Transcript The Concept of Limiting Reactant

Chapter 3
Stoichiometry
Chapter 3
Chemical Stoichiometry
 Stoichiometry – The study of quantities of
materials consumed and produced in chemical
reactions.
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Section 3.1
Counting by Weighing



Objects behave as though they were all identical.
Atoms are too small to count.
Need average mass of the object.
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Section 3.1
Counting by Weighing
EXERCISE!
A pile of marbles weigh 394.80 g. 10 marbles weigh
37.60 g. How many marbles are in the pile?
Avg. Mass of 1 Marble =
37.60 g
= 3.76 g / marble
10 marbles
394.80 g = 105 marbles
3.76 g
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Section 3.2
Atomic Masses
 The atomic mass of each element is listed directly
below the element’s symbol in the periodic table
and represent the average mass of the isotopes
that compose the element, weight according to
the natural abundance of each isotope.
Section 3.2
Atomic Masses




12C
is the standard for atomic mass, with a mass of
exactly 12 atomic mass units (u).
The masses of all other atoms are given relative to this
standard.
Elements occur in nature as mixtures of isotopes.
Carbon = 98.89% 12C
1.11% 13C
< 0.01% 14C
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Section 3.2
Atomic Masses
Average Atomic Mass for Carbon


Even though natural carbon does not contain a single
atom with mass 12.01, for stoichiometric purposes, we
can consider carbon to be composed of only one type of
atom with a mass of 12.01.
This enables us to count atoms of natural carbon by
weighing a sample of carbon.
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Section 3.2
Atomic Masses
Average Atomic Mass for Carbon
98.89% of 12 u + 1.11% of 13.0034 u =
exact number
(0.9889)(12 u) + (0.0111)(13.0034 u) =
12.01 u
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Section 3.2
Atomic Masses
Schematic Diagram of a Mass Spectrometer
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Section 3.2
Atomic Masses
EXERCISE!
An element consists of 62.60% of an isotope with
mass 186.956 u and 37.40% of an isotope with mass
184.953 u.
 Calculate the average atomic mass and identify the
element.
 (0.6260)(186.956u)+(0.3740)(184.953)
186.2 u
Rhenium (Re)
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Section 3.3
The Mole
The mole
 A mole (mol) is a unit of measure for an amount
of a chemical substance.
 A mole is the amount of material containing in
 6.022 ×1023 particles
 This number is called Avogadro's number.
Section 3.3
The Mole





The number equal to the number of carbon atoms in
exactly 12 grams of pure 12C.
1 mole of something consists of 6.022 × 1023 units of
that substance (Avogadro’s number).
1 mole of anything is 6.022 ×1023 units of that thing.
The mass of 1mole of an element is equal to its atomic
mass.
1 mole C = 6.022 × 1023 C atoms = 12.01 g C
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Section 3.3
The Mole
Molar mass
 The atomic mass of any substance express in
grams is the molar mass of that substance.
 The atomic mass of iron is 55.85amu.
 Therefore, the molar mass of iron is 55.85g/mol.
 Since oxygen occurs naturally as a diatomic, O2
the molar mass is 2 times 16.00g = 32.00g/mol.
Section 3.3
The Mole
EXERCISE!

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Section 3.3
The Mole
Converting from moles to grams

Section 3.3
The Mole
Convert from moles to gram

Section 3.3
The Mole
Convert from grams to moles

Section 3.3
The Mole
Convert from grams to moles

Section 3.3
The Mole
Converting from moles to molecule




0.200 moles of H2O contain how many molecules
We multiple by 6.022 x 1023
0.200 x 6.022x 1023 mol-1
1.20 x 1023 molecules
Section 3.3
The Mole
Conversion chart
 Grams to moles Moles to atom and Molecule
Section 3.3
The Mole








Calculate how many grams are in
4.56 moles of KCIO
5.63 moles NaOH
4.5 moles KI
Calculate how many moles are in
75.6 grams of KBr
200 grams of KCIO4
750 g of Na2CO3
Section 3.3
The Mole
 .0.65 moles of H2O contains How Many Molecules
 0.43 moles of Nitrogen contain How many atoms
 Calculate the number of molecules in o.750 gram
of Fe
 Calculate the number of molecules in 1.058g of
H2O
Section 3.4
Molar Mass

Mass in grams of one mole of the substance:
Molar Mass of N = 14.01 g/mol
Molar Mass of H2O = 18.02 g/mol
(2 × 1.008 g) + 16.00 g
Molar Mass of Ba(NO3)2 = 261.35 g/mol
137.33 g + (2 × 14.01 g) + (6 × 16.00 g)
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Section 3.4
Molar Mass
Moles and molar mass

