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Mark S. CracoliceEdward I. Peters

Mark S. Cracolice • The University of Montana

Chapter 10Quantity Relationshipsin Chemical Reactions

Conversion Factors from EquationsThe coefficients in a chemical equation give us the conversion factors to convert from the number of moles of one substance,

to the number of moles of other substance in a reaction.

4 NH3(g) + 5 O2 (g) 4 NO(g) + 6 H2O (g)

2

3Omol5

NHmol4

NOmol4NHmol4 3

O23

Hmol6NHmol4•

NOmol4Omol5 2

OHmol6Omol5

2

2

OHmol6NOmol4

2

Conversion Factors from EquationsExample:How many moles of water are formed by the reaction of 3.20

moles of ammonia NH3 with oxygen.

Step 1: Write and balance the chemical equation:

Conversion Factors from Equations4 NH3(g) + 5 O2 (g) 4 NO(g) + 6 H2O (g)

Step2 Find what is given : 3.20 mol NH3

 

Step 3 Find what is to be found: ? mol H2O Step 4 Find the path : mol NH3 mol H2O Step 5 : Find the right conversion factor

3

2NHmol4Hmol6 O

Conversion Factors from Equations

4 NH3(g) + 5 O2 (g) 4 NO(g) + 6 H2O (g)

Step 6 Set up the calculation:

3.20 mol NH3 x (6 mol H2O / 4 mol NH3) = 4.80 mol H2O

Note that 6 and 4 are exact numbers; they do not affect the significant figures in the final answer.

Conversion Factors from EquationsExample 2 How many moles of oxygen are required to burn

2.40 moles of ethane, C2H6?Step 1: Write the chemical equation

2 C2H6(g) + 7 O2 (g) 4 CO2(g) + 6 H2O (l) • conversion conversion factor • Number of moles C2H6 number of moles O2: 7 mol O2/ 2 mol C2H6

• Number of moles C2H6 number of moles H2O : 6 mol H2O / 2 mol C2H6

• Number of moles C2H6 number of moles CO2 : 4 mol CO2 / 2 mol C2H6

• Number of moles O2 number of moles CO2 : 4 mol CO2/ 7 mol O2

• Number of moles O2 number of moles H2O : 6 mol H2O / 7 mol O2 • Number of moles CO2 number of moles H2O 6 mol H2O / 4 mol CO2:

Conversion Factors from Equations2 C2H6(g) + 7 O2 (g) 4 CO2(g) + 6 H2O (l)

Step2 Find what is given : 2.40 mol C2H6

 

Step 3 Find what is to be found: ? mol O2

 Step 4 Find the path : mol C2H6 mol O2

 Step 5 : Find the right conversion factor 7 mol O2/ 2 mol C2H6

Conversion Factors from Equations

2 C2H6(g) + 7 O2 (g) 4 CO2(g) + 6 H2O (l)

Step 6 Set up the calculation:

2.40 mol C2H6 x (7 mol O2/ 2 mol C2H6) = 8.40 mol O2

Note that 7 and 2 are exact numbers; they do not affect the significant figures in the final answer.

Conversion Factors from EquationsExample 3 Ammonia is formed from its elements. How many

moles of hydrogen are needed to produce 4.20 moles NH3. Chemical equation :

N2 + 3 H2 2 NH3

 

Conversion Factors from Equations

N2 + 3 H2 2 NH3

 Given 4.20 moles NH3

Wanted mol H2

Path 4.20 moles NH3 mol H2

Factor 3 mol H2/ 2 moles NH3

Calculation 4.20 moles NH3 x (3 mol H2/ 2 moles NH3) = 6.30 mol NH3

Mass–Mass StoichiometryHow to Solve a Stoichiometry Problem:

The Stoichiometry Path

Step 1: Change the mass of the given species to moles.

Step 2: Change the moles of the given species to the moles of the wanted species.

Step 3: Change the moles of the wanted species to mass.

