Warm-up
What is the mass of one mole of NH3?
How many molecules is this?
1
Stoichiometry
Calculating the quantities of
reactants & products in a
chemical reaction.
2
Stoichiometry
Puts together 3 concepts:
1. Writing chemical formulas
2. The mole
3. Balanced chemical equations
3
Stoichiometry
Anitec
AgNO3 + KBr  AgBr(s) + KNO3
$200/kg
x 500 kg
=$100,000
photographic
emulsion
4
Stoichiometry
Analogy: Baking a cake (recipe)
It won’t come out right if the
ingredients aren’t in the right
proportion.
5
Interpreting a
Chemical Equation
N2(g) + 3H2(g)  2NH3(g)
2
atoms
6
atoms
8
atoms
# atoms conserved
6
Interpreting a
Chemical Equation
N2(g) + 3H2(g)  2NH3(g)
1
3
molecule molecules
2
molecules
# molecules not conserved
7
Interpreting a
Chemical Equation
N2(g) + 3H2(g)  2NH3(g)
1
mole
3
moles
2
moles
# moles not conserved
8
Interpreting a
Chemical Equation
N2(g) + 3H2(g)  2NH3(g)
1 mol x 3 mol x
2 mol x
28 g/mol 2 g/mol
17 g/mol
28 g + 6 g
=
34 g
mass conserved !!!
9
Interpreting a
Chemical Equation
N2(g) + 3H2(g)  2NH3(g)
2 mol x
3 mol x
1 mol x
22.4 L/mol
22.4 L/mol 22.4 L/mol
22.4 L
67.2 L
44.8 L
volume not conserved
10
You Try It !
2CO(g) + O2(g)  2CO2(g)
Interpret this equation in terms of:
1. Number of atoms
2. Number of moles
3. Masses of all chemicals
4. Volume of gases at STP
5. What is conserved?
11
Mole-Mole Calculations
N2(g) + 3H2(g)  2NH3(g)
 Conversion factors:
1 mol N2 = 3 mol H2 = 2 mol NH3
How many moles of H2 react
with 1 mole of N2 ?
12
Mole-Mole Calculations
N2(g) + 3H2(g)  2NH3(g)
1 mol N2 = 3 mol H2 = 2 mol NH3
How many moles of H2 react
with 0.6 mole of N2 ?
3 mol H2
0.6 mol N2 x
1 mol N2
= 1.8 mol H2
13
Mole-Mole Calculations
N2(g) + 3H2(g)  2NH3(g)
You try it.
How many moles of H2 are needed
to make 8.3 moles of NH3 ?
14
Mole-Mole Calculations
H
4Al + 3O2  2Al2O3
You try it:
How many moles of Al and O2 are
needed to prepare 0.56 moles of Al2O3 ?
15
Mole-Mole Calculations
16
Gram-Gram Calculations
Since you cannot measure moles
directly in the lab, many calculations
are done with mass (grams).
17
N2(g) + 3H2(g)  2NH3(g)
6.2 g
?g
How many grams of NH3 can be
prepared from 6.2 g of N2 ?
6.2 g N2
1 mol N2
2 mol NH3
x
x
1
28 g N2
1 mol N2
17
g
NH
3 = 7.5 g NH
x
3
1 mol NH3
18
4Al(s) + 3O2(g)  2Al2O3(s)
You try it.
How many grams of
oxygen gas are needed to
prepare 57 g of Al2O3 ?
Label the substances in
each conversion factor.
19
New “Road Map”
Need balanced Particle
Particle
chemical equation
Mass
Volume
Mole A
Mole B
Stoichiometric
step
Mass
Volume
20
2H2O(l)  2H2(g) + O2(g)
How many molecules of oxygen are
produced by decomposing 37 g of water ?
Particle
Particle
Mass
Volume
Mole A
Mole B
Mass
Volume
21
Practice
NaN3 is used in air bags.
2NaN3(s)  2Na(s) + 3N2(g)
How many liters of N2 (at STP) can
be produced from 60.0 g NaN3.
22
Stoichiometry
23
Limiting Reagent
How many PB&J
sandwiches can you make?
Big jar of peanut butter
2 gallons of jelly
4 slices of bread
24
Limiting Reagent
BREAD: “Limiting reagent”
determines how much product can
be made because it runs out first.
PB and J: “Excess reagents” since
each of these will be left over.
25
Limiting Reagent
2Na + Cl2  2NaCl
If 6.7 moles of Na reacts with
2.4 moles of Cl2
what is the limiting reagent?
how many moles of NaCl
are produced?
how much excess reagent remains?
26
2Na + Cl2  2NaCl
6.7 mol
2.4 mol
? mol
2
mol
NaCl
6.7 mol Na x
= 6.7 mol NaCl
2 mol Na
excess
2
mol
NaCl
2.4 mol Cl2 x
= 4.8 mol NaCl
1 mol Cl2
limiting
produced
2
mol
Na
2.4 mol Cl2 x
= 4.8 mol Na
used up
1 mol Cl2
Excess: 6.7 mol – 4.8 mol = 1.9 mol Na
27
Limiting Reagent: Method
1. Calculate amount of product from
each reactant using stoichiometry.
2. Least amount of product determines
the limiting reagent, and this is the
amount of product made.
3. Use the limiting reagent to determine
the amount of excess reagent used up.
