Transcript Chemical Stoichiometry

Chemical Quantities
Chapter 9
Chemical Stoichiometry
Stoichiometry - The study of quantities of
materials consumed and produced in
chemical reactions.
Stoichiometry is used to determine how
much stomach acid an antacid tablet can
neutralize.
Chemical Equations
A balanced chemical equation is like a
recipe. One needs to know what the
ingredients are and what relative
amounts of ingredients are needed for
both recipes and chemical equations.
Chemical Equation
A representation of a chemical reaction:
C2H5OH(l) + __O2(g)  __CO2(g) + __H2O(g)
reactants
products
Is this equation balanced?
C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(g)
Chemical Equation
C2H5OH (l) + 3O2(g)  2CO2(g) + 3H2O(g)
Microscopic:
1 molecule of ethanol reacts with 3 molecules of
oxygen to produce 2 molecules of carbon dioxide
and 3 molecules of water.
Macroscopic:
1 mole of ethanol reacts with 3 moles of oxygen to
produce 2 moles of carbon dioxide and 3 moles of
water.
Chemical Equation
C2H5OH (l) + 3O2(g)  2CO2(g) + 3H2O(g)
Microscopic:
1 dozen molecules of ethanol reacts with 3 dozen molecules of
oxygen to produce 2 dozen molecules of carbon dioxide and
3 dozen molecules of water.
Macroscopic:
6.02 x 1023 molecules of ethanol reacts with 3(6.02 x 1023)
molecules of oxygen to produce 2(6.02 x 1023) molecules of
carbon dioxide and 3(6.02 x 1023 ) molecules of water.
Chemical Equation
C2H5OH (l) + 3O2(g)  2CO2(g) + 3H2O(g)
Macroscopic:
46.0 g of ethanol reacts with 96.0g of oxygen to
produce 88.0 g of carbon dioxide and 54.0 g of
water.
142.0 g of reactants = 142.0 g of products.
Atoms and mass are conserved in a chemical
reaction, but moles and molecules are not!!!
Moles & Molecules
__C3H8(g) + __O2(g) ---> __CO2(g) + __HOH(g)
C3H8(g) + 5O2(g) ---> 3CO2(g) + 4HOH(g)
How many moles and molecules of each
substance are there?
Mole Ratio
Mole ratio -- a conversion factor based upon a
balanced equation and used to determine relative
amounts of reactants and products.
Mole ratios can exist between a reactant and a
product, between two reactants, or between two
products.
C3H8(g) + 5O2(g) ---> 3CO2(g) + 4HOH(g)
5molO2
3molCO2
3molCO2
5molO2
Types of Stoichiometry
Problems
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Mole to Mole
Gram to Mole
Mole to Gram
Gram to Molecules
Molecules to Gram
Calculating Masses of
Reactants and Products
1.
2.
3.
4.
Balance the equation.
Convert mass to moles.
Set up mole ratios.
Use mole ratios to calculate moles of
desired reactant or product.
5. Convert moles to grams, if necessary.
Mole To Mole Problems
What number of moles of oxygen would be
used in burning 5.8 moles of propane,
C3H8?
C3H8(g) + 5O2(g) ---> 3CO2(g) + 4HOH(g)
(5.8 mol C3H8)(5 mol O2/1 mol C3H8) =
29 mol O2
Gram to Mole & Gram to Gram
__Al(s) + __I2(s) ---> __AlI3(s)
2Al(s) + 3I2(s) ---> 2AlI3(s)
How many moles and how many grams of
aluminum iodide can be produce from 35.0
g of aluminum?
Gram to Mole & Gram to Gram
2Al(s) + 3I2(s) ---> 2AlI3(s)
(35.0 g Al) (1 mol/26.98 g)(2 mol AlI3/2 mol
Al) = 1.30 mol AlI3
(35.0 g Al) (1 mol/26.98 g)(2 mol AlI3/2 mol
Al)(407.68 g/1 mol) = 529 g AlI3
Grams to Molecules
__LiOH(s) + __CO2(g) ---> __Li2CO3(s) + __HOH(l)
2LiOH(s) + CO2(g) ---> Li2CO3(s) + HOH(l)
How many molecules of water would be formed
from 1.00 x 103 g of LiOH?
(1.00 x 103 g LiOH)(1 mol/23.95 g)(1 mol HOH/
2 mol LiOH)(6.02 x 1023 molecules/1 mol)
= 1.26 x 1025 molecules HOH
Stoichiometric Quantities
Stoichiometric Quantities -- quantities of
reactants mixed in exactly the amounts that
result in their all being used up at the same
time.
