Chemistry Chapter 9Chemistry Chapter 9
Unit 6Unit 6
StoichiometryStoichiometry
The arithmetic of equationsThe arithmetic of equations
�� Equations are recipes. Equations are recipes.
�� They tell chemists what amounts of reactants They tell chemists what amounts of reactants
to mix and what amounts of products to to mix and what amounts of products to
expect.expect.
��What is Stoichiometry?What is Stoichiometry?
�� The calculation of quantities in chemical The calculation of quantities in chemical
reactionsreactions
A quick exampleA quick example�� If we consider chemical equations to be recipes:If we consider chemical equations to be recipes:
�� When you bake cookies, you probably follow a recipeWhen you bake cookies, you probably follow a recipe
�� The recipe tells you what ingredients to mix and in The recipe tells you what ingredients to mix and in what ratiowhat ratio
�� The end result of the combination of ingredients is The end result of the combination of ingredients is cookiescookies
�� If you want more cookies you can double or triple the If you want more cookies you can double or triple the reciperecipe
�� If you want fewer cookies, the recipe can be cut in If you want fewer cookies, the recipe can be cut in half or quarteredhalf or quartered
�� The ingredients are the reactants and the cookies are The ingredients are the reactants and the cookies are the productsthe products
�� In a way, the cookie recipe gives the same kind In a way, the cookie recipe gives the same kind of information that a balanced chemical equation of information that a balanced chemical equation doesdoes
Stoichiometry: A Closer LookStoichiometry: A Closer Look
�� Lets examine Lets examine stoichiometrystoichiometry a little closer a little closer
by looking at the production of ammonia by looking at the production of ammonia
from it’s elements: Write the equation from it’s elements: Write the equation
below:below:
NN2(g)2(g) + 3H+ 3H2(g)2(g) →→ 2NH2NH3(g)3(g)
NN2(g)2(g) + 3H+ 3H2(g)2(g) →→ 2NH2NH3(g)3(g)
�� Valuable information can be derived from Valuable information can be derived from
this equation:this equation:
�� Write Write sentencessentences to describe the to describe the
equation equation --
1.1. In terms of PARTICLESIn terms of PARTICLES
One molecule of nitrogen gas reacts with One molecule of nitrogen gas reacts with
three molecules of hydrogen gas to three molecules of hydrogen gas to
produce two molecules of ammonia gas.produce two molecules of ammonia gas.
NN2(g)2(g) + 3H+ 3H2(g)2(g) →→ 2NH2NH3(g)3(g)
2.2. In terms of MOLESIn terms of MOLES
One mole of nitrogen gas reacts with One mole of nitrogen gas reacts with
three moles of hydrogen gas to produce three moles of hydrogen gas to produce
two moles of ammonia gas. two moles of ammonia gas.
The coefficients of a balanced chemical equation indicate The coefficients of a balanced chemical equation indicate
the relative number of moles of reactants to the moles the relative number of moles of reactants to the moles
of products in a reaction. This is the most important of products in a reaction. This is the most important
information obtained from a balanced chemical information obtained from a balanced chemical
equation.equation.
NN2(g)2(g) + 3H+ 3H2(g)2(g) →→ 2NH2NH3(g)3(g)
3.3. In terms of MASS In terms of MASS
Reactants:Reactants:
N = 14 x 2 = 28gN = 14 x 2 = 28g
H = 1 x 6 = +H = 1 x 6 = + 6g6g
34g total34g total
Products:Products:
N = 14 x 2 = 28gN = 14 x 2 = 28g
H = 1 x 6 = +H = 1 x 6 = + 6g6g
34g total34g total
Balanced chemical equations must obey the law of conservation ofBalanced chemical equations must obey the law of conservation ofmass. Remember that mass is related to number of atoms. Even mass. Remember that mass is related to number of atoms. Even though the number of moles is different, the mass of reactants though the number of moles is different, the mass of reactants
and products is equal.and products is equal.