28
Limiting Reagent
2Cu + S  Cu2S
80.0 g
25.0 g
?g
How many grams of Cu2S is produced
when 80.0 g Cu is reacted with 25.0 g S ?
29
Convert Cu & S to Product
2Cu + S  Cu2S
80.0 g
25.0 g
?g
Starting with Cu:
1
mol
Cu
S
1
mol
Cu
2
x
80.0g Cu x
2 mol Cu
63.5 g Cu
159
g
Cu
S
2
x
= 100. g Cu2S
1 mol Cu2S
30
Convert Cu & S to Product
2Cu + S  Cu2S
80.0 g
25.0 g
?g
Starting with S:
1
mol
Cu
S
1
mol
S
2
25.0 g S x
x
1 mol S
32.1 g S
x 159 g Cu2S = 124 g Cu S
2
1 mol Cu2S
31
Determining Limiting
Reagent & Amount of Product
2Cu + S  Cu2S
80.0g
25.0 g
100. g
124 g
Thus Cu is limiting and 100 g Cu2S
can be produced.
S is in excess. How much excess?
32
Amount of Excess Reagent
2Cu + S  Cu2S
80.0g
25.0 g 100. g
Convert limiting to excess reagent:
80.0g Cu x 1 mol Cu x 1 mol S
2 mol Cu
63.5 g Cu
32.1 g S
x
= 20.2 g S
4.8 g excess S
1 mol S
33
Give It a Shot
CH4 + 2O2  2H2O +CO2
How many grams of CO2 can be
prepared by reacting 95.0 L of O2
(at STP) with 25.0 g of CH4?
How much of the excess reagent is
left over?
34
Percent Yield
Calculating the amount of product from
a reaction based on stoichiometry gives
the “theoretical yield.”
Analogy: test grade
35
Percent Yield
Real chemical reactions do not give
100 % theoretical yield.
Impurities
Side reactions
Loss of product during handling
This results in an “actual yield.”
36
Percent Yield
from experimental data
actual yield (g)
x 100
% yield =
theoretical yield (g)
from stoichiometry
37
Percent Yield
2KClO3  2KCl + 3O2
What is the percent yield if
12.5 g of O2 are produced by
decomposing 43.0 g KClO3 ?
38
Percent Yield: You Try It !!
2Na + Cl2  2NaCl
What is the percent yield
of NaCl if 30.5 g of NaCl
are produced by reacting
15.2 g Na with excess Cl2 ?
39
40
Warm-up
2H2O(l)  2H2(g) + O2(g)
How many moles of each
substance are reacting in this
chemical equation?
41
Warm-up
For the reaction:
2H2 + O2  2H2O
•How many moles of water can be
made from 7.6 moles of oxygen
and excess hydrogen?
42
Warm-up
For the reaction:
2H2 + O2  2H2O
•How many grams of water can be
made from 12 g of oxygen?
43
Warm-up
For 4.8 moles of the gas SO3,
calculate the :
•mass in grams
•number of molecules
•liters of gas at STP
44
Warm-up
2KClO3  2KCl + 3O2
How many moles of oxygen can be
made from 5.2 moles of KClO3?
How many grams of oxygen can
be made from 27.3 g KClO3?
45
Warm-up
For the reaction:
2Na + Cl2  2NaCl
How many formula units of NaCl can
be made from 56 L of Cl2 at STP?
46
Warm-up
N2(g) + 3H2(g)  2NH3(g)
How many grams of hydrogen will
react with 103 g of nitrogen?
47
Warm-up
N2(g) + 3H2(g)  2NH3(g)
How many liters of hydrogen will react
24
with 3.16 x 10 molecules of nitrogen?
Particle
Particle
Mass
Volume
Mole A Mole B
Mass
Volume
48
Warm-up
How many grams of copper(II) chloride
are needed to react with 42 g Al in a
single replacement reaction?
49
Warm-up
2CO(g) + O2(g)  2CO2(g)
How many liters of CO are needed
to produce 27.2 g CO2?
Particle
Particle
Mass
Volume
Mole A Mole B
Mass
Volume
50
Lab Warm-up
Predict the products and balance:
CuCl2(aq) + Fe(s)  ?
+2
•if Fe is in one of the products
+3
•if Fe is in one of the products
•what is the Fe to Cu mole ratio
in each case?
51
Warm-up
For the reaction:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
•How many mL water is produced by
reacting 24.0L CH4 and 1.02 x 1024 mlc O2?
•How much excess reagent is left over?
Careful: water is a liquid!
52
Warm-up
Write a balanced equation for the reaction
of aluminum with aqueous zinc nitrate.
For this reaction, if 38.0 g Al are mixed
with 322 g of zinc nitrate, how many
grams of zinc are produced.
How much excess reagent remains?
53
Warm-up
CuCl2(aq) will react with Al(s) in a single
replacement reaction.
For this reaction, if 15 g Al are mixed
with 95 g CuCl2, it is found that 35 g Cu
are produced.
How much excess reagent is left?
What is the percent yield?
54
Lab 12 Analysis
CuCl2 + Fe  Cu + FeCl2
3CuCl2 + 2Fe  3Cu + 2FeCl3
A.
B.
C.
D.
DATA TABLE
Mass empty dry beaker_________
Mass iron nails (before rxn)______
Mass iron nails (after rxn)_______
Mass beaker & dry product______
4. g Fe = ?
6. g Cu = ?
5. g Fe  mol Fe 7. g Cu  mol Cu
55