Chapter 8 Quantities In Reactions. Homework Assigned Problems (odd numbers only) “Problems” 17...
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Transcript of Chapter 8 Quantities In Reactions. Homework Assigned Problems (odd numbers only) “Problems” 17...
Chapter 8Quantities
In Reactions
Homework• Assigned Problems (odd numbers only)
• “Problems” 17 to 73• “Cumulative Problems” 75-95• “Highlight Problems” 97-99
Calculations Using Balanced Equations: Stoichiometry
• Stoichiometry is the study of the quantitative relationships among reactants and products in a chemical reaction
• These chemical calculations can be used to determine the amount of one reactant needed to completely react with another
• Or, to determine the amount of reactant needed to produce a desired amount of product
• To calculate chemical quantities in reactions involves knowing how to interpret a balanced chemical equation
Calculations Using Balanced Equations: Law of Conservation of Mass
• As a chemical reaction proceeds:–Reactants are consumed and new
materials with new chemical properties are produced
–Bonds are broken, formed, or atoms are rearranged which produces new substances
–No material is lost or gained as original substances change to new substances
Calculations Using Balanced Equations: Law of Conservation of Mass
• Law of Conservation of Mass–Quantity of matter does not change
during a chemical reaction–The sum of the masses of products is
equal to the sum of masses of reactants
–Atoms are neither created nor destroyed in chemical reactions
–Only a balanced equation obeys this law
Mole Relationships in Chemical Equations: Conservation of Mass
• A balanced equation has the same number of atoms on each side of the arrow
(g)OH (g)CO (g)O (g)CH 2224 22
Information Available from a Balanced Chemical Equation
• “1 mole of methane gas reacts with 2 moles of oxygen gas to produce 1 mole of carbon dioxide and 2 moles of water vapor.”
• Multiplying each of the molar masses by the coefficient will give the total mass of reactants and products
Making Molecules:Mole-Mole Conversions
A balanced chemical equations tell us:– The formulas and symbols of the reactants and
products– The physical state of each substance– If special conditions such as heat are required– The number of molecules, formula units, or atoms
of each type of molecule involved in the reaction • Number can be in terms of single atoms, or
moles of atoms– The relative number of moles of each reactant and
product
Making Molecules:Mole-Mole Conversions
• In a balanced equation, conversion from moles of one substance to another will be determined by the values of the coefficients
• Balancing an equation will generate the coefficients that equal the number of moles of each reactant and product
• To determine how many moles of methanol would be produced if 0.295 moles of hydrogen gas is consumed:
• Requires a mole to mole relationship between methanol and hydrogen gas
)OH(CH (g)H CO(g) 32 2
Making Molecules:Mole-Mole Factors
• Converting the given mole amount of hydrogen gas enables you to find the number of moles methanol produced
• Coefficients of the balanced equation can be used to make mole to mole relationships between the different reactants and products
)((g)(g) OHCH H CO 32 2
Given: 0.295 mol H2 Find: mol CH3OH
2 mol H2 = 1 mol CH3OH
Making Molecules:Mole-Mole Factors
• The unit path begins with moles of H2 and ends with moles of CH3OH
• From any mole-mole relationship, two mole-mole conversion factors can be made:
)((g)(g) OHCH H CO 32 2
mol H2 mol CH3OH
2
3
3
2
H mol 2
OHCH mol 1
OHCH mol 1
H mol 2or
2
3
H mol 2
OHCH mol 1
Mole-molefactor
2 mol H2 = 1 mol CH3OH
Making Molecules:Mole-Mole Factors
• Write all of the possible mole-mole factors for the following chemical equation
)((g)(g) OH O H 222 2 2