NN2(g)2(g) + 3H+ 3H2(g)2(g) →→ 2NH2NH3(g)3(g)
4.4. In terms of VOLUME (for gases only!)In terms of VOLUME (for gases only!)
Remember that one mole of any gas at Remember that one mole of any gas at
STP is 22.4 L. Therefore, the volume (as STP is 22.4 L. Therefore, the volume (as
well as number of moles, molecules, and well as number of moles, molecules, and
formula units) of gases may not be the formula units) of gases may not be the
same.same.
NN2(g)2(g) + 3H+ 3H2(g)2(g) →→ 2NH2NH3(g)3(g)
�� What was conserved (stayed the same What was conserved (stayed the same
from left to right) in this equation? from left to right) in this equation?
(2 answers)(2 answers)
1. Mass1. Mass
2. Number of Atoms2. Number of Atoms
Another quick exampleAnother quick exampleImagine that you are in charge of manufacturing Imagine that you are in charge of manufacturing for Tiny Tyke Tricycle Company. The business for Tiny Tyke Tricycle Company. The business plan for Tiny Tike requires the production of 128 plan for Tiny Tike requires the production of 128 custom custom –– made tricycles each day. One of your made tricycles each day. One of your responsibilities is to be sure that there are responsibilities is to be sure that there are enough parts available at the start of each day to enough parts available at the start of each day to make these tricycles. To simplify this discussion, make these tricycles. To simplify this discussion, assume that the major components of the assume that the major components of the tricycle are the frame (F), the seat (S), the tricycle are the frame (F), the seat (S), the wheels (W), the handlebars (H), and the pedals wheels (W), the handlebars (H), and the pedals (P). The finished tricycle has a “formula” of (P). The finished tricycle has a “formula” of FSWFSW33HPHP22. The balanced equation for the . The balanced equation for the production of a tricycle is:production of a tricycle is:
F + S + 3W + H + 2P F + S + 3W + H + 2P →→ FSWFSW33HPHP22
F + S + 3W + H + 2P F + S + 3W + H + 2P →→ FSWFSW33HPHP22
�� In a five day workweek, Tiny Tike is scheduled In a five day workweek, Tiny Tike is scheduled
to make 640 tricycles. How many wheels should to make 640 tricycles. How many wheels should
be in the plant on Monday morning to make be in the plant on Monday morning to make
these tricycles?these tricycles?
�� # of tricycles to be made: 640 tricycles# of tricycles to be made: 640 tricycles
�� 1 FSW1 FSW33HPHP2 2 = 3W (from balanced equation)= 3W (from balanced equation)
�� Use the conversion factor: Use the conversion factor: 3W3W
1 FSW1 FSW33HPHP2 2
640 FSW640 FSW33HPHP2 2 x x 3W3W = 1920W= 1920W
1 FSW1 FSW33HPHP2 2
F + S + 3W + H + 2P F + S + 3W + H + 2P →→ FSWFSW33HPHP22
�� Here is another one:Here is another one:
�� How many tricycle seats, wheels, and pedals How many tricycle seats, wheels, and pedals
are needed to make 288 tricycles?are needed to make 288 tricycles?
Seats:Seats:
288288
Wheels:Wheels:
864 864
Pedals:Pedals:
576576
Practice Problems: Practice Problems: 1.1. Try this:Try this:
a.a. Write the equation for the complete Write the equation for the complete combustion of liquid ethanol, Ccombustion of liquid ethanol, C22HH55OH:OH:
b.b. Interpret the equation in terms of Interpret the equation in terms of numbers of moleculesnumbers of molecules
c.c. Interpret the equation in terms of Interpret the equation in terms of molesmoles
d.d. Interpret the equation in terms of massInterpret the equation in terms of mass
e.e. Why can’t this problem be interpreted Why can’t this problem be interpreted in terms of volume?in terms of volume?