Mass–Mass StoichiometryMass-to-Mass Stoichiometry Path

Mass of Moles of Moles of Mass ofGiven Given Wanted Wanted

Molar Mass Equation Molar Mass coefficients

Mass Given × × ×

Mass–Mass StoichiometryExample 1 Calculate the number of grams of carbon dioxide

produced by burning 66.0 g of heptanes C7 H16 (l) Step 1 Chemical reaction :

C7H16(l) + 11 O2 (g) 7 CO2(g) + 8 H2O (l)  Step2 Find what is given : 66.0 g C7H16

Step 3 Find what is to be found: ? g of CO2

Mass–Mass StoichiometryC7H16(l) + 11 O2 (g) 7 CO2(g) + 8 H2O (l)

Step 4 Find the path : g of C7H16 mol C7H16 mol CO2 g of CO2

 Step 5 : Find the right conversion factors

g of C7H16 mol C7H16 : 1 mol C2H6/ 100.20 g C7H16

mol C7H16 mol CO2 7 mol CO2/ 1 mol C7H16

mol CO2 g of CO2 44.01 g CO2/ 1 mol CO2

 

Mass–Mass StoichiometryC7H16(l) + 11 O2 (g) 7 CO2(g) + 8 H2O (l)

 Step 6 Set up the calculation: 

66.0 g C7H16 x (1 mol C7H16/ 100.20 g C7H16) x (7 mol CO2/ 1 mol C7H16) x (44.01 g CO2/ 1mol CO2) =

203 g of CO2

Percent YieldThe actual yield of a chemical reaction is

usually less than the ideal yieldpredicted by a stoichiometry calculation because:

• reactants may be impure

• the reaction may not go to completion

• other reactions may occur

Actual yield is experimentally determined.

Percent YieldPercent yield expresses the ratio of

actual yield to ideal yield:

% yield = × 100%

Percent YieldExample Calculate the theoretical yield of carbon dioxide and the

percent yield when burning of 66.0 grams of C7 H16 produced 181 grams of CO2 .

C7H16(l) + 11 O2 (g) 7 CO2(g) + 8 H2O (l)

Theoretical yield of CO2 66.0 g x ( 1mol C7H16/ 100.20 g C7H16) x

( 7 mol CO2 / 1 mol. C7H16) x ( 44.01 CO2 / 1 mol. CO2 ) =

= 203 g CO2

Percent yield = ( 181g/203g) x 100 = 89.2 %

Percent YieldExample:Consider the reaction of hydrogen and nitrogen that forms

ammonia with a 92.2% yield. What quantity of ammonia will be produced by reacting 125 g of hydrogen with excess nitrogen?

Solution:Use 92.2% yield as a PER expression:

3 H2 + N2 2 NH3

125 g H2 × × × ×

= 649 g NH3

Limiting Reactants

Three atoms of carbon react with two molecules of oxygen:C(g) + O2(g) CO2(g)

Limiting ReactantsThe reaction will stop when two molecules

of oxygen are used up.Two carbon dioxide molecules will form;One carbon atom will remain unreacted:

C + O2 CO2

Start 3 2 0

Used (+) orProduced (–) – 2 – 2 + 2

Finish 1 0 2

Limiting ReactantsLimiting Reactant

The reactant that is completely used up.(Oxygen)

Excess ReactantThe reactant initially present in excess.

(some will remain unreacted)(Carbon)

Limiting Reactants• The reactant that is completely used up by the reaction, is

called the limiting reactant. Other reactants have some excess which will remain unreacted.

• There are two approaches to solving limiting reactant problems: the comparison of moles method and the smaller amount method.

Limiting Reactants: Compare MolesComparison-of-Moles Method

How to Solve a Limiting Reactant Problem:

1. Convert the number of grams of each reactant to moles.2. Identify the limiting reactant by comparing the theoretical

mole ratio of reactants to the actual mole ratio.3. Calculate the number of moles of each species that reacts or

is produced using the limiting reactant number of moles.4. Calculate the number of moles of each species that remains

after the reaction.5. Change the number of moles of each species to grams.

Limiting Reactants: Compare MolesExample Calculate the mass of antimony(III) iodide that can be

produced by the reaction of 129 g antimony, Sb (Z=51), and 381 g iodine. Also find the number of grams of the element that will be left.

