed that iron metal was the limiting reactant since it was present in lesser amount in grams initially (10.0 g of Fe and 15.0 g of Cl2). But it turned out that Cl2 was the limiting reactant. The molar masses of the substances and the reaction stoichiometry come into play also, so we can't automatically assume which substance is the limiting reactant until we go through the steps as we did above.
Step 4: Finally! Convert moles of FeCl3 to grams.
Grams of FeCl3 = 0.141031788 mol X 162.204 g/mol = 22.9 g rounded to 3 significant figures.
Practice makes perfect! Working stoichiometry problems properly will strengthen your skills in working many other
types of chemistry problems as well. 
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