Section 3.4
Molar Mass
Moles and molar mass

Section 3.4
Molar Mass
Moles and molar mass

Section 3.4
Molar Mass
CONCEPT CHECK!
Calculate the number of copper atoms in a 63.55 g
sample of copper.
6.022×1023 Cu atoms
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Section 3.4
Molar Mass
CONCEPT CHECK!
Which of the following 100.0 g samples contains the
greatest number of atoms?
a) Magnesium
b) Zinc
c) Silver
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Section 3.4
Molar Mass
EXERCISE!
Rank the following according to number of atoms
(greatest to least):
a) 107.9 g of silver
b) 70.0 g of zinc
c) 21.0 g of magnesium
b)
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a)
c)
29
Section 3.4
Molar Mass
EXERCISE!
Consider separate 100.0 gram samples of each of the
following:
H2O, N2O, C3H6O2, CO2
 Rank them from greatest to least number of
oxygen atoms.
H2O, CO2, C3H6O2, N2O
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Section 3.5
Learning to Solve Problems
Conceptual Problem Solving

Where are we going?
 Read the problem and decide on the final goal.
 How do we get there?
 Work backwards from the final goal to decide where
to start.
 Reality check.
 Does my answer make sense? Is it reasonable?
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Section 3.6
Percent Composition of Compounds

Mass percent of an element:
mass of element in compound
mass % =
× 100%
mass of compound

For iron in iron(III) oxide, (Fe2O3):
mass % Fe =
2(55.85 g)
111.70 g
=
n 100% = 69.94%
2(55.85 g)+3(16.00 g) 159.70 g
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Section 3.6
Percent Composition of Compounds
 What are the mass % of carbon and oxygen in
carbon dioxide co2
 Find the mass of individual atoms
 Look up the atomic masses for carbon and oxygen
 C is 12.01g/mol
 O is 16.00g/mol
 Find the number of grams of each component
make up one mole of Co2
Section 3.6
Percent Composition of Compounds
 One mole of Co2 contain 1 mole of carbon atom
and 2 mole of oxygen atoms.
 12.01 g (1mol) of C
 32.00g (2 mol2 x 16.00 gram per mole) of O
 The mass of one mole of Co2 is
 12.01g +32.00 g = 44.01g
 Find the mass % of each atom
Section 3.6
Percent Composition of Compounds






Find mass %of each atom
Mass % of carbon = (12.01g/ 44.01g ) ×100
Mass of C = 27.29 %
Mass of oxygen =(32.00g/44.01g) × 100
Mass of O = 72.71 %
27.29 + 72.71 = 100.00
Section 3.6
Percent Composition of Compounds
Exercise
 Calculate mass% of C10H14O
 Calculate the Mass % of C2H5OH
 Calculate the mass % of C
Section 3.7
Determining the Formula of a Compound
Formulas

Empirical formula = CH
 Simplest whole-number ratio
 Molecular formula = (empirical formula)n
[n = integer]
 Molecular formula = C6H6 = (CH)6
 Actual formula of the compound
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Section 3.7
Determining the Formula of a Compound
Analyzing for Carbon and Hydrogen

Device used to determine the mass percent of each
element in a compound.
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Section 3.7
Determining the Formula of a Compound

Section 3.7
Determining the Formula of a Compound
Empirical Formula

Section 3.7
Determining the Formula of a Compound

Section 3.7
Determining the Formula of a Compound





Empirical formula mass = 141.94g/mol
Molar mass = 283.88g/mol
Molar mass / empirical mass
283.88/141.94 =2
(P2O5)2 P4O10
Section 3.7
Determining the Formula of a Compound
EXERCISE!
The composition of adipic acid is 49.3% C, 6.9% H,
and 43.8% O (by mass). The molar mass of the
compound is about 146 g/mol.
 What is the empirical formula?
C3H5O2
 What is the molecular formula?
C6H10O4
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Section 3.7
Determining the Formula of a Compound
Exercise
 A compound that contains only carbon, Hydrogen
, Oxygen and Nitrogen is 63.68% C, 12.38% N ,
9.80% H, 14.14% Oxygen calculate the empirical
formula.
 A compound that contain 48.64%C 8.16% H,
 42.20g Oxygen Find the empirical formula
Section 3.7
Determining the Formula of a Compound

A representation of a chemical reaction:
C2H5OH + 3O2
reactants

2CO2 + 3H2O
products
Reactants are only placed on the left side of
the arrow, products are only placed on the
right side of the arrow.
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Section 3.8
Chemical Equations
C2H5OH + 3O2