 2 Sb + 3 I2 2 SbI3

Calculate the number of moles:

Moles of Sb = 129 g Sb x (1 mol Sb/ 121.8 g Sb) = 1.06 mol Sb

Moles of I2 = 381 g I2 x (1 mol I2 / 253.8 g I2 ) = 1.50 mol I2

Limiting Reactants: Compare Moles2 Sb + 3 I2 2 SbI3

Theoretical mole ratio:mol of I2 / mol of Sb = 3 mol of I2 / 2 mol of Sb =

1.5 mol of I2 / 1 mol of Sb Actual mole ratio:mol of I2 / mol of Sb = 1.5 mol of I2 / 1.06 mol of Sb

=1.42 mol of I2 / 1 mol of Sb

Actual mole ratio of iodine over antimony is smaller than theoretical mole ratio, so iodine is the limiting reactant and antimony is in excess.

Limiting Reactants: Compare Moles2 Sb + 3 I2 2 SbI3

The number of mole Sb used = 1.50 mol of I2 x ( 2 mol of Sb/ 3 mol of I2 ) = 1.00 mol of Sb

mol of Sb in excess = (1.06 - 1.00) mol of Sb = 0.06 mol of Sb

mass of Sb in excess = 0.06 mol of Sb x (121.8 g Sb/mol Sb)= 7 g Sb

Limiting Reactants: Compare Moles2 Sb + 3 I2 2 SbI3

The number of mole SbI3 produced =1.50 mol of I2 x ( 2 mol of SbI3 / 3 mol of I2 ) = 1.00 mol of SbI3

mass of SbI3 produced = 1.00 mol of SbI3 x (502.5 g SbI3 /mol

SbI3)= 502.5 g SbI3

Limiting Reactants: Smaller AmountHow to Solve a Limiting Reactant Problem by Smaller

Amount Method

1. Calculate the amount of product that can be formed by the initial amount of each reactant.a) The reactant that yields the smaller amount of product is

the limiting reactant.b) The smaller amount of product is the amount that will be

formed when all of the limiting reactant is used up.2. Calculate the amount of excess reactant that is used by the

total amount of limiting reactant.3. Subtract from the amount of excess reactant present initially

the amount that is used by all of the limiting reactant. The difference is the amount of excess reactant that is left.

Limiting Reactants: Smaller AmountExample Calculate the mass of antimony(III) iodide that can be

produced by the reaction of 129 g antimony, Sb (Z=51), and 381 g iodine. Also find the number of grams of the element that will be left.

 2 Sb + 3 I2 2 SbI3

Solution:Step 1 is to calculate the amount of product that can be formed

by the initial amount of each reactant.

Limiting Reactants: Smaller Amount

2 Sb + 3 I2 2 SbI3

First, assume that antimony is the limiting reactant. Grams of SbI3 produced =

0.129g Sb x(1 mol Sb/121.8 gSb) x (2mol SbI3/2mol Sb) x

(502.5 g SbI3 /molSbI3) =

= 532 g SbI3

Limiting Reactants: Smaller Amount2 Sb + 3 I2 2 SbI3

Second, assume that iodine is the limiting reactant.

grams of SbCl3 produced =

381g I2 x ( mol I2 / 253.8 g I2 ) x (2mol SbI3/3mol I2) x (502.5 g SbI3 / mol SbI3) =

• = 503 g SbI3

• The reactant iodine, which yields the smaller amount (503 g) of SbI3 , is the limiting reactant.

• The amount of antimony(III) iodide produced is 503 g.

Limiting Reactants: Smaller Amount2 Sb + 3 I2 2 SbI3

The amount of antimony required is calculated as follows Grams of antimony required =

381 g I2 x (1 mol I2 /253.8 g I2 ) x (2 mol Sb/ 3 mol I2) x (121.8 g Sb/mol Sb) = 122 g Sb

• The amount of antimony in excess = 129 g (initial) – 122g (used) = 7 g Sb (left)

EnergyEnergy

The ability to do work or transfer heat.