How often do you think this occurs in reality?
Almost never!!!!
Limiting Reactant
The limiting reactant is the reactant
that is consumed first, limiting the
amounts of products formed.
Almost all stoichiometric situations are
of the limiting reactant type.
The reactants that are left over and
unreacted are said to be in excess.
Figure 9.1: A mixture of 5CH4 and 3H20 molecules
undergoes the reaction CH4(g) + H20(g) ---> 3H2 + CO(g)
Double Cheeseburger Problem
At the local “Burger Barn” a worker finds the
following inventory:
22 hamburger patties
15 hamburger buns
7 slices of onion
18 slices of cheese
How many double cheeseburgers with onion
and cheese can be made to sell?
Double Cheeseburger Problem
2 h.b. patties + 1 h.b. bun + 2 slices cheese +
1 slice onion ---> 1 double cheeseburger
What is the limiting reactant?
Onion
How many double cheeseburgers with onion
and cheese can be made to sell?
7 double cheeseburgers
Double Cheeseburger Problem
2 h.b. patties + 1 h.b. bun + 2 slices cheese +
1 slice onion ---> 1 double cheeseburger
What materials are in excess and by how
much?
8 hamburger patties
8 hamburger buns
4 slices cheese
Solving a Stoichiometry Problem
1.
2.
3.
4.
Balance the equation.
Convert masses to moles.
Determine which reactant is limiting.
Use moles of limiting reactant and mole
ratios to find moles of desired product.
5. Convert from moles to grams.
Limiting Reactant Problem
If 56.0 g of Li reacts with 56.0 g of N2, how
many grams of Li3N can be produced?
__Li(s) + __N2(g) ---> __Li3N(s)
6 Li(s) + N2(g) ---> 2 Li3N(s)
(56.0 g Li) (1 mol/6.94g)(1 mol N2/6 mol Li)
(28.0 g/1 mol) = 37.7 g N2
Since there were 56.0 g of N2 and only 37.7 g
used, N2 is the excess and Li is the Limiting
Reactant.
Limiting Reactant Problem
6 Li(s) + N2(g) ---> 2 Li3N(s)
(56.0 g Li)(1 mol/6.94g)(2 mol Li3N/6 mol Li)
(34.8 g/1 mol) = 93.6 g Li3N
How many grams of nitrogen are left?
56.0g N2 given - 37.7 g used = 18.3 g excessN2
Double Cheeseburger Yield
At the local “Burger Barn” a worker finds the
following inventory:
22 hamburger patties
15 hamburger buns
7 slices of onion
18 slices of cheese
We found that seven double cheeseburgers
could be made from these ingredients.
Double Cheeseburger Yield
If a worker eats one slice of onion, how many
double cheeseburgers can actually be made?
6 double cheeseburgers
The number of cheeseburgers that could have
been made (7) is the theoretical yield.
The number of cheeseburgers that actually
were made (6) is the actual yield.
Double Cheeseburger Yield
actual _ yield
100%
%Yield 
theoretical _ yield
6
%Yield  100% 
7
% Yield = 85.7 %
% Yield
Values calculated using stoichiometry are
always theoretical yields!
Values determined experimentally in the
laboratory are actual yields!
Limiting Reactant & % Yield
If 68.5 kg of CO(g) is reacted with 8.60 kg of
H2(g), what is the theoretical yield of methanol
that can be produced?
__H2(g) + __CO(g) ---> __CH3OH(l)
2 H2(g) + CO(g) ---> CH3OH(l)
(68.5 kg CO)(1 mol/28.0 g)(2 mol H2/1 mol CO)
(2.02 g/1mol) = 9.88 kg H2
Limiting Reactant & % Yield
2 H2(g) + CO(g) ---> CH3OH(l)
Since only 8.60 kg of H2 were provided, the H2 is
the limiting reactant, and the CO is in excess.
(8.60 kg H2)(1000 g/1 kg)(1 mol/2.02 g)(1 mol CH3OH/2
mol H2)(32.0 g/1 mol) = 6.85 x 104 g CH3OH
Limiting Reactant & % Yield
2 H2(g) + CO(g) ---> CH3OH(l)
If in the laboratory only 3.57 x 104 g of
CH3OH is produced, what is the % yield?
actual _ yield
100%
%Yield 
theoretical _ yield
3.57 x10 4 g
100%
%Yield 
4
6.85 x10 g
% Yield = 52.1 %