Transcript Chapter 3: Stoichiometry

Chapter 3: Stoichiometry
●
“measuring elements”
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Must account for ALL atoms in a chemical reaction
→
+
2 H2
+
O2
→
2 H 2O
Chapter 3: Stoichiometry
→
+
2 CO
+
+
2 CO2
→
+
CH4
→
O2
4 Cl2
→
+
CCl4
+
4 HCl
Chapter 3: Stoichiometry
C 2H 4
3 O2
+
2 Al
+
2 NH4NO3
H2O
6 HCl
→
2 N2
→
2CO2
→
2 AlCl3
+
O2
+
+
+
4
3 H2
2 H 2O
Chapter 3: Stoichiometry
Three basic reaction types:
●
Combination Reactions
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Decomposition Reactions
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Combustions (in air)
Chapter 3: Stoichiometry
Combination Reactions
Two or more reactants combine to form a single product
C (s) + O2 (g)
→
CO2 (g)
Decomposition Reactions
A single reactant breaks into two or more products
2 KClO3 (s)
→
2 KCl (s)
+
3 O2 (g)
Chapter 3: Stoichiometry
Combustions in Air = reactions with oxygen
Write the balanced reaction equation for the combustion of magnesium to
magnesium oxide:
Chapter 3: Stoichiometry
Combustions of Hydrocarbons in Air
= reactions with oxygen to form carbon dioxide and water
(complete combustion)
Write the balanced reaction equation for the combustion of C2H4 gas
C2H4 (g)
Chapter 3: Stoichiometry
C2H4 (g)
+
3 O2 (g)
→
→
+
2 CO2 (g)
+
2 H2O (g)
+
How many C2H4 molecules are in the flask?
If you know the weight of one molecule of C2H4
and the total weight of gas in the flask, you can
calculate the number of molecules in the flask
●
Chapter 3: Stoichiometry
Molecular weight / Formula weight:
=> sum of all atomic weights in molecular formula
for molecular compounds
MW of C2H4
FW of Mg(OH)2
for ionic compounds
Chapter 3: Stoichiometry
Ca(NO3)2
Type of compound:
Ions:
Total number of oxygen atoms:
Name:
Chapter 3: Stoichiometry
Ca(NO3)2
Percentage of oxygen, by mass:
(1) total mass of Ca(NO3 )2 in amu
(2) mass of oxygen in compound, in amu
(3) percentage of oxygen
Chapter 3: Stoichiometry
Number of individual molecules are difficult to deal with
=> definition of a “package” of molecules or particles
.
... .. .. . .
. . .
. .. ... .... .....
.. . .. . .
..
1 dozen eggs
=
12 individual eggs
1 soda sixpack
=
6 individual bottles of soda
1 mole of molecules
=
6.02 x 1023 individual molecules
Avogadro's Number
Chapter 3: Stoichiometry
1 dozen eggs
=
12 individual eggs
How many moles of eggs are in an egg carton that holds a dozen eggs?
12individual eggs ×
Chapter 3: Stoichiometry
1 dozen eggs
=
12 individual eggs
How many moles of eggs are in an egg carton that holds a dozen eggs?
Chapter 3: Stoichiometry
Ca(NO3)2
How many moles of oxygen are in 2.4 moles of Ca(NO3)2 ?
Chapter 3: Stoichiometry
Molar Mass
= mass of one mole of a substance in grams
FW or MW of substance in amu's = mass of 1mole of substance in grams
FW of Ca(NO3)2 = 164.1 amu
Molar Mass of Ca(NO3)2 = 164.1 g/mol
MW of O2 = 2 x 16.0 amu = 32 amu
Molar Mass of O2 = 32 g/mol
Chapter 3: Stoichiometry
What is the mass in grams of 0.527 moles of CH3OH?
(1) determine molar mass of CH3OH
32 g/mol
(2) use MM to convert moles into grams
16.9 g CH3OH
Chapter 3: Stoichiometry
mol
→
←
Molar Mass
gram
How many hydrogen atoms are in 4.5 g of CH3OH?
= 0.14 mol CH3OH
= 0.56 mol H atoms
= 3.4 x 1023 H atoms
Chapter 3: Stoichiometry
Quantitative Information from Balanced Equations
How many grams of CO2 would be produced by the combustion
of 2 moles of CO?
