Transcript Chapter 9 Stoichiometry

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Section 8.1
The Arithmetic of Equations

OBJECTIVES:

Calculate the amount of reactants required, or
product formed, in a non chemical process.
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Section 8.1
The Arithmetic of Equations

OBJECTIVES:

Interpret balanced chemical equations in terms of
interacting moles, representative particles, masses,
and gas volume at STP.
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Cookies?

When baking cookies, a recipe is usually used,
telling the exact amount of each ingredient
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If you need more, you can double or triple the
amount
Thus, a recipe is much like a balanced
equation
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Stoichiometry
Greek for “measuring elements”
 The calculations of quantities in chemical
reactions based on a balanced equation.
 We can interpret balanced chemical equations
several ways.

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1. In terms of Particles
Element- made of atoms
 Molecular compound (made of only nonmetals) = molecules
 Ionic Compounds (made of a metal and nonmetal parts) = formula units (ions)
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2H2 + O2  2H2O

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Two molecules of hydrogen and one molecule of oxygen
form two molecules of water.
2 Al2O3 Al + 3O2
2 formula units
Al2O3
and 3 molecules
form
4 atoms Al
O2
2Na + 2H2O  2NaOH + H2
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Look at it differently
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2H2 + O2  2H2O
2 dozen molecules of hydrogen and 1 dozen molecules of
oxygen form 2 dozen molecules of water.
2 x (6.02 x 1023) molecules of hydrogen and 1 x (6.02 x
1023) molecules of oxygen form 2 x (6.02 x 1023)
molecules of water.
2 moles of hydrogen and 1 mole of oxygen form 2 moles
of water.
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2. In terms of Moles
2 Al2O3 Al + 3O2
 2Na + 2H2O  2NaOH + H2
 The coefficients tell us how many moles of
each substance
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3. In terms of Mass
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The Law of Conservation of Mass applies
We can check using moles
2H2 + O2  2H2O
2 moles H2
2.02 g H2
1 mole H2
=
4.04 g H2
1 mole O2
32.00 g O2
1 mole O2
=
32.00 g O2
36.04g gHH
+O
36.04
+O
2
2
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In terms of Mass

2H2 + O2  2H2O
2 moles H2O
18.02 g H2O
1 mole H2O
= 36.04 g H2O
2H2 + O2  2H2O
36.04 g H2 + O2
= 36.04 g H2O
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4. In terms of Volume

2H2 + O2  2H2O
At STP, 1 mol of any gas = 22.4 L

(2 x 22.4 L H2) + (1 x 22.4 L O2)  (2 x 22.4 L H2O)
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NOTE: mass and atoms are always conservedhowever, molecules, formula units, moles, and
volumes will not necessarily be conserved!
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Practice:

Show that the following equation follows the
Law of Conservation of Mass:

2 Al2O3 Al + 3O2
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Section 8.2
Chemical Calculations

OBJECTIVES:
 Construct mole ratios from balanced chemical
equations, and apply these ratios in mole-mole
stoichiometric calculations.
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Section 8.2
Chemical Calculations

OBJECTIVES:

Calculate stoichiometric quantities from balanced
chemical equations using units of moles, mass,
representative particles, and volumes of gases at
STP.
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Mole to Mole conversions

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2 Al2O3 Al + 3O2
each time we use 2 moles of Al2O3 we will also make 3
moles of O2
2 moles Al2O3
3 mole O2
or
3 mole O2
2 moles Al2O3
These are possible conversion factors
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Mole to Mole conversions
How many moles of O2 are produced when
3.34 moles of Al2O3 decompose?
 2 Al2O3 Al + 3O2

3.34 mol Al2O3
3 mol O2
2 mol Al2O3
= 5.01 mol O2
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Practice:
2C2H2 + 5 O2  4CO2 + 2 H2O
• If 3.84 moles of C2H2 are burned, how many moles
(9.6 mol)
of O2 are needed?
•How many moles of C2H2 are needed to produce
(8.95 mol)
8.95 mole of H2O?
•If 2.47 moles of C2H2 are burned, how many
moles of CO2 are formed?
(4.94 mol)
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How do you get good at this?
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Mass-Mass Calculations
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We do not measure moles directly, so what can we do?
We can convert grams to moles
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Then do the math with the mole ratio
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Use the Periodic Table for mass values
Balanced equation gives mole ratio!
Then turn the moles back to grams
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Use Periodic table values
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For example...
If 10.1 g of Fe are added to a solution of
Copper (II) Sulfate, how much solid copper
would form?
 2Fe + 3CuSO4  Fe2(SO4)3 + 3Cu
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Answer = 17.2 g Cu
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More practice...
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How many liters of CO2 at STP will be
produced from the complete combustion of
23.2 g C4H10 ?
Answer = 35.8 L CO2
What volume of Oxygen would be
required?
Answer = 58.2 L O2
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Volume-Volume Calculations
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How many liters of CH4 at STP are required to
completely react with 17.5 L of O2 ?
CH4 + 2O2  CO2 + 2H2O
17.5 L O2
1 mol O2
22.4 L O2
1 mol CH4
2 mol O2
22.4 L CH4
1 mol CH4
= 8.75 L CH4
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Avogadro told us:
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Equal volumes of gas, at the same temperature and
pressure contain the same number of particles.
Moles are numbers of particles
You can treat reactions as if they happen liters at a
time, as long as you keep the temperature and
pressure the same.
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Shortcut for Volume-Volume:
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How many liters of H2O at STP are produced by
completely burning 17.5 L of CH4 ?
CH4 + 2O2  CO2 + 2H2O
17.5 L CH4
2 L H2O
1 L CH4
= 35.0 L H2O
Note: This only works for Volume-Volume
problems.
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Section 8.3
Limiting Reagent & Percent Yield

