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This is a Sequential Activity
Group Activity: Stoichiometry
Conservation of Mass at Work
Definition
The word Stoichiometry comes from the Greek
stoicheion, which means to measure the
elements
A good definition of the term’s meaning in
the study of chemistry is the “quantitative
study of reactants and products in a chemical
reaction” (from Chemistry
by Raymond Chang)
Stoichiometry allows one to calculate how much of
a given product a reaction is expected to produce
based on how much of the reactants are available
Given the mass, volume and density, or the number
of moles of reactants, one can calculate the mass,
volume (if the density is known) or moles of product
Molar
Ratios
Calculations using stoichiometry depend on
the molar relationships in chemical equations; this
is why a properly balanced chemical equation is so
important
A properly balanced chemical equation shows the
molar ratios of each of the species present, whether
they are reactants or products
Take the combustion of propane as an example:
C_{3}H_{8} + 5O_{2} -->
3CO_{2} + 4H_{2}O
The ratios found in this equation are as
follows: 1 mol propane:5 mol oxygen
(each mole of C_{3}H_{8} requires
five moles of O_{2} to burn completely) 1 mol propane:3 mol carbon dioxide
(each mole of completely burned
C_{3}H_{8} produces three moles of
CO_{2}) 1 mol propane:4 mol water
(each mole of completely burned
C_{3}H_{8} produces four moles of
H_{2}O) 5 mol oxygen:3 mol carbon dioxide
(for every five moles of O_{2} consumed,
three moles of CO_{2} are produced) 5 mol oxygen:4 mol water
(for every five moles of O_{2} consumed,
four moles of H_{2}O are produced) 3 mol carbon dioxide:4 mol water
(for every three moles of CO_{2} produced,
4 moles of H_{2}O are produced)
Each of these ratios can be written in reverse
order and can be thought of as a fraction or a
conversion factor
The key to using stoichiometry is to calculate
using moles and to convert whatever information you
are given into moles, whether it be mass or volume
and density
It is easy to calculate the number of moles
needed or moles produced if the starting information
is given in moles: just find the appropriate molar
ratio and multiply by the moles in the starting
information (be sure to write the ratio with the
quantity you start with on the bottom of the ratio)
It only requires one additional step to find the
mass of a substance that is needed or produced given
the number of moles of one of the substances: just
convert the answer you find in moles into grams using
the molar mass of the substance
Given the mass of one of the substances you can
find the mass of reactants needed or products
produced by first converting the given mass into a
number of moles, finding the number of moles of the
other substance using the proper molar ratio, then
converting that number of moles into grams
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To add another layer of complexity, say you are
given a volume of a substance and asked to find the
volume needed/produced; for this type of problem
you always need to know the density of both the
substance whose volume is known and the substance
whose volume is the answer to the problem
This type of problem is solved by converting
the volume into a mass using the formula m =
D·V (make sure your units all match in this
equation!); next find the number of moles; next use
that number to find the desired number of moles
using the proper molar ratio; next use the number
of moles to find the mass; finally, use the mass to
find the volume using formula V = m/D
The theoretical yield is the amount of a
given product one would expect would be produced
based solely on the molar ratios and the amount of
each starting material
The starting materials may be present in amounts
that do not match the molar ratios in the chemical
equation
When this is the case one of the reactants is a
limiting reactant (or limiting reagent) and the other
reactant(s) is/are said to be in excess
An excess reactant is one that is not completely
used up when the reaction is complete
The limiting reactant is the one that will be
used up before any of the others
Because of this, the moles of the limiting
reactant are what determine the theoretical yield of
a reaction
To find out which reactant is the limiting
reactant you use the molar ratios given in the
balanced chemical equation
Take the combustion of propane again: if you have
10 mol C_{3}H_{8} and 1 mol
O_{2} you know that oxygen is the limiting
reactant because each mole of propane requires five
moles of oxygen to burn
If you are given masses, the problem is not any
harder; you just have to convert the mass of each
reactant into a number of moles and multiply that
number by the proper ratio to find out how many moles
of the other reactant are needed
If you have more moles of O_{2} than this
calculation shows you will need then
C_{3}H_{8} is the limiting reactant
If you have fewer moles of O_{2} than
this calculation shows you will need then
C_{3}H_{8} is the excess reactant and
O_{2} is the limiting reactant
To find the theoretical yield of a reaction, use
the number of moles of the limiting reactant
only in calculating the amount of a given
product produced
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Finding Molar Ratios
Balance each of the following chemical equations.
