1. Stoichiometry
By Mr. M

2. Stoichiometry Consider: 4NH3 + 5O2  6H2O + 4NO
4/15/2018
Consider: 4NH3 + 5O2  6H2O + 4NO
Recall that many conversion factors exist:
4 mol NH3/5 mol O2, 6 mol H2O/4 mol NH3, etc
In words, this tells us that for every 4 moles of NH3, 5 moles of O2 are required, etc.
“Stoichiometry” refers to the relative quantities of moles. It also refers to calculations that make use of mole ratios.
Recall also that molar masses provide factors:
1 mol NH3 / 17 g NH3, 32 g O2 / 1 mol O2
Is 4 g NH3 / 5 g O2 a conversion factor?
No. The equation tells us moles not grams.
Notice that stoichiometry requires precision

3. Stoichiometry questions (1)
4/15/2018
Stoichiometry questions (1)
Consider : 4NH3 + 5O2  6H2O + 4NO
How many moles of H2O are produced if mol of O2 are used?
How many moles of NO are produced in the reaction if 17 mol of H2O are also produced?
6 mol H2O
5 mol O2 x
mol H2O =
# mol H2O=
0.176 mol O2
4 mol NO
6 mol H2O x
11.33 mol NO =
# mol NO=
17 mol H2O
Notice that a correctly balanced equation is essential to get the right answer

4. Stoichiometry questions (2)
4/15/2018
Stoichiometry questions (2)
Consider : 4NH3 + 5O2  6H2O + 4NO
How many grams of H2O are produced if 1.9 mol of NH3 are combined with excess oxygen?
How many grams of O2 are required to produce 0.3 mol of H2O?
# g H2O=
6 mol H2O
4 mol NH3 x
18.02 g H2O
1 mol H2O x
51.4 g H2O =
1.9 mol NH3
# g O2=
5 mol O2
6 mol H2O x
32 g O2
1 mol O2 x
8 g O2 =
0.3 mol H2O

5. Stoichiometry questions (3)
4/15/2018
Stoichiometry questions (3)
Consider : 4NH3 + 5O2  6H2O + 4NO
How many grams of NO is produced if 12 g of O2 is combined with excess ammonia?
# g NO=
1 mol O2
32 g O2 x
4 mol NO
5 mol O2 x
30.01 g NO
1 mol NO x
12 g O2
9.0 g NO =

6. Converting grams to grams
4/15/2018
Converting grams to grams
Notice that we cannot directly convert from grams of one compound to grams of another. Instead we have to go through moles.
Many stoichiometry problems follow a pattern: grams(x)  moles(x)  moles(y)  grams(y)
We can start anywhere along this path depending on the question we want to answer
Notice that we cannot directly convert from grams of one compound to grams of another. Instead we have to go through moles.
Many stoichiometry problems follow a pattern: grams(x)  moles(x)  moles(y)  grams(y)
We can start anywhere along this path depending on the question we want to answer
Notice that we cannot directly convert from grams of one compound to grams of another. Instead we have to go through moles.
Many stoichiometry problems follow a pattern: grams(x)  moles(x)  moles(y)  grams(y)
We can start anywhere along this path depending on the question we want to answer
Notice that we cannot directly convert from grams of one compound to grams of another. Instead we have to go through moles.
Many stoichiometry problems follow a pattern: grams(x)  moles(x)  moles(y)  grams(y)
We can start anywhere along this path depending on the question we want to answer

7. Moving along the stoichiometry path
4/15/2018
Moving along the stoichiometry path
We always use the same type of information to make the jumps between steps:
grams (x)  moles (x)  moles (y)  grams (y)
Molar mass of x
Molar mass of y
Mole ratio from balanced equation
Given: 4NH3 + 5O2  6H2O + 4NO
a) How many moles of H2O can be made using 0.5 mol NH3? b) what mass of NH3 is needed to make 1.5 mol NO? c) how many grams of NO can be made from 120 g of NH3?

8. Answers 4NH3 + 5O2  6H2O + 4NO a) b) c) 6 mol H2O 4 mol NH3 x
4/15/2018
Answers
4NH3 + 5O2  6H2O + 4NO a) b) c)
6 mol H2O
4 mol NH3 x
0.75 mol H2O =
# mol H2O=
0.5 mol NH3
# g NH3=
4 mol NH3
4 mol NO x
17.04 g NH3
1 mol NH3 x
25.6 g NH3 =
1.5 mol NO
# g NO=
1 mol NH3
17.04 g NH3 x
4 mol NO
4 mol NH3 x
30.01 g NO
1 mol NO x
120 g NH3
211 g NO =

9. Limiting Reactant: Cookies
1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen
If we had the specified amount of all ingredients listed, could we make 4 dozen cookies?
What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies?
What if we only had one egg, could we make 3 dozen cookies?

10. Limiting Reactant
Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant. That reactant is said to be in excess (there is too much). The other reactant limits how much product we get. Once it runs out, the reaction stops. This is called the limiting reactant.

11. Limiting Reactant
To find the correct answer, we have to try all of the reactants. We have to calculate how much of a product we can get from each of the reactants to determine which reactant is the limiting one.
The lower amount of a product is the correct answer.
The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it!
Be sure to pick a product! You can’t compare to see which is greater and which is lower unless the product is the same!

12. Limiting Reactant: Example
10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced?
2 Al + 3 Cl2  2 AlCl3
Start with Al:
Now Cl2:
10.0 g Al 1 mol Al mol AlCl g AlCl3
27.0 g Al mol Al mol AlCl3
= 49.4g AlCl3
35.0g Cl mol Cl mol AlCl g AlCl3
71.0 g Cl mol Cl mol AlCl3
= 43.9g AlCl3
Limiting Reactant

13. Percent Yield actual yield x 100% 2) theoretical yield
22/11/99 1)
Yield: the amount of product
Theoretical yield: the amount of product we expect, based on stoichiometric calculations
Actual yield: amount of product from a procedure or experiment (this is given in the question)
Percent yield:
x 100%
actual yield
theoretical yield 2)
Not all product is recovered (e.g. spattering)
Reactant impurities (e.g. weigh out 100 g of chemical which has 20 g of junk)
A side reaction occurs (e.g. MgO vs. Mg3N2)
The reaction does not go to completion

14. Sample problem
22/11/99
What is the % yield of H2O if 138 g H2O is produced from 16 g H2 and excess O2?
Step 1: write the balanced chemical equation
2H2 + O2  2H2O
Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated:
1 mol H2
2.02 g H2 x
2 mol H2O
2 mol H2 x
18.02 g H2O
1 mol H2O x
# g H2O=
16 g H2
143 g =
Step 3: Calculate % yield
actual
theoretical
138 g H2O
143 g H2O =
% yield =
x 100%
x 100%
96.7% =