1. Chapter 8 Stoichiometry: Moles meet Reactions!
Calculating quantities from chemical reactions.

2. Write the equation for combination reaction between sodium and chlorine.
Na Cl NaCl 2 2
The above equation can mean:
1. 2 Na atoms + 1 Cl2 molecule yield
2 NaCl formula units.
2. 2 doz. Na atoms + 1 doz. Cl2 molecules yield
2 doz. NaCl formula units.
3. 2 mol Na + 1 mol Cl2 yields 2 mol NaCl
Mole Ratio!!
The COEFFICIENTS in the balanced equation tell ______
______________________________________________
the
relative number of moles of reactants and products.

3. Mol – Mol Conversions using Mole Ratio
From the coefficients in the balanced equation we can write mole ratios:
2 mol Na = 1 mol Cl2
2 mol Na = 2 mol NaCl (equals 1:1 ratio)
1 mol Cl2 = 2 mol NaCl
These mole ratios can be used as conversion factors to solve problems.

4. mole ratio = 1mol Cl2: 2mol NaCl
Problem - 1
We can now use the appropriate mole ratio to solve each problem:
Ex. 1 How many moles of NaCl will form from 2.7 moles of Cl2 ?
mole ratio = 1mol Cl2: 2mol NaCl
2 mol NaCl
2.7 mol Cl2
x =
5.4
mol NaCl
1 mol Cl2
G:M:M:G

5. mole ratio = 2 mol Na : 1 mol Cl2
Problem - 2
How many moles of Na are needed to react with 4.5 mol of Cl2?
mole ratio = 2 mol Na : 1 mol Cl2
2 mol Na
4.5 mol Cl2
9.0
mol Na
x =
1 mol Cl2
G:M:M:G

6. Mole- Gram Conversions
We already know that 1 mol = gfm
2Na + 1Cl2  2 NaCl
Calculate the gfm’s:
1 mol Na = ________ g 23
1(23)
1 mol Cl2 = ________ g 70
2(35)
1 mol NaCl = ________ g 58
1(23) + 1(35)

7. Now we will use BOTH the mole ratios and the gfms to solve problems.1 2 3 4

8. 1. How many moles of Cl2 are needed to form 50.0 g of NaCl?
mol NaCl
mol Cl2 1
50.0 g NaCl x x
= mol Cl2
g NaCl 2
mol NaCl 58
= .431 mol Cl2
G:M:M:G

9. 2. How many grams of NaCl are formed from 3.1 moles of Na?
mol NaCl
g NaCl 58
3.1 mol Na x x
= g NaCl 2
mol Na 1
mol NaCl
= 180 g NaCl
G:M:M:G

10. How many grams of Na are needed to react with 24.5g of Cl2?1
mol Cl2
g Na 2
mol Na 23
24.5 g Cl2 x x x 1
mol Cl2
g Cl2 1
mol Na 70
= g Na
16.1
G:M:M:G

11. 4. How many grams of Cl2 are needed to form 44.0 g of NaCl?1
mol NaCl 1
mol Cl2 70
g Cl2
44.0 g NaCl x x x 2
mol NaCl 58
g NaCl 1
mol Cl2
= g Cl2
26.6
G:M:M:G

12. Limiting Reagent Problems
Note - A reagent is the same as a reactant.
the reactant that limits the
amount of product formed.
Limiting Reagent -
The Limiting Reagent is the reactant that determines how much product can be produced. Because you run out of this reactant no more product is made.

13. Limiting Reagent Example 1: When 10.0 grams of hydrogen combines
with 20.0 grams of oxygen, how many grams of water will be produced?
Write the balanced equation:
2H2 + O2  2H2O
H O N Cl Br I F
Calculate the gfms:
H2 =
O2 =
H2O =
2(1)
2(16)
2(1) + 1(16)
= 2 g
= 32 g
= 18 g

14. G:M:M:G = g H2O = g H2O 22.5 mol H2 1 2 mol H2O 18 g H2O 10.0gH2 2
We don’t know which reactant will limit the amount of water that is formed, so we must figure out the amount of product for each of the given amounts of reactants.
In essence we perform 2 separate problems.
mol H2 1 2
mol H2O 18
g H2O
10.0gH2 x x x 2
g H2 2
mol H2 1
mol H2O
= g H2O
90.0 1
mol O2 2
mol H2O 18
g H2O
20 g O2 x x x 32
g O2 1
mol O2 1
mol H2O
= g H2O
22.5
G:M:M:G

15. When the two given amounts of reactants are
combined in a chemical reaction, the lesser
amount calculated above will be the amount that
actually forms. The reactant that yields the
LEAST amount of product is called the “limiting
reagent.”
The amount of product formed in the above reaction:
22.5 g
Theoretical Yield
The limiting reagent in the above reaction is:
Oxygen
Lets say students performed this reaction and they obtained 20.2 g of water. This is the Actual Yield.

16. % yield
is the ratio of actual yield (lab) to the theoretical yield (calculated)
% yield =
Actual / Theoretical x 100
% yield = / X 100 =
= 89.78%

17. 2Na + 1Cl2  2 NaCl G:M:M:G = g NaCl = g NaCl mol Cl2 1 2 mol NaCl 58
If 20.0 g of chlorine is combined with 20.0 g of
sodium, how many g of NaCl will form?
In essence we perform 2 separate problems.
mol Cl2 1 2
mol NaCl 58
g NaCl
20.0 gCl2 x x x 70
g Cl2 1
mol Cl2 1
mol NaCl
Limiting Reagent
= g NaCl
33.1
Theoretical Yield 1
mol Na 2
mol NaCl 58
g NaCl
20.0 g Na x x x 23
g Na 2
mol Na 1
mol NaCl
= g NaCl
50.4
2Na + 1Cl2  2 NaCl
G:M:M:G

18. While creating NaCl in the lab, 27. 8g on salt were produced
While creating NaCl in the lab, 27.8g on salt were produced. What is the % yield of salt?
% Yield = Actual / Theoretical x 100
% yield = / 33.1 x 100 =
= 69.09%