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Stoichiometry
Chapter 9
StoichiometryChapter 9
Version 1.0
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Objectives
Define stoichiometry
Solve stoichiometric problems involving
mole-ratio method
Mass-mass relation
Mass-volume relation
Volume-volume relation
Define excess reagent, limiting reagent, and
percent yield
Differentiate theoretical from actual yield 2
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A Short ReviewA Short Review
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The molar mass of an element is its
atomic mass in grams. It contains 6.022 x 1023 atoms
(Avogadros number) of the element.
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The molar mass of an element or compound isthe sum of the atomic masses of all its atoms.
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Formula Weight
Formula weightFormula weight: the sum of the atomicweights in atomic mass units (amu) of all
atoms in a compounds formula:
Ionic Compounds
Sodium chloride (NaCl) 23.0 amu Na + 35.5 amu Cl = 58.5 amu
Aspirin (C9H8O4) 9(12.0 amu C) + 8(1.0 amu H) +4(16.0 amu O) = 180.0 amu
Water (H2O) 2(1.0 amu H) + 16.0 amu O = 18.0 amu
Nickel(II) chloride hydrate(NiCl26H2O)
58.7 amu Ni + 2(35.5 amu Cl) +12(1.0) amu H) + 6(12.0 amu O) = 237.7 amu
Molecular Compounds
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Formula Weight
Formula weightFormula weight can be used for both ionicand molecular compounds; it tells nothing
about whether a compound is ionic or
molecular. Molecular weightMolecular weight should be used only for
molecular compounds.
For comparison, we use formula weight forionic compounds and molecular weight for
molecular compounds.
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grams of a substancemolar mass =number of moles of the substance
grams of a monoatomic elementmolar mass =number of moles of the element
23number of moleculesnumber of moles =
6.022 x 10 molecules/mole
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Avogadros
Number ofParticles
6 x 1023
Particles
Molar Mass
1 MOLE
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2 2Al + Fe2O3 Fe + Al2O3
For calculations of mole-mass-volume
relationships.
The chemical equation mustbe balanced.
2 mol 2 mol1 mol 1 mol
The equation is balanced.
The number in front of a formula
represents the number of moles of the
reactant or product.
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Stoichiometry: The area of chemistry
that deals with the quantitativerelationships between reactants and
products.
Mole Ratio: a ratio between the molesof any two substances involved in a
chemical reaction.
The coefficients used in mole ratioexpressions are derived from the
coefficients used in the balanced
equation.
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Stoichiometry
Stoichiometry:Stoichiometry: the study of mass relationshipsin chemical reactions
following is an overview of the types of
calculations we study
M o l e s o f AG r a m s o f AG r a m s o f BM o l e s o f B
F r o m m o l e s t o m o l e s ,u s e t h e c o e f f i c i e n t s i n
t h e b a l a n c e d e q u a t i o na s a c o n v e r s i o n f a c t o r
F r o m g r a m s t o m o l e s ,u s e m o l a r m a s s ( g / m o l )
a s a c o n v e r s i o n f a c t o r
F r o m m o l e s t o g r a m s ,u s e m o l a r m a s s ( g / m o l )
a s a c o n v e r s i o n f a c t o r
Y o u a r e g i v e n o n e o f t h e s e A n d a s k e d t o f i n d o n e o f t h e s e
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ExamplesExamplesExamplesExamples
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2
23 m
1 mol
olH
N
N2 + 3H2 2NH31 mol 2 mol3 mol
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1 mol 2 mol3 mol
N2 + 3H2 2NH3
2
32 mo
3 molH
l NH
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The mole ratio is used to convert thenumber of moles of one substance to
the corresponding number of moles of
another substance in a stoichiometryproblem.
The mole ratio is used in the solution of
every type of stoichiometry problem.
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Stoichiometry
Problem: how many grams of nitrogen, N2, arerequired to produce 7.50 g of ammonia, NH3
first find how many moles of NH3 are in7.50 g of NH3
next find how many moles of N2 arerequired to produce this many moles of NH3
7.50 g NH3 x1 mol NH3
17.0 g NH3= mol NH3
7.50 g NH3 x1 mol NH3
17.0 g NH3
x1 mol N2
2 mol NH3= mol N2
N2 (g) + 2NH3 ( g)3H2 (g)
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Stoichiometry
Finally convert moles of N2 to grams of N2and now do the math
x28.0 g N2
1 mol N2= 6.18 g N27.50 g NH3 x
1 mol NH3
17.0 g NH3
x1 mol N2
2 mol NH3
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Identify the starting substance from the
data given in the problem statement.