Practice Problem 1Practice Problem 1
��Write the equation for the complete Write the equation for the complete
combustion of liquid ethanol, combustion of liquid ethanol,
CC22HH55OH:OH:
CC22HH55OHOH(l)(l) + 3O+ 3O2(g)2(g) →→ 3H3H22OO(l)(l) + 2CO+ 2CO2(g)2(g)
CC22HH55OHOH(l)(l) + 3O+ 3O2(g)2(g) →→ 3H3H22OO(l)(l) + 2CO+ 2CO2(g)2(g)
�� Interpret the equation in terms of Interpret the equation in terms of
numbers of molecules.numbers of molecules.
�� 1 molecule of liquid ethanol reacts with 1 molecule of liquid ethanol reacts with
3 molecules of oxygen gas to yield 3 3 molecules of oxygen gas to yield 3
molecules of liquid water and 2 molecules of liquid water and 2
molecules of carbon dioxide gas.molecules of carbon dioxide gas.
CC22HH55OHOH(l)(l) + 3O+ 3O2(g)2(g) →→ 3H3H22OO(l)(l) + 2CO+ 2CO2(g)2(g)
�� Interpret the equation in terms of Interpret the equation in terms of
molesmoles
�� 1 mole of liquid ethanol reacts with 3 1 mole of liquid ethanol reacts with 3
moles of oxygen gas to yield 3 moles of moles of oxygen gas to yield 3 moles of
liquid water and 2 moles of carbon liquid water and 2 moles of carbon
dioxide gas.dioxide gas.
�� 4 moles total reactants4 moles total reactants
�� 5 moles total products5 moles total products
CC22HH55OHOH(l)(l) + 3O+ 3O2(g)2(g) →→ 3H3H22OO(l)(l) + 2CO+ 2CO2(g)2(g)
�� Interpret the equation in terms of Interpret the equation in terms of massmass
�� Reactants:Reactants:
•• 1 molecule of ethanol = 46g1 molecule of ethanol = 46g
•• 3 molecules of oxygen = 96g3 molecules of oxygen = 96g
•• 142g total reactants142g total reactants
�� Products:Products:
•• 3 molecules of water 3 molecules of water = 54g= 54g
•• 2 molecules of carbon dioxide = 88g2 molecules of carbon dioxide = 88g
•• 142g total products142g total products
CC22HH55OHOH(l)(l) + 3O+ 3O2(g)2(g) →→ 3H3H22OO(l)(l) + 2CO+ 2CO2(g)2(g)
��Why can’t this problem be interpreted in Why can’t this problem be interpreted in
terms of volume? terms of volume?
�� Ethanol is in liquid form. The volume terms of Ethanol is in liquid form. The volume terms of
writing is based on gases.writing is based on gases.
Practice Problem 2Practice Problem 2
��Write the balanced equation for the single Write the balanced equation for the single
–– replacement reaction of solid potassium replacement reaction of solid potassium
metal reacting with liquid water:metal reacting with liquid water:
2K2K(s)(s) + H+ H22OO(l)(l) →→ HH2(g)2(g) + K+ K22OO(aq)(aq)
2K2K(s)(s) + H+ H22OO(l)(l) →→ HH2(g)2(g) + K+ K22OO(aq)(aq)
�� Interpret this equation in terms of:Interpret this equation in terms of:
•• Interaction particlesInteraction particles--
•• molesmoles
•• massmass
•• volume (if allowed)volume (if allowed)
2K2K(s)(s) + H+ H22OO(l)(l) →→ HH2(g)2(g) + K+ K22OO(aq)(aq)
�� Interaction particles:Interaction particles:
�� Two atoms of solid potassium react with one Two atoms of solid potassium react with one
molecule of liquid water to produce one molecule of liquid water to produce one
molecule of hydrogen gas and one formula molecule of hydrogen gas and one formula
unit of aqueous potassium oxide.unit of aqueous potassium oxide.