2CO2 + 3H2O
The equation is balanced.
All atoms present in the reactants are
accounted for in the products.
1 mole of ethanol reacts with 3 moles of
oxygen to produce 2 moles of carbon dioxide
and 3 moles of water.
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Section 3.8
Chemical Equations


The balanced equation represents an overall ratio of
reactants and products, not what actually “happens”
during a reaction.
Use the coefficients in the balanced equation to decide
the amount of each reactant that is used, and the
amount of each product that is formed.
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Section 3.9
Balancing Chemical Equations
Writing and Balancing the Equation for a Chemical Reaction
1. Determine what reaction is occurring. What are the reactants,
the products, and the physical states involved?
2. Write the unbalanced equation that summarizes the reaction
described in step 1.
3. Balance the equation by inspection, starting with the most
complicated molecule(s). The same number of each type of atom
needs to appear on both reactant and product sides. Do NOT
change the formulas of any of the reactants or products.
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Section 3.9
Balancing Chemical Equations
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Section 3.9
Balancing Chemical Equations
Balancing chemical equation
 Al + O2 →Al2O3
 2Al + O2→Al2O3
 There are three atoms of Oxygen on one side and
2 atoms on the other side. We can balance the O
by placing the coefficient 3/2 in front of O2
 2Al + 3/2O2 →Al2O3
 Multiply both side of the equation by 2
 4Al + 3O2 →2Al2O3
Section 3.9
Balancing Chemical Equations
CONCEPT CHECK!
Which of the following are true concerning balanced chemical
equations? There may be more than one true statement.
I. The number of molecules is conserved.
II. The coefficients tell you how much of each substance you
have.
III. Atoms are neither created nor destroyed.
IV. The coefficients indicate the mass ratios of the
substances used.
V. The sum of the coefficients on the reactant side equals the
sum of the coefficients on the product side.
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Section 3.9
Balancing Chemical Equations
Exercise







NH3 + O2 →NO + H2O
C+ O2 → CO
CO + O2 → CO2
H2+Br2→HBr
K +H2O → KOH + H2
Mg + O2 →Mgo
O3 → O2
Section 3.9
Balancing Chemical Equations
Exercise







H2O2 →H2O + O2
N2 + H2 →NH3
Zn + AgCl → ZnCl2 + Ag
S8 + O2 → SO2
NaOH +H2SO4 →Na2SO4+H2O
Cl2 + NaI → NaCl + I2
KOH + H3PO4 →K3PO4+ H2O
Section 3.9
Balancing Chemical Equations
Notice




The number of atoms of each type of element must be
the same on both sides of a balanced equation.
Subscripts must not be changed to balance an equation.
A balanced equation tells us the ratio of the number of
molecules which react and are produced in a chemical
reaction.
Coefficients can be fractions, although they are usually
given as lowest integer multiples.
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Section 3.10
Stoichiometric Calculations:
Amounts of Reactants and Products
Stoichiometric Calculations


Chemical equations can be used to relate the masses of
reacting chemicals.
Stoichiometry is the quantitative study of reactant and
products in a chemical reaction.
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Section 3.10
Stoichiometric Calculations:
Amounts of Reactants and Products
Stoichiometric calculations
 A basic question raised in the chemical laboratory
Is “ How much starting material will be formed
from specific amount of starting material (
reactant)?” or in some cases, we might ask the
reserve question: “ How much starting material
must be used to obtain a specific amount of
product?”.
Section 3.10
Stoichiometric Calculations:
Amounts of Reactants and Products
Calculating Masses of Reactants and Products in Reactions
1. Balance the equation for the reaction.
2. Convert the known mass of the reactant or product to
moles of that substance.
3. Use the balanced equation to set up the appropriate
mole ratios.
4. Use the appropriate mole ratios to calculate the
number of moles of the desired reactant or product.
5. Convert from moles back to grams if required by the
problem.
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Section 3.10
Stoichiometric Calculations:
Amounts of Reactants and Products
Stoichiometric calculations
 C6H12O6 + 6CO2 → 6CO2 + 6H2O
 If 856 g of C6H12O6 is consumed by a person over a
certain period, what is the mass of CO2 produced.
Section 3.10
Stoichiometric Calculations:
Amounts of Reactants and Products
Step 1: The balanced equation is given in the problem.
Step 2: To convert grams of C6H12O6 to moles of C6H12O6, we
write
Step 3: From the mole ratio, we see that
1 mol C6H12O6
≏ 6 mol CO2.
Therefore, the number of moles of CO2 formed is
Section 3.10
Stoichiometric Calculations:
Step 4: Finally,
the number
of grams of CO2 formed is given by
Amounts
of Reactants
and Products
After some practice, we can combine the conversion steps
into one equation:
Section 3.10
Stoichiometric Calculations:
Amounts of Reactants and Products
 C3H8(g) + 5O2(g) → 3CO2(g) +4H2O
 What mass of oxygen will react with 96.1 g of
propane?
 Step 1: Write the balance chemical equation
 Step 2: What are the moles of propane
 To find the moles of propane, we need to know
the molar mass of propane
 Molar mass of propane 36+8 = 44
Section 3.10
Stoichiometric Calculations:
Amounts of Reactants and Products