2 mol H2
1 mol O2
2 mol H 2
2 mol H2O1 mol O2
2 mol H2O
1 mol O2
2 mol H2
2 mol H2O2 mol H 2 2
2
O mol 1
OH mol 2
2 mol H2 = 2 mol H2O 1 mol O2 = 2 mol H2O2 mol H2 = 1 mol O2
Using Mole-Mole Factors in Calculations:Calculating Moles of a Product
• Calculations based on balanced equations require the use of mole to mole (conversion) factors– Equation must be balanced
– Identify the known (given) and needed (find) substances
– Make the conversion factor based on:
KNOWN is whatof tcoefficien
SOUGHTis whatof tcoefficienmolesofnumbergiven )(KNOWN
Using Mole-Mole Factors in Calculations:Calculating Moles of a Product
• Using mole-mole factors from a BALANCED chemical equation–You can convert moles of one
compound to moles of another compound using the correct mole-mole factor
moles A moles B
grams A grams B
MM of A MM of B
moles B moles A
Stoichiometry
Mole-molefactor
Using Mole-Mole Factors in Calculations:Calculating Moles of a Product
• Calculate the moles of CO2 formed when 4.30 moles of C3H8 reacts with (the required) 21.5 moles of O2
• Balance the equation• Plan to convert the given amount of moles to the
needed amount of moles• Use coefficients to state the relationships and
mole-mole factors• Set up the problem using the mole-mole factor
and canceling units
O(g)H (g)CO (g)O (g)HC 22283 5 3 4
Using Mole-Mole Factors in Calculations:Calculating Moles of a Product
83
2
2
83
HC mol 1
CO mol 3
CO mol 3
HC mol 1and
moles A moles B
grams A grams B
MW of A MW of B
moles A moles B
Stoichiometry
83HC mol 4.302CO mol
C3H8(g) O2 (g) CO2 (g) H2O(g)5 3 4
Given: Find:
1 mol C3H8 = 3 mol CO2
Mole-molefactor
Mole-mole factorMole-mole relation
MM of A MM of B
4.30 mol C3H8 3 mol CO 2
1 mol C3H8
Using Mole-Mole Factors in Calculations:Calculating Moles of a Product
• Set up the problem using the mole-mole factor that cancels given moles and provides needed moles
C3H8(g) O2 (g) CO2 (g) H2O(g)5 3 4
83HC mol 4.30 :given 2CO mol :find
12.9 mol CO 2
Mole-mole factor83
2
HC mol 1
CO mol 3
Making Molecules: Mass-to-Mass Conversions
• From the balanced equation–It is also possible to start with a known
mass of one substance–Then convert to moles of another
substance–Start with a given amount (of grams) of
a substance–Then use mole-mole factor to find the
sought-after mass of another substance
Making Molecules: Mass-to-Mass Conversions
• The most common type of stoichiometric calculation is the mass-to-mass calculation
• In this type of problem: • The mass of one substance involved in a
chemical reaction is given• Find the mass of another substance involved in
the reaction• If a chemist only has so many grams of a certain
chemical – Can calculate how many grams of another substance can
be produced– Can calculate how many grams of another reactant are
required to react with it
Mass-to-Mass Conversions
• To convert the grams of one substance to grams of another substance:
• Find the mole-mole factor using the coefficients in the balanced equation– You can only relate (moles-moles) of two compounds, not
grams-to-grams– Ratios ONLY apply to moles, NOT grams– Must convert grams to moles, then use mole-mole factor
moles A moles B
grams A grams B
MW of A MW of B
moles A moles B
StoichiometryMM of A MM of B
moles A
moles BMole-molefactor
Mass-to-Mass Conversions
• Mass-to-mass conversions begin with a given mass of substance A
• By use of the balanced equation, find the mass of another (substance B)1. Convert grams of A to moles of A2. Convert the moles of A to moles of B by use of mole-mole
ratio3. Convert moles of B to mass of B
moles A moles B
grams A grams B
MM of A MM of B
moles B moles A
Stoichiometry
Mass-to-Mass Conversions(Mass of Product from Mass of Reactant)
• Calculate the mass of carbon dioxide produced when 96.1 g of propane react with sufficient oxygen.