2K2K(s)(s) + H+ H22OO(l)(l) →→ HH2(g)2(g) + K+ K22OO(aq)(aq)
��Moles:Moles:
�� Two moles of solid potassium metal reacts Two moles of solid potassium metal reacts
with one mole of liquid water to produce one with one mole of liquid water to produce one
mole of hydrogen gas and one mole of mole of hydrogen gas and one mole of
aqueous potassium oxide.aqueous potassium oxide.
�� 3 moles total reactants3 moles total reactants
�� 2 moles total products2 moles total products
2K2K(s)(s) + H+ H22OO(l)(l) →→ HH2(g)2(g) + K+ K22OO(aq)(aq)
��Mass:Mass:
�� Reactants:Reactants:
•• K = 39g (2) = 78gK = 39g (2) = 78g
•• HH22O = 18g (1) = +O = 18g (1) = + 18g18g
96g total96g total
�� Products:Products:
•• HH22 = 1g (2) = 2g= 1g (2) = 2g
•• KK22O = 94g (1) = +O = 94g (1) = + 94g94g
96g total96g total
2K2K(s)(s) + H+ H22OO(l)(l) →→ HH2(g)2(g) + K+ K22OO(aq)(aq)
�� volume (if allowed)volume (if allowed)
�� This is not allowed because not all are gases.This is not allowed because not all are gases.
Practice Test A Objective 1Practice Test A Objective 1
1.1. Write the balanced equation for the Write the balanced equation for the
formation of gaseous fluorine trioxide formation of gaseous fluorine trioxide
from its elements. Include the from its elements. Include the
“adjectives”.“adjectives”.
FF2(g)2(g) + 3O+ 3O2(g)2(g) →→ 2FO2FO3(g)3(g)
Practice Test A Objective 1Practice Test A Objective 1
2.2. Interpret this equation in terms of the Interpret this equation in terms of the
number of representative particles.number of representative particles.
FF2(g)2(g) + 3O+ 3O2(g)2(g) →→ 2FO2FO3(g)3(g)
�� 1 molecule of fluorine gas reacts with 3 1 molecule of fluorine gas reacts with 3
molecules of oxygen gas to produce 2 molecules of oxygen gas to produce 2
molecules of fluorine trioxide gas.molecules of fluorine trioxide gas.
Practice Test A Objective 1Practice Test A Objective 1
3.3. Interpret this equation in terms of number Interpret this equation in terms of number
of moles.of moles.
FF2(g)2(g) + 3O+ 3O2(g)2(g) →→ 2FO2FO3(g)3(g)
�� 1 mole of fluorine gas reacts with 3 moles of 1 mole of fluorine gas reacts with 3 moles of
oxygen gas to produce 2 moles of fluorine oxygen gas to produce 2 moles of fluorine
trioxide gas.trioxide gas.
Practice Test A Objective 1Practice Test A Objective 1
4.4. Interpret this equation in terms of mass.Interpret this equation in terms of mass.
FF2(g)2(g) + 3O+ 3O2(g)2(g) →→ 2FO2FO3(g)3(g)
�� Reactants:Reactants:
•• F = 19g (2) = 38gF = 19g (2) = 38g
•• O = 16g (6) = +O = 16g (6) = + 96g96g
134g total134g total
�� Products:Products:
•• F = 19g (2) = 38gF = 19g (2) = 38g
•• O = 16g (6) = +O = 16g (6) = + 96g96g
134g total134g total
Practice Test A Objective 1Practice Test A Objective 1
5.5. Interpret this equation in terms of Interpret this equation in terms of volume.volume.
FF2(g)2(g) + 3O+ 3O2(g)2(g) →→ 2FO2FO3(g)3(g)
�� Reactants:Reactants:
•• FF22 = 1 mol x 22.4 L/mol = 22.4 L= 1 mol x 22.4 L/mol = 22.4 L
•• OO22 = 3 mol x 22.4 L/mol = + = 3 mol x 22.4 L/mol = + 67.2 L67.2 L
89.6 L total89.6 L total
�� Products:Products:
•• FOFO33 = 2 mol x 22.4 L/mol = 44.8 L total= 2 mol x 22.4 L/mol = 44.8 L total
Chemical CalculationsChemical Calculations
Section 9.2Section 9.2
MoleMole--Mole Mole
CalculationsCalculations
The heart of the The heart of the stoichiometrystoichiometry
problem.problem.
