1. Stoichiometry: the study of the quantitative relationships that exist between the amounts of reactants and products in a chemical reaction.
Stoichiometry is based on the Law of Conservation of Mass which says that mass can not be created or destroyed in a chemical reaction.
In equation form: Mass reactants = Mass products
If g of HgO react, we should obtain g of Hg and 32.0 g of O2.
This leads to the concept of a balanced chemical equation!
2 HgO (s) → Hg (l) O2 (g)
Notice that the balanced equation has no 2 to 1 ratio between the mass of HgO and the mass of O2. The coefficients in chemical equations are based on relative numbers of atoms not on relative masses.

2. CH4 (g) + H2O (g) → CO (g) + 3 H2 (g)
There is a 1 to 1 ratio of molecules between CH4 and H2O.
CH4 is methane and H2O is water.
1 methane molecule and 1 water molecule
1 to 1 ratio between CH4 and H2O.
2 methane molecules and 2 water molecules
2 to 2 is also a 1 to 1 ratio between CH4 and H2O.
7 methane molecules and 7 water molecules
7 to 7 is also a
1 to 1 ratio between CH4 and H2O.

3. CH4 (g) + H2O (g) → CO (g) + 3 H2 (g)
6.022X1023 methane molecules and 6.022X1023 water molecules
6.022X1023 to 6.022X1023 is also a 1 to 1 ratio between CH4 and H2O
Realizing that 6.022X1023 molecules is also 1 mole leads to the idea that
1 mole methane to 1 mole water is also a 1 to 1 ratio.
0.5 mole to 0.5 mole is also a 1 to 1 ratio.
We will use this language from now on with chemical equations.
The balanced chemical equation above now reads:
1 mole of methane reacts with 1 mole of water
to produce 1 mole of carbon monoxide and 3 moles of hydrogen.

4. ( ) ( ) ( ) ( ) ( ) ( ) 1 to 1 to 1 to 3 Any Others?
CH4 (g) H2O (g) → CO (g) H2 (g)
Therefore the Stoichiometry of the above reaction is:
1 to to to 3
These coefficients can be used to make conversion factors if we write them in the form of ratios (called mole ratios). ( )
1 molCH4
1 mol H2O ( )
1 molCH4
3 mol H2 ( )
1 mol CO
1 mol H2O ( )
1 molCH4
1 mol H2O ( )
1 molCH4
3 mol H2 ( )
1 mol CO
1 mol H2O
Any Others?

5. Write all the possible mole ratios for each of the reactions below.
2 HgO (s) → Hg (l) O2 (g)
4 NH3 (g) NO (g) → N2 (g) H2O (g)

6. Mole ratios from balanced chemical equations allow us to convert from the moles of one substance in a reaction to the needed number of moles of another substance in a reaction.
For example using the reaction of iron and oxygen below:
4 Fe (s) O2 (g) → Fe2O3 (s)
If we use 1.00 mole of oxygen, how many moles of iron can be used? ( )
1.00 mol O2 ( )
3 mol O2
4 mol Fe
= mol Fe
This mole ratio converts from oxygen into iron!

7. How many moles of Fe2O3 could be produced if 0.623 mole of O2 are used?
4 Fe (s) O2 (g) → Fe2O3 (s) ( )
0.623 mol O2 ( )
3 mol O2
2 mol Fe2O3
= mol Fe2O3
Using the same chemical equation, how many moles of Fe would be required
to react with 2.67 mole of O2?

8. The molar mass problems from chapter 7 can be combined with the stoichiometry concepts from chapter 9 to make extended problems.
If 1.50 mole of iron react with excess oxygen, how many grams of Fe2O3 can be produced?
4 Fe (s) O2 (g) → Fe2O3 (s) ( )
1.50 mol Fe ( )
4 mol Fe
2 mol Fe2O3 ( )
159.7 g Fe2O3
1 mol Fe2O3
= g Fe2O3
Mole Ratio
Molar Mass
Mole
Ratio
Molar
Mass
Mol A
Mol B
grams B

9. When magnesium burns in air, it combines with oxygen to form magnesium oxide.
What mass in grams of magnesium oxide can be produced when 1.46 moles of magnesium burns in air?
Click to see Solution
First write the chemical equation based on the word equation that was given in the problem.
2 Mg (s) O2 (g) → MgO (s)
Use the chemical equation to create the needed mole ratio (from Mg to MgO).
2 mol MgO 2 mol Mg
Use the periodic table to find the molar mass of MgO
1 mol MgO = g MgO
Start with what was given and convert the units “mol Mg” to “g MgO”
1.46 mol Mg
( )
( )
2 mol MgO 2 mol Mg
( )
40.31g MgO 1 mol MgO
= g MgO

10. These problems can be extend one more step if we start with mass of Fe and ask about mass of Fe2O3.
4 Fe (s) O2 (g) → Fe2O3 (s)
If 22.3 g of iron react with excess oxygen, how many grams of Fe2O3 can be produced?
The thought Process is:
Molar
Mass
Mole
Ratio
Molar
Mass
Mass Fe
Mol Fe
Mol
Fe2O3
grams
Fe2O3 ( )
22.3 g Fe ( )
55.85 g Fe
1 mol Fe ( )
4 mol Fe
2 mol Fe2O3 ( )
159.7 g Fe2O3
1 mol Fe2O3
= g Fe2O3