2 CO
+
→
O2
2 CO2
→
+
2 moles CO
+ 1 mole O2
2 x 28 g
+
32 g
→
→
2 moles CO2
2 x 44 g
= 88 g
Chapter 3: Stoichiometry
Quantitative Information from Balanced Equations
2 CO
+
CO2
O2
→
2
You can write a series of stoichiometric factors for this reaction:
2mol CO
1 mol O2
1 mol O2
2mol CO
1 mol O2
2 mol CO2
Chapter 3: Stoichiometry
How many grams of H2O are formed from the complete combustion
of 2 g of C2H4?
18 g/mol
28 g/mol
C2H4 (g)
1 mol
+
3 O2 (g)
→
2 CO2 (g)
+
2 H2O (g)
2 mol
= 2.6 g H2O
Chapter 3: Stoichiometry
Summary
1) determine equation for the reaction
2) balance equation
3) formulate problem:
how much of A
=>
gets converted into how much of B
4) determine MW/FW of substances involved
5) determine stoichiometric factors from balanced equation
Chapter 3: Stoichiometry
Limiting Reactants
+
+
2 of these will
be left over
“in excess”
Limiting Reactant
Chapter 3: Stoichiometry
Limiting Reactants
Limiting Reactant
- limits the amount of product that can be formed
- reacts completely (disappears during the reaction)
- other reactants will be left over, i.e. in excess
Chapter 3: Stoichiometry
Limiting Reactants
N2
Available:
3 mol N2
+
,
3 H2
→ 2 NH3
6 mol H2
How much H2 would we need to completely react 3 mol N2:
= 9 mol H2
How much NH3 can we form with the available reagents?
= 4 mol NH3
Chapter 3: Stoichiometry
Limiting Reactants
N2
Available:
3 mol N2
+
,
3 H2
→ 2 NH3
6 mol H2
How much N2 is left over (in excess)?
= 1 mol N2
Chapter 3: Stoichiometry
Limiting Reactants
2 Al + 3 Cl2 → 2 AlCl3
Available: 0.5 mol Al , 2.5 mol Cl2
How much Cl2 would we need to completely react 0.5 mol Al:
= 0.75 mol Cl2
How much AlCl3 can we form with the available reagents?
= 0.5 mol AlCl3
Chapter 3: Stoichiometry
Theoretical Yield
2 Al
+
3 Cl2
→ 2 AlCl3
Available: 0.5 mol Al , 2.5 mol Cl2
What mass do 0.5 mol AlCl3 correspond to?
= 67 g AlCl3
The maximum mass of product that can be formed is the
theoretical yield
Chapter 3: Stoichiometry
Theoretical Yield
2 Al
+
3 Cl2
→ 2 AlCl3
Available: 0.5 mol Al , 2.5 mol Cl2
Fritz does the reaction with the available reagents he only ends up with
34g. What is the % yield of the reaction?
% yield =
actual yield
× 100 %
theoretical yield
% yield 
34 g
 100 % = 51 %
67 g
Chapter 3: Stoichiometry
Summary
Determine availabe
quantity of reactants
in moles
Determine % yield
of the reaction
Determine if one of
the reactants is a
limiting reactant
Compare with
actual
amount of product
recovered
(actual yield)
Determine
the maximum
# of moles of product
that can be formed
Convert into
grams of
product
(theoretical
yield)
Chapter 3: Stoichiometry
Consider the combustion of methanol:
2 CH3OH + 3 O2
→
2 CO2 + 4 H2O
What is the theoretical yield of water if 44 g of methanol are reacted with
128 g of oxygen?
 1.4 mol CH 3OH
 4.0 mol O2
Chapter 3: Stoichiometry
2 CH3OH + 3 O2
available
(initial) moles
1.4 mol
→
2 CO2 + 4 H2O
4.0 mol O2
= 2.1 mol O2
Chapter 3: Stoichiometry
2 CH3OH + 3 O2
→
2 CO2 + 4 H2O
(3) how many product moles can be formed with limiting reactant ?
= 1.4 mol CO2
= 2.8 mol H2O
Chapter 3: Stoichiometry
2 CH3OH + 3 O2
→
2 CO2 + 4 H2O
(4) What is the mass of H2O formed (theoretical yield )?
 50g H2O