Think WANT OVER GIVENThink WANT OVER GIVEN
�� Example Problem:Example Problem:
�� Rewrite the balanced equation for the Rewrite the balanced equation for the
production of ammonia from its elements:production of ammonia from its elements:
NN2(g)2(g) + 3H+ 3H2(g)2(g) →→ 2NH2NH3(g)3(g)
NN2(g)2(g) + 3H+ 3H2(g)2(g) →→ 2NH2NH3(g)3(g)
�� How many moles of ammonia are How many moles of ammonia are
produced when 0.6 mol of nitrogen react produced when 0.6 mol of nitrogen react
with hydrogen? with hydrogen?
0.6 mol N0.6 mol N22 x x 2 mol NH2 mol NH33 = 1.2 mol NH= 1.2 mol NH33
1mol N1mol N22
Example ProblemExample Problem
��Write the equation of aluminum oxide from Write the equation of aluminum oxide from
its elements.its elements.
4Al4Al(s)(s) + 3O+ 3O2(g)2(g) →→ 2Al2Al22OO3(s)3(s)
Example ProblemExample Problem
4Al4Al(s)(s) + 3O+ 3O2(g)2(g) →→ 2Al2Al22OO3(s)3(s)
1.1. How many moles of aluminum are needed to form 3.7 How many moles of aluminum are needed to form 3.7 moles of aluminum oxide?moles of aluminum oxide?
�� 3.7 mol 3.7 mol AlAl22OO3 3 x x 4mol Al4mol Al = 7.4 mol Al= 7.4 mol Al
2mol 2mol AlAl22OO33
2.2. How many moles of oxygen are required to react How many moles of oxygen are required to react completely with 14.8 moles of aluminum?completely with 14.8 moles of aluminum?
�� 14.8 mol Al x 14.8 mol Al x 3 mol O3 mol O22 = 11.1 mol O= 11.1 mol O22
4 mol Al4 mol Al
3.3. How many moles of aluminum oxide are formed when How many moles of aluminum oxide are formed when 0.78 moles of oxygen reacts with aluminum?0.78 moles of oxygen reacts with aluminum?
�� 0.78 mol O0.78 mol O22 x x 2 mol 2 mol AlAl22OO33 = 0.52 mol Al= 0.52 mol Al22OO33
3 mol O3 mol O22
MassMass--Mass CalculationsMass Calculations
�� Rewrite the equation for the production of Rewrite the equation for the production of
ammonia: ammonia:
NN2(g)2(g) + 3H+ 3H2(g)2(g) →→ 2NH2NH3(g)3(g)
�� Calculate the number of grams of ammonia Calculate the number of grams of ammonia
produced by the reaction of 5.4 grams of hydrogen produced by the reaction of 5.4 grams of hydrogen
with an excess of nitrogen.with an excess of nitrogen.
�� 5.4g H5.4g H22 x x 1 mol H1 mol H22 = 2.7 mol H= 2.7 mol H22
2g H2g H22
�� 2.7 mol H2.7 mol H22 x x 2 mol NH2 mol NH3 3 = 1.8 mol NH= 1.8 mol NH33
3 mol H3 mol H22
�� 1.8 mol NH1.8 mol NH3 3 x x 17.0g NH17.0g NH33 = 31g NH= 31g NH33
1 mol NH1 mol NH33
Example Problem Example Problem
�� Write the balanced chemical equation for the Write the balanced chemical equation for the decomposition for potassium chlorate.decomposition for potassium chlorate.
�� Calculate the number of grams of reactant Calculate the number of grams of reactant necessary to produce 50 grams of potassium necessary to produce 50 grams of potassium chloride.chloride.