11. What mass of aluminum can be produced from the decomposition of 647
What mass of aluminum can be produced from the decomposition of g of aluminum oxide?
Click to see Solution
First write the chemical equation based on the word equation that was given in the problem.
2 Al2O3 (s) → Al (s) O2 (g)
Use the chemical equation to create the needed mole ratio (from Al2O3 to Al).
4 mol Al 2 molAl2O3
Use the periodic table to find the molar mass of Al2O3 and Al
1 mol Al2O3 = g Al2O and mol Al = g Al
Start with what was given and convert the units “g Al2O3” to “g Al” ( )
647.2 g Al2O3 ( )
102.0 g Al2O3
1 mol Al2O3 ( )
2 mol Al2O3
4 mol Al ( )
26.98 g Al
1 mol Al
= g Al

12. Problem Type Recognition Diagram
Particles
Moles
Mass
Avogadro’s
Number
Molar
Type A: particles A to moles A (or moles A to particles A)
Type B: moles A to mass A (or mass A to moles A)
Type C: particles A to mass A (or mass A to particles A)
Molar
Mass A
Mass B
Mole
Ratio
Moles A
Moles B
Type 1: moles A to moles B
Type 2: moles A to mass B
Type 3: mass A to moles B
Type 4: mass A to mass B
If you do not like these names, make up your own!

13. ( ) ( ) ( ) ( ) Click to see Solution
How many grams of CO2 can be made when 38.5 g of C2H6 react with excess O2?
C2H O2
CO2
H2O
First: balance the equation so that you can find any needed mole ratios.
2 C2H O2
4 CO2
H2O
Second: excess O2 means that you can ignore how much O2 is used.
Third: complete the problem like previous examples. ( )
38.5 g C2H6 ( )
30.07 g C2H6
1 mol C2H6 ( )
2 mol C2H6
4 mol CO2 ( )
44.01 g CO2
1 mol CO2
= 113 g CO2

14. BLT Sandwich Recipe:
If your kitchen has the following list of items, how many BLT sandwiches could you make that follow the recipe?
2 pieces of bread
5 slices of bacon
3 slices of tomato
1 piece of lettuce
Kitchen List:
Bread: 52 pieces
Bacon: 70 slices
Tomato: 30 slices
Lettuce: 25 pieces
(52 bread)(1 sandwich/2 bread) = 26 sandwiches
(70 bacon)(1 sandwich/5 bacon) = 14 sandwiches
(30 tomato)(1 sandwich/3 tomato) = 10 sandwiches
(25 lettuce)(1 sandwich/1 lettuce) = 25 sandwiches
Since tomato ran out after 10 sandwiches, 10 sandwiches is the most that can be made using the recipe.
The item that “runs out” first is called the limiting reactant because it limits the amount of product that can be made.

15. Determine the amount of P4O10 produced when 25
Determine the amount of P4O10 produced when 25.0 g of P4 reacts with 50.0 g of O2.
Click to see Solution
First write the chemical equation (balanced!).
P4 (s) O2 (g) → P4O10 (s) 5
Second identify important equivalence statements (molar masses and mole ratios).
1 mole P4 = g P4
1 mole O2 = g O2
5 mole O2 = 1 mole P4O10
1 mole P4 = 1 mole P4O10
Third, set up the calculations (like type 4) but start two separate problems (one starting with each reactant given in the question).
Fourth, evaluate your answers. The smallest amount of product produced by the two reactants is the theoretical yield (the maximum amount of product that can be produced in the reaction).

16. Determine the amount of P4O10 produced when 25
Determine the amount of P4O10 produced when 25.0 g of P4 reacts with 25.0 g of O2.
Continued:
P4 (s) O2 (g) → P4O10 (s)
Click to see Solution ( )
25.0 g P4 ( )
123.9 g P4
1 mol P4 ( )
1 mol P4
1 mol P4O10 ( )
283.9 g P4O10
1 mol P4O10
= g P4O10 ( )
25.0 g O2 ( )
32.00 g O2
1 mol O2 ( )
5 mol O2
1 mol P4O10 ( )
283.9 g P4O10
1 mol P4O10
= g P4O10
So, how much P4O10 could we actually make?

17. ( ) Percent Yield Actual Yield Theoretical Yield Percent Yield = X 100
When we perform a reaction, we often fail to collect all of the product that could have been produced because of difficulties with separation of the product from excess reactants or other products.
The amount of product we should produce based on the starting amounts of reactants is called the theoretical yield and is the maximum amount of product that we could produce from that reaction.
The actual yield is the amount of purified product that is actually collected-available for use for another purpose.
The percent yield is the actual yield divided by the theoretical yield times 100.

18. When a student reacted 25. 0 g of P4 with 25. 0 g of O2, a total of 36
When a student reacted 25.0 g of P4 with 25.0 g of O2, a total of 36.8 g of P4O10 was collected. What is the percent yield for the reaction?
Click to see Solution
Note: this is the same set of starting conditions that was used in the previous example.
After completing the limiting reactant problem (previous example) to determine the theoretical yield for this reaction, we can used the formula for percent yield.
Actual Yield = 36.8 g P4O Theoretical Yield = 44.4 g P4O10 (see previous work)
Actual Yield
Theoretical Yield
Percent Yield =
X 100
36.8 g P4O10
44.4 g P4O10
Percent Yield =
X 100
Percent Yield = %