Write down all the molar ratios that are in each equation.
First, write the ratios for the first reactant and each of
the other substance. Second, write the ratios for the
second substance with each of the others, except the first
one since you have already written it. Work your way
through each equation in this manner.
H_{2} +
O_{2} -->
H_{2}O
K +
H_{2}O -->
KOH +
H_{2}
P_{4}O_{10}
+ H_{2}O -->
H_{3}PO_{4}
Calculating Moles Needed/Produced
The following equations are already balanced. You are
given an amount in moles for one of the substances in the
equation. Find out how many moles of each of the other
substances you would need (reactants) or would be produced
(products).
Cu + 2AgNO_{3} --> 2Ag
+
Cu(NO_{3})_{2} 1.78
mol Ag
2C + O_{2} -->
2CO 1 mol
O_{2}
2KNO_{3} -->
2KNO_{2} +
O_{2} 1.3 x
10^{3} mol KNO_{3}
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Calculating Mass Needed/Produced
The following equations are already balanced. You are
given an amount in grams for one of the substances in the
equation. Find out how the mass of each of the other
substances you would need or would be produced. You will
(obviously) need to find the molar masses of each of the
reactants and products.
2C + O_{2} -->
2CO 14 g C
2KNO_{3} -->
2KNO_{2} +
O_{2} 29.8
g KNO_{2}
3Fe + 4H_{2}O -->
Fe_{3}O_{4} +
4H_{2} 167.54
g Fe
Stoichiometry with Limiting Reagents
Titanium (IV) oxide (TiO_{2}) is used as a pigment in paints
and as a whitening and coating agent for paper. It can be made by
reacting O_{2} with TiCl_{4}:
TiCl_{4} + O_{2} --> TiO_{2} + 2Cl_{2}
If 4.5 mol of TiCl_{4} react with 3.5 mol O_{2}, identify
both the limiting and the excess reactants. How many moles of the excess
reactant would remain if all of the limiting reactant is used up? How
many moles of each product will be formed if the reaction goes to
completion?
Chemical Ratios (Stoichiometry)
The word Stoichiometry (stoy'-ki-AH-me-tree) comes from the Greek
stoicheion, which means to measure the elements. It is a technique for
finding out how much of a chemical product you will get from a chemical reaction. It
can also be used to figure out how much of a starting material you need to get a
certain amount of a chemical product. It is arguably one of the most important
pieces of chemical knowledge.
It works like this. A balanced chemical reaction is used to figure out the
mathematical relationships among the reactants and products. These relationships
take the form of ratios. Take an equation used in the text at the beginning of this
group activity: 2C + O_{2} 2CO. This
equation shows that there is a 2:1 molar ratio between C and O_{2}. That is, it takes
one mole of O_{2} to completely burn up two moles of C. This is a direct logical conclusion from the idea that it takes one whole molecule of O_{2} to react with two atoms of C. Similarly, there is a 1:1 molar ratio between C and CO and a 1:2 molar ratio between O_{2} and CO.
Answer the following questions using equations you balanced earlier in this activity. Use a separate sheet of paper to work out the solutions, showing all work. One is done as an example.
Given 10 mol C, how many moles of O_{2} are required to completely react with all of the C?
1 mol O_{2}
10 mol C × ---------- = 5 mol O_{2}
2 mol C
Given 5.4 mol O_{2} how many moles of O_{3} can be produced?
How much ammonia can be manufactured from 1,000 mol N_{2} assuming you have enough H_{2}?
How many moles of HCl do you need to completely react with 7.7 × 10^{-2} mol of zinc to make ZnCl_{2} and H_{2}?
How many moles of fluorine gas are needed to make 2.5 mol of SF_{6} assuming no shortage of sulfur?
You have 4.6 mol NH_{3}. How many moles of O_{2} do you need to completely react with all of the ammonia?