�� How many grams of oxygen gas are produced How many grams of oxygen gas are produced when you decompose 15 grams of potassium when you decompose 15 grams of potassium chlorate?chlorate?
�� How many MOLES of each product are How many MOLES of each product are produced when 10 grams of potassium chlorate produced when 10 grams of potassium chlorate are heated?are heated?
�� How many grams of each product are produced How many grams of each product are produced when 2.5 moles of reactant are used? when 2.5 moles of reactant are used?
Example ProblemExample Problem
��Write the balanced chemical equation for Write the balanced chemical equation for
the decomposition for potassium chlorate.the decomposition for potassium chlorate.
2KClO2KClO33 →→ 3O3O22 + 2KCl+ 2KCl
2KClO2KClO33 →→ 3O3O22 + 2KCl+ 2KCl
�� Calculate the number of grams of reactant Calculate the number of grams of reactant
necessary to produce 50 grams of necessary to produce 50 grams of
potassium chloride.potassium chloride.
50g 50g KClKCl x x 1 mol 1 mol KClKCl x x 2 mol KClO2 mol KClO33 x x 122.5g KClO122.5g KClO3 3 = 82.21g KClO= 82.21g KClO33
74.5g 74.5g KClKCl 2 mol 2 mol KClKCl 1 mol KClO1 mol KClO33
2KClO2KClO33 →→ 3O3O22 + 2KCl+ 2KCl
�� How many grams of oxygen gas are How many grams of oxygen gas are
produced when you decompose 15 grams produced when you decompose 15 grams
of potassium chlorate?of potassium chlorate?
15g KClO15g KClO3 3 x x 1mol KClO1mol KClO3 3 x x 3mol O3mol O22 x x 32g O32g O2 2 = 5.88g O= 5.88g O22
122.5g KClO122.5g KClO3 3 2mol KClO2mol KClO33 1mol O1mol O22
2KClO2KClO33 →→ 3O3O22 + 2KCl+ 2KCl
�� How many MOLES of each product are How many MOLES of each product are
produced when 10 grams of potassium produced when 10 grams of potassium
chlorate are heated?chlorate are heated?
10g KClO10g KClO3 3 x x 1mol KClO1mol KClO3 3 x x 3mol O3mol O22 = 0.12mol O= 0.12mol O22
122.5g KClO122.5g KClO3 3 2mol KClO2mol KClO33
10g KClO10g KClO3 3 x x 1mol KClO1mol KClO3 3 x x 2mol 2mol KClKCl = 0.08mol = 0.08mol KClKCl
122.5g KClO122.5g KClO3 3 2mol KClO2mol KClO3 3
2KClO2KClO33 →→ 3O3O22 + 2KCl+ 2KCl
�� How many grams of each product are How many grams of each product are produced when 2.5 moles of reactant are produced when 2.5 moles of reactant are used? used?
2.5 mol KClO2.5 mol KClO3 3 x x 3 mol O3 mol O22 = 3.75 mol O= 3.75 mol O22
2 mol KClO2 mol KClO33
3.75 mol O3.75 mol O22 x x 32g O32g O22 = = 120g O120g O22
1 mol O1 mol O22
2.5 mol KClO2.5 mol KClO3 3 x x 2 mol 2 mol KClKCl = 2.5 mol = 2.5 mol KClKCl
2 mol KClO2 mol KClO33
2.5 mol 2.5 mol KClKCl x x 74.5g 74.5g KClKCl = = 186.25g 186.25g KClKCl
1 mol 1 mol KClKCl
Other Other StoichiometricStoichiometric
CalculationsCalculations
The The StoichiometryStoichiometry Road MapRoad Map
��With your knowledge of conversion factors With your knowledge of conversion factors
and the problemand the problem--solving diagram, you can solving diagram, you can
solve a variety of solve a variety of stoichiometricstoichiometric problems.problems.
And Have lots of fun doing it! And Have lots of fun doing it!
Example ProblemExample Problem
�� How many molecules of oxygen are How many molecules of oxygen are
produced when a sample of 29.2 grams of produced when a sample of 29.2 grams of
water is decomposed by electrolysis?water is decomposed by electrolysis?