Convert the quantity of the starting
substance to moles, if it is not already
done.1 mole
moles = gramsmolar mass
Step 1. Determine the number of moles of starting
substance.
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The number of moles of each substance in the
balanced equation is indicated by thecoefficient in front of each substance. Use
these coefficients to set up the mole ratio.
moles of desired substance in the equationmole ratio =
moles of starting substance in the equation
Step 2. Determine the mole ratio of the desired
substance to the starting substance.
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moles of desired substance
in the equationmoles of desired substance = moles of starting substance
moles of starting substance
in the equation
Step 2. Determine the mole ratio of the desired
substance to the starting substance.
Multiply the number of moles of starting
substance (from Step 1) by the mole ratio
to obtain the number of moles of desired
substance.
h f ll i i h l f bCl
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moles of desired substance in the equationmoles of desired substance = moles of starting substance
moles of starting substance in the equation
In the following reaction how many moles of PbCl2are formed if 5.000 moles of NaCl react?
2NaCl(aq) + Pb(NO3)2(aq) PbCl2(s) + 2NaNO3(aq)
5.000 moles NaCl2moles of PbCl = 22.500 mol PbCl2
1 mol PbCl
2 mol NaCl
=
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Step 3. Calculate the desired substance in the
units specified in the problem.
If the answer is to be in moles, the
calculation is complete
If units other than moles are wanted,
multiply the moles of the desired substance
(from Step 2) by the appropriate factor to
convert moles to the units required.
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Step 3. Calculate the desired substance in the
units specified in the problem.
molar mass1. To calculate : grams =gr moles x
1 moams
l
22 2
2
18.02 g H O90.10 grams H O = 5.000 mol H O
1 mol H
O
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236.022 x 10 atoms
2. To calculate : atoms = moles1 mo
atomsl
2324 6.022 x 10 Na atoms3.011 x 10 Na atoms = 5.000 moles Na atoms
1 mol Na atoms
Na
Step 3. Calculate the desired substance in the
units specified in the problem.
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236.022 x 10 molecules
3. To calculate : molecules =mol moles x1 mol
ecules
2324 2
2 2
2
6.022 x 10 H O molecules3.011 x 10 molecules H O = 5.000 moles H O
1 mol H O
Step 3. Calculate the desired substance in the
units specified in the problem.
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Stoichiometry
Problem 1. What mass of aluminum oxide isrequired to prepare 27 g of aluminum?
Problem 2. How many grams each of CO22
and NH33
are produced from 0.83 mol of
urea?
Al2O3 ( s)electrolysis Al(s) + O2 ( g)
(NH2 )2CO(aq) + 2NH3 (aq) + CO2 (g)H2O
Urea
urease
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Mole-MoleMole-MoleCalculationsCalculations
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Phosphoric Acid
Phosphoric acid (H3PO4) is one of the most
widely produced industrial chemicals in the
world.
Most of the worlds phosphoric acid is
produced by the wet process which involves
the reaction of phosphate rock, Ca5(PO4)3,F
with sulfuric acid (H2SO4).
Ca5(PO4)3F(s) + 5H2SO4 3H3PO4 + HF + 5CaSO4
C l l t th b f l f h h i id
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Mole Ratio
Calculate the number of moles of phosphoric acid
(H3PO4) formed by the reaction of 10 moles of sulfuric
acid (H2SO4).
Ca5(PO4)3F + 5H2SO4 3H3PO4 + HF + 5CaSO4
Step 1 Moles starting substance: 10.0 mol H2SO4
Step 2 The conversion needed is
moles H2SO4 moles H3PO4
1 mol 5 mol 3 mol 1 mol 5 mol
3 42 4
2 4
3 mol H PO10 mol H SO x =
5 mol H SO3 4
6 mol H PO
C l l t th b f l f lf i id
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Step 2 The conversion needed is
moles Ca5(PO4)3F moles H2SO4
Calculate the number of moles of sulfuric acid
(H2SO4) that react with 10 moles of Ca5(PO4)3F.
