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Chapter 12 Stoichiometry
12.1 The Arithmetic of Equations - In this section you will learn to use a balanced equation to calculate amounts of
reactants needed or amounts of products produced.
1. Using Everyday Equations - everyday equations are called recipes
A balanced chemical equation provides the same kind of quantitative information that a recipe does. You can double
a recipe to make more, cut it in half to make less, etc. A balanced equation shows us how to make any quantity of any
chemical the same way recipes shows us how to make any food.
Consider the making of tricycles in a production plant that needs to make 640. tricycles/week. The equation for making
tricycles is below
How many petals will you need to order each week?
640. tricycles X
Solution - 1 tricycle has 2 petals
2 petals/tricycle = 1280 petals
How many wheels will you need to order each week?
640. tricycles X ________________/tricycle =
Solution - 1 tricycle has __________________________
_____________________ ? wheels
2. Using Balanced Chemical Equations - A balanced chemical equation tells you what amounts of reactants to mix and
what amounts of product to expect.
Chemists use balanced chemical equations as a basis to calculate how much reactant is needed or how much product
is formed in a reaction.
The calculation of quantities in chemical reactions is a subject of chemistry called stoichiometry. These quantities are
typically measured in grams or moles, but could be expressed in liters and molecules as well.
We track the amounts of reactants and products using ratios of moles of each substance from the balanced chemical
equation.
3. Interpreting Chemical Equations - Consider the Haber process which turns N2 from the atmosphere into ammonia
(NH3 - used for fertilizer)
A balanced chemical equation can be interpreted in terms of different quantities, including numbers of atoms,
molecules, moles, mass, or volume.
a. number of atoms - At the atomic level, this reaction show that 2 N atoms combine with 6 H atoms to yield the
product which is also composed of 2 N atoms and 6 H atoms.
Still considering the Haber process:
N2(g)
+
3 H2(g)
--------------->
2 NH3(g)
b. number of molecules - the balanced equation shows that 1 molecule of nitrogen reacts with 3 molecules of hydrogen
in order to form 2 molecules of ammonia. The ratio of molecules of N2:H2:NH3 is always 1:3:2. This means that if you
could get 10 molecules of nitrogen to react with 30 molecules of hydrogen, you will get 20 molecules of ammonia. Small
numbers of molecules are not possible to count, so we use the mol = 6.02 X 1023 molecules instead
c. Moles - The coefficients in a balanced chemical equation tells us the number of moles of reactants and products. The
equation tells us that 1 mol of N2(g) reacts with 3mol of H2(g) to yield 2 mol of NH3(g). Using this information, we can
calculate the amounts of reactants and products. Note: the total number of moles of reactants does not equal the total
moles of products in the above reaction. Number of moles is not conserved, but mass/numbers of atoms is conserved.
d. Mass - The number and types of atoms does not change in a chemical reaction, and therefore the mass of the
reactants will equal the mass of the products. 1 mol of N2 = 28g, 3 mole of H2 = 6 g, and 2 moles of NH3 = 34 g
e. Volume - When this rx occurs at STP, the balanced equation tells us that 22.4 L N2 will combine with 67.2 L H2 to
form 44.8 L of NH3.
4. Mass Conservation in Chemical Reactions Mass and atoms are conserved in every chemical reaction; however, molecule number, formula units, moles, and
volumes are not typically conserved. In some cases they are, consider the formation of hydrogen iodide
H2(g)
+
I2(g)
------------------>
2 HI (g)
Note: In this rx, molecules, moles, and volume are conserved
but in the majority of chemical reactions they are not.
Practice Problems
12.1 Hydrogen sulfide, which smells like rotten eggs, is found in volcanic gases. The balanced equation for the
burning of hydrogen sulfide is:
2 H2S(g)
+
Interpret this equation in terms of
3O2(g)
-------------------->
2 SO2(g)
+
2 H2O(g)
a. representative particles and moles & b. masses of reactants and products
a.
b.
3. Interpret the equation for the formation of water from its elements in terms of numbers of molecules and moles, and
volumes of gases at STP ( 1 mole =22.4L).
2H2(g)
+
O2(g) -------------->
2H2O(g)
4. Balance the following equation.
C2H2(g)
+
O2(g) ------------------>
CO2(g)
+
H2O(g)
Interpret the balanced equation in terms of number of moles, volumes of gas at STP, and masses of reactants and
products.