OBJECTIVES:
 Identify and use the limiting reagent in a
reaction to calculate the maximum amount of
product(s) produced, and the amount of excess
reagent.
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Section 8.3
Limiting Reagent & Percent Yield

OBJECTIVES:
 Calculate theoretical yield, actual yield, or
percent yield, given appropriate information.
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“Limiting” Reagent
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If you are given one dozen loaves of bread, a gallon of
mustard, and three pieces of salami, how many salami
sandwiches can you make?
The limiting reagent is the reactant you run out of first.
The excess reagent is the one you have left over.
The limiting reagent determines how much product you
can make
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How do you find out?
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Do two stoichiometry problems.
The one that makes the least product is the limiting
reagent.
For example
Copper reacts with sulfur to form copper ( I )
sulfide. If 10.6 g of copper reacts with 3.83 g S
how much product will be formed?
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If 10.6 g of copper reacts with 3.83 g S. How many grams
of product will be formed?
2Cu + S  Cu2S
10.6 g Cu
3.83 g S
Cu is the Limiting
1 mol
Cu2S it 159.16 g Cu2S
1 mol Cu
Reagent
because
2less
molproduct
Cu
63.55g creates
Cu
1 mol Cu2S
= 13.3 g Cu2S
1 mol S
1 mol Cu2S
32.06g S
1 mol S
159.16 g Cu2S
1 mol Cu2S
= 19.0 g Cu2S
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Another example
If 10.1 g of magnesium and 2.87 L of HCl gas
are reacted, how many liters of gas will be
produced?
 How many grams of solid?
 How much excess reagent remains?
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Limiting Reagent Simulation
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Still another example
If 10.3 g of aluminum are reacted with 51.7 g
of CuSO4 how much copper will be produced?
 How much excess reagent will remain?
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%
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Yield
The amount of product made in a chemical reaction.
 There are three types:
1. Actual yield- what you get in the lab when the chemicals
are mixed
2. Theoretical yield- what the balanced equation tells should
be made
3. Percent yield =
Actual
X 100
Theoretical
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Example
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6.78 g of copper is produced when 3.92 g of Al are
reacted with excess copper (II) sulfate.
2Al + 3 CuSO4  Al2(SO4)3 + 3Cu
What is the actual yield?
What is the theoretical yield?
What is the percent yield?
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% Yield Problem
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AY is 6.78 g of Cu
TY – Use the balanced equation and do a mass to mass
calculation
2Al + 3 CuSO4 Al2(SO4)3 + 3Cu
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3.92 g
3.92 g Al
xg
1 mol Al
3mol Cu
63.55 g Cu
26.98 g Al
2 mol Al
1 mol Cu
13.9 g Cu
% Yield = (AY / TY) x 100 = (6.78g / 13.9 g) x 100 = 48.9 % yield
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Details
Percent yield tells us how “efficient” a
reaction is.
 Percent yield can not be bigger than 100 %.
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The Heat of Reaction
At the start of a reaction the reactants have a great heat content.
Enthalpy is the amount of heat that a substance has at a given
temperature and pressure (see Table 8.1 pg 190)
The heat of a reaction is the heat that is released or absorbed
during a chemical reaction. Heat of Reaction is represented by
The symbol H
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Heat can be added to make a reaction occur. This is an Endothermic
Reaction. H will be positive. The reaction can be represented as
follows:
2 NH3
N2 + 3H2
H = + 92.38 kJ
or
2 NH3 + 92.38 kJ N2 + 3H2
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Cold Pack – Endothermic Reaction
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When heat is released by a chemical reaction, the reaction is
said to be Exothermic. H will be negative
Heat is most often released by a chemical reaction. The released
heat can be represented by the following:
Na + Cl2
2NaCl
H = - 822.4 kJ
or
Na + Cl2
2NaCl + 822.4kJ
How can you calculate the heat of a reaction???
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Glow Stick – Exothermic Reaction
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Calculate the heat of reaction ( H), in kilojoules, for the reaction
Use table 8.1 pg 190
4 CO2(g)
4 (-393.5)
2O2(g) + 4CO(g)
2 (0)
(-1574)
(0)
+ 4 (-110.5)
+ (-442)
H = NRG of Products – NRG of Reactants
–
H = (-442 kJ)
H = 1132 kJ
(-1574)
(+ indicates an endothermic rxn so)
4 CO2(g) + 1132kJ
2O2(g) + 4CO(g)
How much energy is used to make 100g of CO(g)?
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4 CO2(g) + 1132kJ
x kJ
100 g CO
2O2(g) + 4CO(g)
100g
1 mol CO
1132 kJ
28.01 g CO
4 mol CO
1010 kJ
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