Ca5(PO4)3F + 5H2SO4
3H3PO4 + HF + 5CaSO4
Mole Ratio
Step 1 The starting substance is 10.0 mol Ca5(PO4)3F
1 mol 5 mol 3 mol 1 mol 5 mol
2 45 4 3
5 4 3
5 mol H SO10 mol Ca (PO ) F x =
1 mol Ca (PO ) F2 450 mol H SO
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Mole-Mass CalculationsMole-Mass Calculations
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1. The object of this type of problem is
to calculate the mass of onesubstance that reacts with or is
produced from a given number of
moles of another substance in achemical reaction.
2. If the mass of the starting substance
is given, we need to convert it to
moles.
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3. We use the mole ratio to convert
moles of starting substance to molesof desired substance.
4. We can then change moles of desiredsubstance to mass of desired
substance if called for by the
problem.
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ExamplesExamplesExamplesExamples
Calculate the number of moles of H SO necessary to
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2 4
3 4
5 mol H SO=
3 mol H PO
Calculate the number of moles of H2SO4 necessary to
yield 784 g of H3PO4.
Ca5(PO4)3F+ 5H2SO4
3H3PO4 + HF + 5CaSO4Method 1 Step by Step
Step 1 The starting substance is 784 grams of H3PO4.
Step 2 Convert grams of H3PO4 to moles of H3PO4.
Step 3 Convert moles of H3PO4 to moles of H2SO4 by themole-ratio method.
3 4
3 4
1 mol H PO=
98.0 g H PO
3 48.00 mol H PO
2 413.3 mol H SO
3 4784 g H PO
3 48.00 mol H PO
Calculate the number of moles of H SO necessary to
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Mole Ratio
2 4
3 4
5 mol H SO=
3mol H PO
Calculate the number of moles of H2SO4 necessary to
yield 784 g of H3PO4
Ca5(PO4)3F+ 5H2SO4
3H3PO4 + HF + 5CaSO4Method 2 Continuous
grams H3PO4 moles H3PO4 moles H2SO4
The conversion needed is
3 4
784 g H PO 3 4
3 4
1 mol H PO
98.0 g H PO
2 413.3 mol H SO
Calculate the number of grams of H required to form
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36.0
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Method 1 Step by Step
Step 1 The starting substance is 12.0 moles of NH3
Step 2 Calculate moles of H2 by the mole-ratio method.
Step 3 Convert moles of H2 to grams of H2.
Calculate the number of grams of H2 required to form
12.0 moles of NH3.
N2 + 3H2
2NH3
312.0 mol NH 218.0 mol H
2
1
8.0
Mole Ratio
2
3
3 mol H=
2 mol NH
22
2.0
2g
1m
ol
=
Calculate the number of grams of H required to form
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Mole Ratio
2
3
3 mol H
2 mol NH
moles NH3 moles H2 grams H2
Calculate the number of grams of H2 required to form
12.0 moles of NH3.
N2 + 3H2
2NH3Method 2 Continuous
The conversion needed is
312.0 mol NH2
2
2.02 gH=
1mol H
236.0 g H
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Mass-Mass CalculationsMass-Mass Calculations
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Solving mass-mass stoichiometry problems
requires all the steps of the mole-ratiomethod.
1. The mass of starting substance is
converted to moles.
2. The mole ratio is then used to determine
moles of desired substance.
3. The moles of desired substance are
converted to mass of desired substance.
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Mass-mass Relations
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g
u
g
ugu
NN
MmMmmm =
mu = mass of unknown substancemg = mass of given substance
Mmu = molar mass of the given substance
Mmg = molar mass of the given substance
Nu = coefficient of the unknown in the balanced
equation
Ng = coefficient of the given substance in the
balanced equation
Calculate the number of grams of NH formed by the
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Calculate the number of grams of NH3 formed by the
reaction of 112 grams of H2.
N2
+ 3H2
2NH3Method 1 Step by Step
Step 1 The starting substance is 112 grams of H2. Convert
112 g of H2 to moles.
grams moles2
2
1 mol H
2.02 gH
=
2112 g H 255.4 moles H
Step 2 Calculate the moles of NH3 by the mole ratio method.