Balanced Equation: 2HBalanced Equation: 2H22OO(l)(l) →→ 2H2H22(g)(g) + O+ O2(g)2(g)
29.2g H29.2g H22O x O x 1mol H1mol H22OO X X 1mol O1mol O22 X X 6.02 e^6.02 e^2323 molecules Omolecules O22 ==
18g H18g H220 2mol H0 2mol H22O 1 mol 0O 1 mol 022
4.8 X 10^4.8 X 10^2323 molecules Omolecules O22
Example ProblemExample Problem
�� How many liters of oxygen are produced How many liters of oxygen are produced
by the decomposition of 6.54 grams of by the decomposition of 6.54 grams of
potassium chlorate?potassium chlorate?
Balanced Equation: 2KClOBalanced Equation: 2KClO3(s)3(s) →→ 3O3O2(g)2(g) + 2KCl+ 2KCl(s)(s)
6.54g KClO6.54g KClO33 x x 1mol KClO1mol KClO33 x x 3mol O3mol O22 x x 22.4L O22.4L O22 ==
122.5g KClO122.5g KClO33 2mol KClO2mol KClO3 3 1mol O1mol O22
1.79L O1.79L O22
Example ProblemExample Problem
�� Assuming STP, how many liters of oxygen gas Assuming STP, how many liters of oxygen gas
are needed to produce 19.8 L sulfur trioxide are needed to produce 19.8 L sulfur trioxide
according to this balanced equation?according to this balanced equation?
2SO2SO2(g)2(g) + O+ O2(g)2(g) �� 2SO2SO3(g)3(g)
19.8L SO19.8L SO33 x x 1mol SO1mol SO33 x x 1mol O1mol O22 x x 22.4L O22.4L O22 ==
22.4L SO22.4L SO3 3 2mol SO2mol SO3 3 1mol O1mol O22
9.9L O9.9L O22
Example ProblemExample Problem
�� Nitrogen monoxide and oxygen gas combine to form the Nitrogen monoxide and oxygen gas combine to form the
brown gas nitrogen dioxide. How many milliliters of brown gas nitrogen dioxide. How many milliliters of
nitrogen dioxide are produced when 3.4 milliliters of nitrogen dioxide are produced when 3.4 milliliters of
oxygen react with an excess of nitrogen monoxide? oxygen react with an excess of nitrogen monoxide?
Assume STP conditions. Assume STP conditions.
Balanced Equation: 2NO + OBalanced Equation: 2NO + O22 →→ 2NO2NO22
0.0034L O0.0034L O22 x x 1mol O1mol O22 x x 2mol NO2mol NO22 x x 22.4L NO22.4L NO22 x x 1000ml NO1000ml NO2 2 ==
22.4L O22.4L O22 1mol O1mol O22 1mol NO1mol NO22 1L NO1L NO22
Practice ProblemsPractice Problems
See Note Pack and TransparenciesSee Note Pack and Transparencies
And most of all…..And most of all…..
Have FunHave Fun
Section 9.3Section 9.3
Limiting ReagentLimiting Reagent
Limiting ReagentLimiting Reagent
�� The limiting reagent limits or determines The limiting reagent limits or determines
the amount of product that can be formed the amount of product that can be formed
in a reactionin a reaction
�� In contrast, the excess reagent is the In contrast, the excess reagent is the
reagent that is not used up in a reaction reagent that is not used up in a reaction
(there is some left over)(there is some left over)
Ammonia Production (again)Ammonia Production (again)
NN2(g)2(g) + 3H+ 3H2(g)2(g) →→ 2NH2NH3(g)3(g)
�� This represents the most efficient recipe This represents the most efficient recipe
that a chemist can followthat a chemist can follow
��What is the limiting reagent if the reaction What is the limiting reagent if the reaction
is run with 2 moles of nitrogen and 3 is run with 2 moles of nitrogen and 3
moles of hydrogen?moles of hydrogen?