12.2 Chemical Calculations - In this section, you will learn how to use a balanced chemical equation to calculate the
amount of product formed in a chemical reaction
1. Writing and Using Mole Ratios - Consider the Haber process again: N2(g)
+
3 H2(g)
--------------->
2 NH3(g)
The most important interpretation of the balanced equation is that 1 mole of N2 reacts with 3 moles of H2 to produce 2
moles of NH3. With this interpretation, ratios (mole ratios) can be written that relate moles of reactants to moles of
products.
mole ratio is a conversion factor derived from the coefficients of a balanced chemical equation.
In chemical calculations, mole ratios are used to convert between moles of reactant and moles of product, between
moles of reactants, or between moles of products.
For instance, Three mole ratios can be derived from the balanced equation of the Haber process
1 mol N2
3 mole H2
2 mol NH3
1 mol N2
3 mol H2
2mol NH3
The first ratio relates amounts of reactants, the second ratio relates the product to one reactant, and the third ratio
relates the product to the other reactant. Any ratio can be inverted to suit your needs. Try the problem below
a. Mole-Mole Calculations
Problem 12.2 How many moles of ammonia are produced when 0.60 mol of nitrogen reacts with hydrogen?
Solution: Write what you are given, and multiply this by the mole ratio that relates what you're given with what you
need to find (you need to find moles of NH3, so the ratio needed is the middle one)
0.60 mol N2 X
2 mole NH3
1 mole N2
=
1.2 mol NH3
Check : look at the rx and ask yourself if the answer makes sense
How many moles of hydrogen are needed to produce 7.2 moles of ammonia?
How many moles of nitrogen are needed to completely react with 0.500 mol hydrogen?
Practice Problems
11. This equation shows the formation of aluminum oxide, which is found on the surface of aluminum objects exposed
to the air.
4 Al(s)
+
3O2(g) ------------> 2Al2O3(s)
a. write the six mol ratios that can be derived from this equation (there are technically 3 ratios, but each one has it's
inverse)
b. How many moles of aluminum are needed to form 3.7 mol Al2O3?
c. How many moles of oxygen are required to react completely with 14.8 mol Al?
d. How many moles of Al2O3 are formed when 0.78 mol O2 reacts with excess aluminum?
___________________________________________________________________________________
b. Mass-Mass Calculations - We can't weigh substances in moles, our balances use grams!
If we are given a sample's mass in grams, we use the molar mass of the substance to convert it to moles. Then we use a
mole ratio from the balanced equation to find moles of an unknown, which can then be converted to grams by
multiplying by the unknown's molar mass.
Problem 12.3 Calculate the number of grams of NH3 produced by the reaction of 5.40 g of hydrogen with an excess of
nitrogen. The balanced equation is:
N2(g) + 3H2(g) ------------->
2NH3(g)
Solution: 1 Convert 5.40 g H2 to moles of H2 by dividing by its molar mass, multiply this answer by the mole ratio (3 mol
H2 = 2 mol NH3) and multiply by the molar mass of NH3 in order to get grams of NH3.
5.40 g H2
X
1 mol H2
2.0 g H2
X
2 mol NH3
3 mol H2
X
17.0 g NH3
1 mol NH3
=
30.6 g NH3
Practice Problems
13. Acetylene gas (C2H2) is produced by adding water to calcium carbide CaC2 (balanced equation is below)
CaC2(s)
+
2H2O(l)
------------------>
C2H2(g)
+
Ca(OH)2(aq)
How many grams of acetylene are produced by adding water to 5.00 g CaC2?
Determine how many moles of CaC2 are needed to react completely with 49.0 g H2O?
How many grams of oxygen are produced when 29.2 g of water is decomposed by electrolysis according to the balanced
equation: 2H2O(l)
----------------------->
2H2(g)
+
O2(g)
c. Other Stoichiometric Calculations - Mole-Molecule Problems - convert given quantity to moles and then to
molecules (1 mole = 6.02 X1023)
How many molecules of oxygen are produced when 29.2 g of water is decomposed by electrolysis according the
equation above
Practice Problem 15. How many molecules of oxygen are produced by the decomposition of 6.54 g of potassium
chlorate (KClO3) rx: 2 KClO3(s) -----------------> 2 KCl(s)
+
3 O2(g)
16. The last step in the production of nitric acid is the reaction of nitrogen dioxide with water.
3NO2(g)
+
H2O(l) ---------------->
2HNO3(aq)
+
NO(g)
How many grams of nitrogen dioxide must react with water to produce 5.00 X 1022 molecules of nitrogen monoxide?
d. Other Calculations - Mole-Volume problems @ STP
Sample problem 12.5 Nitrogen monoxide and oxygen gas combine to form the brown gas nitrogen dioxide, which
contributes to photochemical smog. How many liters of nitrogen dioxide are produced when 34 L of oxygen reacts with
an excess of nitrogen monoxide? Assume conditions of STP.