3
2
2 mol NH=
3 mol H
255.4 moles H 336.9 moles NH
Calculate the number of grams of NH formed by the
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Calculate the number of grams of NH3 formed by the
reaction of 112 grams of H2.
N2
+ 3H2
2NH3Method 1 Step by Step
Step 3 Convert moles NH3 to grams NH3.
moles grams
336.9 moles NH3
3
17.0 g NH =1 mol NH
3629 g NH
Calculate the number of grams of NH formed by the
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Calculate the number of grams of NH3 formed by the
reaction of 112 grams of H2.
N2
+ 3H2
2NH3
grams H2 moles H2 moles NH3 grams NH3
2
2
1 mol H2.02 g H
2112 gH 32
2 mol NH3 mol H
Method 2 Continuous
3
3
17.0 g NH =1 mol NH
3629 g NH
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Mass-VolumeMass-VolumeCalculationsCalculations
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Mass-volume Relations
Remember:1 mole of any substance occupies 22.4 L
Gases are assumed to behave as ideal
gasesThe reaction must take place at STP
(standard temperature and pressure)
where T = 0o C = 273 KP = 1 atm = 760 mm Hg
As gas not in STP must be converted to49
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Mole-Volume CalculationsMole-Volume Calculations
Mass-Volume CalculationsMass-Volume Calculations
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Mass-volume Relations
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giveng
u
g
gumol1
L4.22
N
N
Mm
1mvol =
L22.4mol1
NNMmvm given
g
uugu =
For unknown volume
For unknown mass
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Mass-volume Relations
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volu = volume of unknown substance
mu = mass of unknown substance
mg = given mass
vg = given volumeMmg = molar mass of the given substance
Mmu = molar mass of the unknown substance
Nu
= coefficient of the unknown in the balanced
equation
Ng = coefficient of the given substance in the
balanced equation
What volume of oxygen (at STP) can be formed
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Step 1 Write the balanced equation
2 KClO3 2 KCl + 3 O2
Step 2 The starting amount is 0.500
mol KClO3. The conversion is
moles KClO3 moles O2 liters O2
What volume of oxygen (at STP) can be formed
from 0.500 mol of potassium chlorate?
What volume of oxygen (at STP) can be formed
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2
3
3 mol O
2 mol KClO
Step 3. Calculate the moles of O2, using the
mole-ratio method.
3(0.500 mol KClO )
What volume of oxygen (at STP) can be formed
from 0.500 mol of potassium chlorate?
Step 4. Convert moles of O2 to liters of O2
2= 0.750 mol O
2(0.750 mol O )22.4 L
1 mol
2= 16.8 L O
2 KClO3 2KCl + 3 O2
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Volume-VolumeVolume-VolumeCalculationsCalculations
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Volume-volume Relations
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g
ugu
N
Nvv =
This can be solved by ratio and proportion
vu= volume of unknown substance
vg= volume of given substance
Nu
= number of moles of unknown substance from
the balanced equation
ng = number of moles of given substance from the
balanced equation
For reacting gases at constant temperature and
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H2(g) + Cl2(g) 2HCl(g)
1 mol H2 1 mol Cl2 2 mol HCl22.4 L
STP
22.4 L
STP
2 x 22.4 L
STP
1 volume 1 volume 2 volumes
Y volume Y volume 2Y volumes
g g p
pressure: Volume-volume relationships are the same
as mole-mole relationships.
What volume of oxygen will react with 150 L of
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yg
hydrogen to form water vapor? What volume of water
vapor will be formed?
2H2(g) + O2(g)
2H2O(g)
Assume that both reactants and products are measured at
STP. Calculate by using reacting volumes:
2 mol 1 mol 2 mol
2 x 22.4 L 22.4 L 2 x 22.4 L
2 volumes 1 volume 2 volumes
What volume of oxygen will react with 150 L of
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yg
hydrogen to form water vapor? What volume of water
vapor will be formed?
2H2(g) + O2(g) 2H2O(g)
( ) 22
22 OL75
Hvol2Ovol1HL150 =
( ) OHL150Hvol2
OHvol2HL150 2
2
22 =
What volume of nitrogen will react with 600. mL of
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g
hydrogen to form ammonia? What volume of
ammonia will be formed?