Section 3.10
Stoichiometric Calculations:
Amounts of Reactants and Products
Calculating Masses of Reactants and Products in Reactions
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Section 3.10
Stoichiometric Calculations:
Amounts of Reactants and Products
EXERCISE!
(Part I)
Methane (CH4) reacts with the oxygen in the air to
produce carbon dioxide and water.
Ammonia (NH3) reacts with the oxygen in the air to
produce nitrogen monoxide and water.
 Write balanced equations for each of these
reactions.
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Section 3.10
Stoichiometric Calculations:
Amounts of Reactants and Products
EXERCISE!
(Part II)
Methane (CH4) reacts with the oxygen in the air to
produce carbon dioxide and water.
Ammonia (NH3) reacts with the oxygen in the air to
produce nitrogen monoxide and water.
 What mass of ammonia would produce the same
amount of water as 1.00 g of methane reacting
with excess oxygen?
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Section 3.10
Stoichiometric Calculations:
Amounts of Reactants and Products
Let’s Think About It

Where are we going?
 To find the mass of ammonia that would produce the
same amount of water as 1.00 g of methane reacting
with excess oxygen.
 How do we get there?
 We need to know:
 How much water is produced from 1.00 g of methane and
excess oxygen.
 How much ammonia is needed to produce the amount of
water calculated above.
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Section 3.11
The Concept of Limiting Reactant
Limiting Reactants


Limiting reactant – the reactant that runs out first and
thus limits the amounts of products that can be formed.
Determine which reactant is limiting to calculate
correctly the amounts of products that will be formed.
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Section 3.11
The Concept of Limiting Reactant
Limiting Reactants
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Section 3.11
The Concept of Limiting Reactant
A. The Concept of Limiting Reactants
 Stoichiometric mixture
 N2(g) + 3H2(g)
2NH3(g)
Section 3.11
The Concept of Limiting Reactant
A. The Concept of Limiting Reactants
 Limiting reactant mixture
 N2(g) + 3H2(g)  2NH3(g)
Section 3.11
The Concept of Limiting Reactant
A. The Concept of Limiting Reactants
 Limiting reactant mixture
 N2(g) + 3H2(g)
2NH3(g)
 Limiting reactant is the reactant that runs out first.
 H2
Section 3.11
The Concept of Limiting Reactant
Limiting Reactants



The amount of products that can form is limited by the
methane.
Methane is the limiting reactant.
Water is in excess.
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Section 3.11
The Concept of Limiting Reactant
 Suppose 25.0 kg of Nitrogen and 5.00 kg of
Hydrogen are mixed and reacted to form
ammonia. How do we calculate the mass of
ammonia produced when this reaction is run to
completion until one of the reactants is
completely consumed.
 N2(g) + 3H2(g) → 2NH2 (g)
Section 3.11
The Concept of Limiting Reactant