2NO(g)
+
O2(g) ------------------>
2NO2(g)
Solution: Convert liters of oxygen to moles (1 mole = 22.4L), use the proper balanced equation mole ratio to get moles
of nitrogen dioxide, and then convert back to liters (22.4L = 1mole)
17. The equation for the combustion of carbon monoxide is: 2CO(g)
liters of oxygen are required to burn 3.86 L of carbon monoxide?
+
O2(g) ----------->
2CO2
: How many
18. Phosphorous and hydrogen can be combined to form phosphine (PH3), the rx is: P4(s) + 6H2(g) --------> 4PH3(g)
How many liters of phosphine are formed when 0.42 L of hydrogen reacts with phosphorous at STP?
19. Consider this rx:
CS2(l)
+ 3O2(g)
----------------> CO2(g)
+
2SO2(g)
Calculate the volume of sulfur dioxide produced when 27.9 mL O2 reacts with carbon disulfide?
How many deciliters of carbon dioxide are produced when 0.38 L SO2 is formed?
12.3 Limiting Reagents and Percent Yield - the amount of product made in a chemical rx may be limited by the amount
of one or more reactants.
1. Limiting Reagent
In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms.
The limiting reagent is the reagent that determines the amount of product that can be formed by a reaction.
The excess reagent is the reagent that is not completely used up.
Consider the Haber process again:
N2(g) + 3H2(g)
----------------->
2NH3(g)
This states that 1 molecule (moles) of N2 reacts with 3 molecules (moles) of H2 to form 2 molecules (moles) NH3
What would happen if 2 moles of N2 were reacted with 3 moles H2?
Would all of the H2 get used up?
Would all the N2 get used up?
In this case, the hydrogen is the limiting reagent, while the nitrogen is the excess reagent - only two moles of NH3 will be
produced while 1 mole of N2 will be left un-reacted.
Limiting Reagent problems can always be recognized when the problem gives the mass of both reactants. These masses
must be converted to moles in order to find out which quantity will limit the production of the product(s).
Sample Problem 12.7 Copper reacts with sulfur to form copper (I) sulfide according to the following balanced equation
2 Cu(s)
+ S(s) -------------> Cu2S(s)
What is the limiting reagent when 80.0 g Cu react with 25.0 g S
Solution: Convert both quantities to moles and compare their ratio to the balanced reaction ratio
2. Percent Yield - In the same way teachers grade tests, chemists calculate percent yield in order to compare their
results with the ideal amount (called theoretical yield).
The percent yield is a measure of the efficiency of a reaction carried out in the laboratory.
The theoretical yield is the maximum amount of product that could be formed from given amounts of reactants.
In contrast, the amount of product that actually forms when the reaction is carried out in the laboratory is called the
actual yield.
The percent yield is the ratio of the actual yield to the theoretical yield expressed as a percent.
Factors during a rx that lower percent yield are: (1) rx not going to completion, (2) contamination from impure
reactants may produce side reactions that diminish actual yield, (3) some product may be lost when it is filtered,
purified, or massed- this is called transfer loss, and (4) the masses of reactants and products may not have been
determined accurately.
Sample problem 12.9 (Calculating theoretical yield) Calcium Carbonate, which is found in seashells, is decomposed by
heating. The balanced equation for this rx is: CaCO3(s) ----------------------->
CaO(s)
+
CO2(g)
What is the theoretical yield of CaO if 24.8 g CaCO3 is heated?
Solution: Convert g CaCO3 to moles, use mole ratio to convert to moles of CaO, and convert to grams with molar mass
29. When 84.8 g of iron(III) oxide reacts with an excess of carbon monoxide, iron is produced. What is the theoretical
yield of Fe?
Fe2O3(s)
+
3CO2(g)
----------------------->
2Fe(s)
+
3CO2(g)
30. When 5.00 g of copper reacts with excess silver nitrate, silver metal and copper(II) nitrate are produced. What is the
theoretical yield of silver in this reaction? (you need to write and balance the rx first)
Sample problem 12.10 - Calculating Percent yield of a Reaction What is the percent yield if 13.1 g of CaO is actually
produced when 24.8 g CaCO3 is heated?
Solution: Calculate the theoretical yield based on 24.8 g CaCO3 (You did this already in 12.9 above). Use percent yield
formula above
Sample Problem
31. If 50.0 g of silicon dioxide is heated with an excess of carbon, 27.9 g of silicon carbide is produced. What is the
percent yield of this reaction? rx SiO2(s) + 3C(s) -------------->
SiC(s) +
2CO(g)