N2(g) + 3H2(g) 2NH3(g)
2600. ml H 2
2
1 vol N3 vol H
2= 200. mL N
2600. ml H3
2
2 vol NH3 vol H
3= 400. mL NH
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Limiting-Reactant andLimiting-Reactant andYield CalculationsYield Calculations
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Limiting Reactant/ReagentLimiting Reactant/ReagentLimiting Reactant/ReagentLimiting Reactant/Reagent
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It is called the limiting reactant
because the amount of it present isinsufficient to react with the amounts
of other reactants that are present.
The limiting reactant limits the amountof product that can be formed.
The limiting reactant/reagent is one of
the reactants in a chemical reaction.
How many bicyclesFrom eight wheels four
From four frames fourFrom three pedal assemblies
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can be assembled
from the parts
shown?
bikes can be constructed.bikes can be constructed.three bikes can be constructed.
The limiting part is thenumber of pedal
assemblies.
9.2
H + Cl 2HCl
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H2 + Cl2 2HCl
+
7 molecules H2 can
form 14 molecules HCl
4 molecules Cl2 can form
8 molecules HCl 3 molecules of H2 remain
H2 is in excess
Cl2 is the limiting
reactant9.3
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Steps Used to Determine theSteps Used to Determine theLimiting ReactantLimiting ReactantSteps Used to Determine theSteps Used to Determine theLimiting ReactantLimiting Reactant
1 Calculate the amount of product (moles or
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1. Calculate the amount of product (moles or
grams, as needed) formed from each
reactant.2. Determine which reactant is limiting. (Thereactant that gives the least amount of
product is the limiting reactant; the other
reactant is in excess.3. Calculate the amount of the other reactant
required to react with the limiting reactant,
then subtract this amount from the startingquantity of the reactant. This gives the
amount of the substance that remains
unreacted.
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ExamplesExamplesExamplesExamples
How many moles of HCl can be produced by reacting
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4.0 mol H2 and 3.5 mol Cl2? Which compound is the
limiting reagent?
Step 1 Calculate the moles of HCl that can form
from each reactant.
24.0 mol H
2
2 mol HCl1 mol H
=
8.0 mol HCl
23.5 mol Cl
2
2 mol HCl
1 mol Cl
=
7.0 mol HCl
H2 + Cl2 2HCl
Step 2 Determine the limiting reactant.
The limiting reactant is Cl2because it
producesless HCl than H2.
How many grams of silver bromide (AgBr) can be
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formed when solutions containing 50.0 g of MgBr2
and 100.0 g of AgNO3 are mixed together? How
many grams of the excess reactant remain unreacted?
Step 1 Calculate the grams of AgBr that can form
from each reactant.
MgBr2(aq) + 2AgNO3 (aq) 2AgBr(s) + Mg(NO3)2(aq)
The conversion needed is
g reactant mol reactant mol AgBr g
AgBr( )250.0 g MgBr 102 g AgBr
2
2
1 mol MgBr
184.1 g MgBr
2
2 mol AgBr
1 mol MgBr
187.8 g AgBr
1 mol AgBr
=
( )3100.0 g AgNO 110.5 g AgBr3
3
1 mol AgNO
169.9 g AgNO
3
2 mol AgBr
2 mol AgNO
187.8 g AgBr
1 mol AgBr
=
How many grams of silver bromide (AgBr) can be
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formed when solutions containing 50.0 g of MgBr2
and 100.0 g of AgNO3 are mixed together? How
many grams of the excess reactant remain unreacted?
Step 2 Determine the limiting reactant.
MgBr2(aq) + 2AgNO3 (aq) 2AgBr(s) + Mg(NO3)2(aq)
( )250.0 g MgBr 102 g AgBr2
2
1 mol MgBr
184.1 g MgBr
2
2 mol AgBr
1 mol MgBr
187.8 g AgBr
1 mol AgBr
=
( )3100.0 g AgNO 110.5 g AgBr3
3
1 mol AgNO
169.9 g AgNO
3
2 mol AgBr
2 mol AgNO
187.8 g AgBr
1 mol AgBr
=
The limiting reactant is MgBr2 because it
forms less Ag Br.
How many grams of the excess reactant (AgNO3)
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remain unreacted?
Step 3 Calculate the grams of unreacted AgNO3.
First calculate the number of grams of
AgNO3 that will react with 50 g of MgBr2.