Section 3.11
The Concept of Limiting Reactant
Urea [(NH2)2CO] is prepared by reacting ammonia with carbon
dioxide:
In one process, 637.2 g of NH3 are treated with 1142 g of CO2.
(a) Which of the two reactants is the limiting reagent?
(b) Calculate the mass of (NH2)2CO formed.
(c) How much excess reagent (in grams) is left at the end of the
reaction?
Section 3.11
The Concept of Limiting Reactant
(a) Strategy The reactant that produces fewer moles of product
is the limiting reagent because it limits the amount of
product that can be formed.
How do we convert from the amount of reactant to amount
of product?
Perform this calculation for each reactant, then compare the
moles of product, (NH2)2CO, formed by the given amounts
of NH3 and CO2 to determine which reactant is the limiting
reagent.
Section 3.11
The Concept of Limiting Reactant
Solution We carry out two separate calculations. First, starting
with 637.2 g of NH3, we calculate the number of moles of
(NH2)2CO that could be produced if all the NH3 reacted
according to the following conversions:
Combining these conversions in one step, we write
Section 3.11
The Concept of Limiting Reactant
Second, for 1142 g of CO2, the conversions are
The number of moles of (NH2)2CO that could be produced if all
the CO2 reacted is
It follows, therefore, that NH3 must be the limiting reagent
because it produces a smaller amount of (NH2)2CO.
Section 3.11
The Concept of Limiting Reactant
(b) Strategy We determined the moles of (NH2)2CO produced
in part (a), using NH3 as the limiting reagent. How do we
convert from moles to grams?
Solution The molar mass of (NH2)2CO is 60.06 g. We use this
as a conversion factor to convert from moles of (NH2)2CO to
grams of (NH2)2CO:
Check Does your answer seem reasonable? 18.71 moles of
product are formed. What is the mass of 1 mole of (NH2)2CO?
Section 3.11
The Concept of Limiting Reactant
(c) Strategy Working backward, we can determine the amount
of CO2 that reacted to produce 18.71 moles of (NH2)2CO. The
amount of CO2 left over is the difference between the initial
amount and the amount reacted.
Solution Starting with 18.71 moles of (NH2)2CO, we can
determine the mass of CO2 that reacted using the mole ratio
from the balanced equation and the molar mass of CO2. The
conversion steps are
Section 3.11
The Concept of Limiting Reactant
Combining these conversions in one step, we write
The amount of CO2 remaining (in excess) is the difference
between the initial amount (1142 g) and the amount reacted
(823.4 g):
mass of CO2 remaining = 1142 g − 823.4 g = 319 g
Section 3.11
The Concept of Limiting Reactant
Notice

We cannot simply add the total moles of all the
reactants to decide which reactant mixture makes the
most product. We must always think about how much
product can be formed by using what we are given, and
the ratio in the balanced equation.
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Section 3.11
The Concept of Limiting Reactant
EXERCISE!
You react 10.0 g of A with 10.0 g of B. What
mass of product will be produced given that the
molar mass of A is 10.0 g/mol, B is
20.0 g/mol, and C is 25.0 g/mol? They react
according to the equation:
A + 3B
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2C
83
Section 3.11
The Concept of Limiting Reactant
Percent Yield

An important indicator of the efficiency of a particular
laboratory or industrial reaction.
Actual yield
 100%  percent yield
Theoretica l yield
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Section 3.11
The Concept of Limiting Reactant
Theoretical Yield The amount of a product formed when the
limiting reagent is complexly consumed.
Actual Yield is the amount of the product actually obtained
from a reaction is always les than the theoretical yield.
Section 3.11
The Concept of Limiting Reactant
Titanium is a strong, lightweight, corrosion-resistant metal that
is used in rockets, aircraft, jet engines, and bicycle frames. It is
prepared by the reaction of titanium(IV) chloride with molten
magnesium between 950°C and 1150°C:
In a certain industrial operation 3.54 × 107 g of TiCl4 are
reacted with 1.13 × 107 g of Mg.
(a) Calculate the theoretical yield of Ti in grams.
(b) Calculate the percent yield if 7.91 × 106 g of Ti are actually
obtained.
Section 3.11
The Concept of Limiting Reactant
(a) Strategy
Because there are two reactants, this is likely to be a limiting
reagent problem. The reactant that produces fewer moles
of product is the limiting reagent.
How do we convert from amount of reactant to amount of
product?
Perform this calculation for each reactant, then compare the
moles of product, Ti, formed.
Section 3.11
The Concept of Limiting Reactant
Solution
Carry out two separate calculations to see which of the two
reactants is the limiting reagent. First, starting with 3.54 × 107
g of TiCl4, calculate the number of moles of Ti that could be
produced if all the TiCl4 reacted. The conversions are
so that
Section 3.11
The Concept of Limiting Reactant
Next, we calculate the number of moles of Ti formed from
1.13 × 107 g of Mg. The conversion steps are
And we write
Therefore, TiCl4 is the limiting reagent because it produces a
smaller amount of Ti.
Section 3.11
The Concept of Limiting Reactant
The mass of Ti formed is
(b) Strategy The mass of Ti determined in part (a) is the
theoretical yield. The amount given in part (b) is the actual yield
of the reaction.
Section 3.11
The Concept of Limiting Reactant
Solution The percent yield is given by
Check Should the percent yield be less than 100 percent?
Section 3.11
The Concept of Limiting Reactant
EXERCISE!
Consider the following reaction:
P4(s) + 6F2(g)
4PF3(g)
 What mass of P4 is needed to produce 85.0 g of
PF3 if the reaction has a 64.9% yield?
46.1 g P4
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