### Stoichiometry

```Chapter 8
Stoichiometry:
Moles meet Reactions!
Calculating quantities from
chemical reactions.
Write the equation for combination
reaction between sodium and chlorine.
2 Na
+
Cl2
2 NaCl
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.
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.
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
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 x --------------- = 9.0 mol Na
1 mol Cl2
G:M:M:G
Mole- Gram Conversions
We already know that 1 mol = gfm
2Na + 1Cl2  2 NaCl
Calculate the gfm’s:
1 mol Na = ________
23 g
1(23)
1 mol Cl2 = ________
70 g
2(35)
1 mol NaCl = ________
58 g
1(23) + 1(35)
Problems
Now we will use BOTH
the mole ratios and the
gfms to solve problems.
1
2
3
4
1. How many moles of Cl2 are
needed to form 50.0 g of NaCl?
1 mol NaCl 1 mol Cl2
50.0 g NaClx -------------- x -------------- = mol Cl2
2 mol NaCl
58 g NaCl
= .431 mol Cl2
G:M:M:G
2. How many grams of NaCl are
formed from 3.1 moles of Na?
2 mol NaCl 58 g NaCl
3.1 mol Na x -------------- x -------------- = g NaCl
2 mol Na
1 mol NaCl
= 180 g NaCl
G:M:M:G
3. How many grams of Na are
needed to react with 24.5g of Cl2?
1 mol Cl2 2 mol Na 23 g Na
24.5 g Cl2 x ------------ x ------------ x -----------1 mol Cl2
1 mol Na
70 g Cl2
= 16.1 g Na
G:M:M:G
4. How many grams of Cl2 are
needed to form 44.0 g of NaCl?
1 mol NaCl 1 mol Cl2
70g Cl2
44.0 g NaClx ------------ x ------------ x -----------58 g NaCl 2mol NaCl 1 mol Cl2
= 26.6g Cl2
G:M:M:G
Limiting Reagent Problems
Note - A reagent is the same as a reactant.
Limiting Reagent -
the reactant that limits the
amount of product formed.
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.
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 = 2(1)
=2g
O2 = 2(16)
= 32 g
H2O = 2(1) + 1(16) = 18 g
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.
1 mol H2 2 mol H2O 18 g H2O
10.0gH2 x ----------- x ------------ x -----------2 g H2
2 mol H2 1 mol H2O
= 90.0 g H2O
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:M:M:G
= 22.5 g H2O
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.
• % yield
• is the ratio of actual yield (lab) to the
theoretical yield (calculated)
% yield =
Actual / Theoretical x 100
% yield = 20.2 / 22.5 X 100 =
= 89.78%
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.
1 mol Cl2 2 mol NaCl 58 g NaCl
20.0 gCl2x ----------- x ------------ x -----------70 g Cl2
1 mol Cl2 1 mol NaCl
Limiting Reagent
Theoretical Yield
= 33.1 g NaCl
1 mol Na 2 mol NaCl 58 g NaCl
20.0 g Nax ----------- x ----------- x -----------23 g Na 2 mol Na 1 mol NaCl
= 50.4 g NaCl
2Na + 1Cl2  2 NaCl
G:M:M:G
While creating NaCl in the lab, 27.8g on salt
were produced. What is the % yield of salt?
% Yield = Actual / Theoretical x 100
% yield = 22.87 / 33.1 x 100 =
= 69.09%
```