MgBr2
(aq) + 2AgNO3
(aq)
2AgBr(s) + Mg(NO3
)2
(aq)
( )250.0 g MgBr 392.3 g AgNO22
1 mol MgBr
184.1 g MgBr
3
2
2 mol AgNO
1 mol MgBr
3
3
169.9 g AgNO
1 mol AgNO
=
The conversion needed is
g MgBr2 mol MgBr2 mol AgNO3 g AgNO3
The amount of MgBr2 that remains is
100.0 g AgNO3
- 92.3 g AgNO3
= 7.7 g AgNO3
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Reaction YieldReaction YieldReaction YieldReaction Yield
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The quantities of products calculated
from equations represent the maximumyield (100%) of product according to the
reaction represented by the equation.
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Many reactions fail to give
a 100% yield of product.
This occurs because of side reactions
and the fact that many reactions are
reversible.
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The theoretical yield of a reaction isthe calculated amount of product that
can be obtained from a given amount
of reactant. The actual yield is the amount of
product finally obtained from a given
amount of reactant.
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The percent yield of a reaction is theratio of the actual yield to the theoretical
yield multiplied by 100.
actual yieldx 100 = percent yield
theoretical yield
Silver bromide was prepared by reacting 200.0 g of
i b id d d f il
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187.8 g AgBr
1 mol AgBr
=
magnesium bromide and an adequate amount of silver
nitrate. Calculate the percent yield if 375.0 g of silver
bromide was obtained from the reaction:MgBr2(aq) + 2AgNO3 (aq) 2AgBr(s) + Mg(NO3)2(aq)
Step 1 Determine the theoretical yield by
calculating the grams of AgBr that can beformed.
The conversion needed is
g MgBr2 mol MgBr2 mol AgBr g AgBr
( )2200.0 g MgBr 408.0 g AgBr2
2
1 mol MgBr
184.1 g MgBr
2
2 mol AgBr
1 mol MgBr
Silver bromide was prepared by reacting 200.0 g of
i b id d d f il
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magnesium bromide and an adequate amount of silver
nitrate. Calculate the percent yield if 375.0 g of silver
bromide was obtained from the reaction:MgBr2(aq) + 2AgNO3 (aq) 2AgBr(s) + Mg(NO3)2(aq)
Step 2 Calculate the percent yield.
actual yieldpercent yield = x 100
theoretical yield
percent yield =375.0 g AgBr
x 100 =408.0 g AgBr
91.9%
must have same units
must have same units
Solve the following problems correctly.
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g p y
1. Given the equation
Al4C3 + 12 H2O 4 Al(OH)3 + 3 CH4
a. How many moles of water are needed to react
with 100 g of Al4C3 ?
b. How many moles of Al(OH)3 will be produced
when 0.600 mol of CH4 is formed?
2. How many grams of zinc phosphate are formed
when 10.0 g of Zn are reacted with phosphoric
acid?
3 Zn + 2 H3PO4 Zn3(PO4)2 + 3 H2 83
Solve the following problems correctly.1 Gi th ti
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1. Given the equation
4 FeS2 + 11 O2 2 Fe2O3 + 8 SO2
a. How many moles ofFe2O3 can be made from 1.00 mol
of FeS2?
b. How many moles of O2
are required to react with
4.5 mol of FeS2?
c. If the reaction produces 1.55 mol ofFe2O3 , how
many mol of SO2
are produced?
d. How many grams of SO2 can be formed from 0.512
mol of FeS2?
e. How many grams of FeS2 are needed to produce 22184
In the following equations, determine which reactant
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is the limiting reactant and which reactant is in excess.
1. 2 Bi(NO3)3 + 3 H2S Bi2S3 + 6 HNO3
50.0 g 6.00 g
2. 3 Fe + 4 H2O Fe3O4 + 4 H2
40.0 g 16.0 g
85
In the following equations, determine which reactant
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is the limiting reactant and which reactant is in excess.
Acetylene (C2H2) can be manufactured bythe reaction of water and calcium carbide,
CaC2:
CaC2 + 2 H2O C2H2 + Ca(OH)2
When 44.5 g of commercial grade (impure)
calcium carbide is reacted, 0.540 mol of C2H2, isproduced. Assuming that all CaC2 was reacted to
C2H2 , what is the percent of CaC2 in the
commercial grade material?86
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