Calculations Involving Reactions
STOICHIOMETRY:
The relationship between the amount of reactants used in a
chemical reaction and the amounts of products produced by
the reaction.
Consider the following chemical reaction:
2 H2(g) + O2(g)  2 H2O(g)
We know that 2 hydrogen molecules react with 1 oxygen
molecule to make 2 molecules of water.
We can also write it as:
200 H2(g) + 100 O2(g)  200 H2O(g)
or even
2 x (6.02 x 1023) H2(g) + 1 x (6.02 x 1023) O2(g)  2 x (6.02 x 1023) H2O(g)
The coefficients can now refer to the amount of molecules, in
moles, which are involved in the reaction.
Therefore, the MOLE RATIO of H2 = 2 is maintained.
O2 1
Consider the equation:
N2(g) + 3 H2(g) 2 NH3(g)
a. How many molecules of N2(g) are required to react with 15
molecules of H2(g)?
b. How many moles of NH3(g) are produced when 18 mol of
H2(g) are reacted?
N2(g) + 3 H2(g) 2 NH3(g)
Exercises p. 124 # 1-5
Stoichiometry Calculations
Involving: Moles, Mass, Gas Volume and Molecules
Stoichiometric calculations allow us to find how much of
“Chemical #2” is involved in a chemical reaction based on
the amount of “Chemical #1” involved.
The relationship between moles of “Chemical #1” and moles
of “Chemical #2” is central to each calculation.
The following diagram shows the kinds of connections
involved:
Example:
The combustion of propane, C3H8(g), proceeds according to
the following equation:
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
a. What mass of CO2(g) is produced by reacting 2.00 mol of
O2(g)?
b. What mass of C3H8(g) is required to produce 100.0 g of
H2O(g)?
c. What mass of H2O(l) is produced if the reaction also
produces 50.0 L of CO2(g) at
STP?
d. What volume of O2(g) at STP is consumed from the air if a
burner produces 10.0 L of CO2(g) at STP?
e. 1.25 x 10-6 g of C3H8(g) is extracted from a powdered rock.
How many molecules of CO2(g) are produced if the gas
sample is burned in the presence of an excess of O2(g)?
Exercises p. 127 #6ac & 7ace
Exercises p. 127 #8,10,11
Hand in #10 – due Friday
Stoichiometry Calculations
Involving: Molar Concentration
Recall:
C= n
V
Where:
n = number of moles
C = concentration in M
V = volume (L)
Stoichiometry calculations are based on problems involving
the relationship between the number of moles (or molecules)
of several species.
Therefore we will largely be using:
n=C×V
Be careful:
When a VOLUME is mentioned, the problem involves
MOLARITY. DO NOT assume “22.4 L”. This is only used
when “at STP” is mentioned.
For example:
Tums is an antacid composed primarily of calcium carbonate
(chalk), and stomach acid is a dilute solution of hydrochloric
acid. The neutralization between CaCO3(s) and stomach acid
is represented by the equation:
CaCO3(s) + 2 HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l)
a) What volume of stomach acid having [HCl] = 0.0010 M is
neutralized by 0.750 g portion of CaCO3(s)?
b) What volume of CO2(g) at STP is produced if 1.25 L of
0.0055 M HCl(aq) reacts with an excess of CaCO3(s)?
CaCO3(s) + 2 HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l)
TITRATION:
A process in which a measured amount of solution is
reacted with a known volume of another solution (one
of the solutions has an unknown concentration) until a
desired EQUIVALENCE POINT is reached.
EQUIVALENCE POINT (STOICHIMETRIC POINT):
The point in a titration where the ratio of the moles of
each species involved exactly equals the ratio of the
coefficients given in the balanced chemical equation.
The equivalence point is recognized by an INDICATOR
(weak acid that changes colour).
For Example:
H3PO4(aq) + 2 KOH(aq)  K2HPO4(aq) + 2 H2O(l)
a) If 19.8 mL of H3PO4(aq) with an unknown concentration
reacts with 25.0 mL of 0.500 M KOH(aq), according to the
above reaction, what is the molarity of the H3PO4(aq)?
b) What volume of 0.200 M KOH(aq) is required to react with
125 mL of 0.250 M H3PO4(aq) in order to produce a solution
of K2HPO4(aq)?