SI (derived) energy unit: Joule

Joule (J):The amount of energy exerted when a force of one newton (force

required to cause a mass of 1 kg to accelerate at a rate of1 m/s2) is applied over a displacement of one meter:

1 joule = 1 newton × 1 meter

1 J = 1 kg • m2/sec2

EnergyNon-SI metric energy unit: Calorie

calorie (historical definition):Amount of energy required to raise the

temperature of 1 g of water by 1°C.

calorie (modern definition)1 cal 4.184 J

A food Calorie (Cal) is a thermochemical kilocalorie.In scientific writing it is capitalized;

in everyday writing, interpret the context.

1 kcal = 4.184 kJ

Thermochemical EquationsEnthalpy

H = E + PVWhere E is the internal energy of the system, P is the pressure of

the system, and V is the volume of the system. P, V, E, H which depend only on the present state of the system, and do not depend on the system’s past or future, are called state functions.

∆H is the enthalpy of reaction.

∆ means “change in”: It is determined by subtracting theinitial quantity from the final quantity

Thermochemical EquationsEnthalpy of Reaction, ∆H

It can be demonstrated that for a reaction studied at constant pressure the heat of reaction is a measure of the change in enthalpy1 for the system.

Heat of reaction = H(products) - H (reactants ) = Δ H

When a system gives off heat (reaction is exothermic)enthalpy of the system goes down and Δ H has a negative value.

When a reaction absorbs heat (reaction is endothermic) enthalpy

increases and Δ H is positive..

Thermochemical EquationsThere are two ways to write a thermochemical EquationMethod 1 The ΔH of the reaction is written to the right of the equation

2 C2H6 (g) + 7O2 (g) 4CO2 (g) + 6H2O (g) ΔH = -2855 kJ Method 2Energy is included in the thermochemical equation as if it were a

reactant or product. 2 C2H6 (g) + 7O2 (g) 4CO2 (g) + 6H2O (g) + 2855 kJ 

Thermochemical Stoichiometry

Since there is a proportional relationship between moles of different substances and heat of reaction, conversion factors can be written between kilojoules and moles of any substance.

These factors are used in solving thermochemical stoichiometry problems.

Thermochemical StoichiometryExample : How many kilojoules of heat are released when 73.0 g C2H6 (g) burn  Equation :2 C2H6 (g) + 7O2 (g) 4CO2 (g) + 6H2O (l) + 3119 kJ

Thermochemical Stoichiometry2 C2H6 (g) + 7O2 (g) 4CO2 (g) + 6H2O (l) + 3119 kJ

Given 73.0 g C2H6 (g) Wanted : kJ

Path g C2H6 (g) mol C2H6 (g) kJ Factors

(1 mol C2H6 / 30.07 g C2H6 )

(3119 kJ/ 2 mol C2H6 (g))

Thermochemical Stoichiometry

2 C2H6 (g) + 7O2 (g) 4CO2 (g) + 6H2O (l) + 3119 kJ

Solve the problem: 73.0 g C2H6 (g) x (1 mol C2H6 / 30.07 g C2H6) x

(3119 kJ/ 2 mol C2H6 (l) ) = = 3.79 x 103 kJ

Thermochemical StoichiometryExample:When propane, C3H8(g), is burned to form gaseous carbon

dioxide and liquid water, 2.22 × 103 kJ of heat is released for every mole of propane burned. What quantity of heat is released when 1.00 × 102 g of propane is burned?

Solution:First, write and balance the thermochemical equation to

determine the stoichiometric relationships.

C3H8 (g) + 5 O2 (g)3 CO2 (g) + 4 H2O (l) ∆H = –2.22x103 kJ

Thermochemical StoichiometryC3H8 (g) + 5 O2 (g)3 CO2 (g) + 4 H2O (l) ∆H = –2.22 × 103 kJ

GIVEN: 1.00 × 102 g C3H8

1 mol C3H8/44.09 g C3H8

PATH: g C3H8 mol C3H8

2.22 × 103 kJ/1 mol C3H8

kJ

1.00 × 102 g C3H8 × × = 5.04 × 103 kJ

HOMEWORKHomework: 1, 7, 23, 35, 45, 47, 67.