HYDROGENHYDROGEN
Example ProblemExample Problem
�� Sodium chloride can be prepared by the reaction Sodium chloride can be prepared by the reaction
of sodium with chlorine gas according to the of sodium with chlorine gas according to the
following equation:following equation:
2Na2Na(s)(s) + Cl+ Cl2(g)2(g) →→ 2NaCl2NaCl
�� Suppose that 6.7 mol of sodium reacts with 3.2 Suppose that 6.7 mol of sodium reacts with 3.2
mol of chlorine.mol of chlorine.
�� How many moles of sodium chloride are produced?How many moles of sodium chloride are produced?
�� What is the limiting reagent?What is the limiting reagent?
Follow along on page 254 of the bookFollow along on page 254 of the book
Example ProblemExample Problem
2Na2Na(s)(s) + Cl+ Cl2(g)2(g) →→ 2NaCl2NaCl
�� Suppose that 6.7 mol of sodium reacts with 3.2 Suppose that 6.7 mol of sodium reacts with 3.2
mol of chlorine.mol of chlorine.
�� How many moles of sodium chloride are produced?How many moles of sodium chloride are produced?
Choose one known and compare it to the required amount of the otChoose one known and compare it to the required amount of the other reactant.her reactant.
6.7 mol Na x 1mol Cl6.7 mol Na x 1mol Cl22 / 2 mol Na = 3.35 mol Cl/ 2 mol Na = 3.35 mol Cl2 2 (Limiting)(Limiting)
3.20 mol Cl3.20 mol Cl22 x 2 mol x 2 mol NaClNaCl / 1 mol Cl/ 1 mol Cl22 = 6.40 mol = 6.40 mol NaClNaCl (Answer)(Answer)
�� What is the limiting reagent?What is the limiting reagent?
The limiting reagent becomes ClThe limiting reagent becomes Cl22 because we do not because we do not
have enough to completely react all of the Nahave enough to completely react all of the Na
Example Problem 2Example Problem 2
��Write the equation for the complete Write the equation for the complete
combustion of combustion of etheneethene, C, C22HH44
�� If 2.7 mol of If 2.7 mol of etheneethene is reacted with 6.3 mol is reacted with 6.3 mol
of oxygen…of oxygen…
�� Calculate the moles of water producedCalculate the moles of water produced
�� What is the limiting reagentWhat is the limiting reagent
Practice Problem 3Practice Problem 3
��Write the equation for the incomplete Write the equation for the incomplete
combustion of combustion of etheneethene, C, C22HH44
�� If 2.7 mol of If 2.7 mol of etheneethene is reacted with 6.3 mol is reacted with 6.3 mol
of oxygen…of oxygen…
�� Calculate the moles of water producedCalculate the moles of water produced
�� What is the limiting reagentWhat is the limiting reagent
Practice Problem 4Practice Problem 4
��When copper reacts with sulfur, solid When copper reacts with sulfur, solid
copper (I) sulfide is produced. Write this copper (I) sulfide is produced. Write this
reaction.reaction.
��What is the limiting reagent when 80.0 What is the limiting reagent when 80.0
grams of copper reacts with 25.0 grams of grams of copper reacts with 25.0 grams of
sulfur?sulfur?
��What is the maximum number of grams of What is the maximum number of grams of
product that can be formed?product that can be formed?
Practice Problem 5Practice Problem 5
�� Hydrogen gas can be produced in the lab by the Hydrogen gas can be produced in the lab by the
reaction of magnesium metal with hydrochloric reaction of magnesium metal with hydrochloric
acid. Write the COMPLETE balanced equation.acid. Write the COMPLETE balanced equation.
�� Identify the limiting reagent when 6 grams of Identify the limiting reagent when 6 grams of HClHCl
react with 5 grams of Mg.react with 5 grams of Mg.
�� How many grams of hydrogen can be How many grams of hydrogen can be
produced?produced?
The EndThe End