Exercises p. 131 #17-19
Exercises p. 131 #20-23
Stoichiometry Calculations
Involving: Excess Quantities
Reactions frequently are carried out where a reactant is
present in excess.
We may do this because:
i. Deliberately adding an excess of one reactant
makes sure the second one completely reacts.
ii. Unavoidably having a reactant in excess because a
limited amount of the other one is present.
1. If 20.0 g of H2(g) is burnt with 100.0 g of O2(g), according to
the reaction,
2H2 + O2  2H2O
Which reactant is present in excess and by how many
grams?
2. If 56.8 g of FeCl2(aq), 14.0 g of KNO3(aq), and 40.0 g of
HCl(aq) are mixed and allowed to react according to the
equation:
3 FeCl2(aq) + KNO3(aq) + 4 HCl(aq)  3 FeCl3(aq) + NO(g) + 2 H2O(l) + KCl(aq)
a. Which chemical is the LIMITING reactant?
b. How many grams of each “excess reactant” are actually
present in excess?
c. How many grams of NO were actually created?
Exercises p. 133 #26, 28
Exercises p. 134 #29 - 31
% Yield and % Purity
Sometimes 100% of the expected amount of products
cannot be achieved in a reaction.
The term “%Yield” is used to express how much (in percent)
is actually obtained.
There are two major reasons this may happen:
1. The reactants may not all react.
o Not all the pure material actually reacts.
o The reactants may be less than 100% pure.
2. Some of the products were lost during procedures.
Eg: Filtration, crystallization, etc.
If you know the expected amount of product based on a
stoichiometry calculation and measure the actual amount of
product formed, you can calculate the %YIELD of the
reaction.
% Yield =
actual yield •100%
theoretical yield
Another way in which less than the “expected” amount of
product is produced is to start with reactants that are not
100% pure. In this case, the % PURITY is calculated.
pure yield
•100%
impure yield
% yield/purity problems fall into three categories:
% PURITY =
Given: mass of reactant and mass of product
When 15.0 g of CH4(g) is reacted with an excess of Cl2(g)
according to the reaction:
CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g)
a total of 29.7 g of CH3Cl(g) is formed. What is the % yield of
the reaction?
If 100.0 g of FeO(s) produces 12.9 g of pure Fe(s) according
to the reaction:
2 FeO(s) + 2 C(s) + O2(g) → 2 Fe(s) + 2 CO2(g)
what is the % purity of the FeO(s) used?
Given: mass of reactant and % yield/purity
PA – RT
Product mass  calculating actual
Reactant mass  calculating theoretical
What mass of K2CO3(s) is produced when 1.50 g of KO2(s) is
reacted with an excess of CO2(g) according to the reaction:
4 KO2(s) + 2 CO2(g) → 2 K2CO3(s) + 3 O2(g)
if the reaction has a 76.0% yield?
What mass of pure sodium carbonate, Na2CO3(s), can be
formed by heating and decomposing 5000g of 79.4%
Na5(CO3)2(HCO3) · 2 H2O(s), according to the reaction:
2Na5(CO3)2(HCO3) · 2 H2O(s) → 5 Na2CO3(s) + CO2(g) + 5 H2O(l)
Given: mass of product and % yield/purity.
PA – RT
Product mass  calculating actual
Reactant mass  calculating theoretical
What mass of CuO(s) is required to make 10.0 g of Cu(s)
according to the reaction:
2 NH3(aq) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3H2O(l)
If the reaction has a 58.0% yield?
What mass of impure zinc metal having a purity of 89.5% is
required to produce 975 mL of hydrogen gas at STP
according to the reaction:
Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)
Exercises p. 137 #33ab
Exercises p. 137 #33cd, 34, 35ab
What is on the exam?
Stoichiometric Calculations involving:
o Energy
o Moles, Mass, Gas Volume, and Molecules
o Molar Concentration (Titrations)
o Excess and Limiting Reactants
Percentage Purity and Percentage Yield
o 3 types of questions you may see