• Balance the equation• Plan to convert the given mass to given moles• Convert the given moles to sought-after moles by the use of mole-
mole factor
• Convert the needed moles to needed mass
C3H8(g) O2 (g) CO2 (g) H2O(g)5 3 4
moles C3H8moles CO2
grams C3H8 grams CO2
MM of C3H8
MM of CO2
moles CO2
moles C3H8
Stoichiometry
Mole-molefactor
Mass-to-mass Conversions: Example 1• Write the equalities
• 1 mol C3H8 = 44.09 g C3H8
• 1 mol CO2 = 44.01 g CO2
• 1 mol C3H8 = 3 mol CO2 to create mole-mole factor
C3H8(g) O2 (g) CO2 (g) H2O(g)5 3 4Find: g of CO2Given: 96.1 g C3H8
moles C3H8moles CO2
grams C3H8 grams CO2
MM of A MM of B
moles B moles A
Stoichiometry
83
2
HC mol 1
CO mol 3
Mass-to-Mass Conversions: Example 1
C3H8(g) O2 (g) CO2 (g) H2O(g)5 3 496.1 g C3H8
2.1796 mol C 3H8
6.539 mol CO2
2CO g 288
83
8383
HC g 44.09
HC mol 1HC g 96.1
2.1796 mol C3H8 3 mol CO2
1 mol C3H8
6.539 mol CO2 44.01 g CO2
1 mol CO2
X g CO2
Mass-to-Mass Conversions: Example 2
• What mass of carbon monoxide and what mass of hydrogen are required to form 6.0 kg of methanol by the following reaction:
)OH(CH (g)H CO(g) 32 2
CO g )(x 2H g )( y 6.0 kg CH3OH
Mass-to-Mass Conversions: Example 2
)(OHCH (g)H (g)CO 32 l 2
6000 g CH3OH
187.27 mol CH3OH
187.27 mol CO
374.53 mol H2
OHCH kg 1
g 1000 OHCH kg 6.0
3
3
6000 g CH3OH 1 mol CH 3OH
32.04 g CH3OH
187.27 mol CH3OH 1 mol CO1 mol CH3OH
187.27 mol CH 3OH 2 mol H 2
1 mol CH 3OH
Mass Calculations Example 2
CO g 5200
2H g 750
187.27 mol CO 28.01 g CO1 mol CO
2
22
H mol 1
H g 2.016 H mol 374.53
Limiting Reactant, Theoretical Yield, and Percent Yield
• The chemical reactants are usually not present in the exact mole-mole ratios as stated in the balanced chemical equation
• Often, one of the reactants is purposely added in an excess amount
• Reasons include:• Increase the rate of reaction• To ensure that one reactant is completely used
up (reacted)• Reactants are not completely converted to
products as stated on paper (theory)
Limiting Reactant, Theoretical Yield, and Percent Yield
• Chemical reactions with two or more reactants will continue until one of the reactants is used up (consumed)
• If one of the reactants is used up, the reaction will stop because there is not enough of the other reactant to react with it
• The reactant used up is called the limiting reactant (reagent)
• This reactant limits the amount of product that can be made
Limiting Reactant
• When you make peanut butter sandwiches
• Required: 2 slices of bread and 1 tbsp. peanut butter per sandwich
• The reaction of nitrogen gas and hydrogen gas forming ammonia gas
• Required: 1 molec. N2 gas and 3 molec. H2 gas
2 slices of br. +1 tbsp p.b. =1sndw.
N2 (g) + 3H2 (g) = 2 NH3 (g)
N2
NH3
NH3
Limiting Reactant
To determine the limiting reactantbetween two reactants:
1. Balance the equation 2. State the mole-mole relationships to
make conversion factors3. Convert the initial masses (reactants)
to moles of each reactant4. Calculate how many moles of product
can be produced by each reactant
Limiting Reactant
5. Convert the moles of product to number of grams of product that each of the reactants would produce
6. Compare the numbers: The reactant producing the least amount of product (grams) is the limiting reactant
Limiting Reactant Problem• Lithium nitride, an ionic compound
containing Li+ and N3- ions, is prepared by the reaction of lithium metal and nitrogen gas. Calculate the mass of lithium nitride formed from 56.0 g of nitrogen gas and 56.0 g of lithium metal.
(s)(g)(s) NLi N Li 32 2656.0 g Li 56.0 g N2
NLi g 3X
Limiting Reactant Problem
6 2Find: g of Li3NGiven: 56.0 g Li Given: 56.0 g N2
moles Li moles Li3N
grams Li grams Li3N
MM of Li MM of Li3N
moles Li3N moles Li
Stoichiometry
1 mol Li = 6.941g Li 1 mol N2 = 28.00 g N2
Li(s) N2(g) Li3N(s)
1 mol Li3N= 34.83 g Li3N
Equalities and Conversion Factors-
Solution Map:
MM of N2
moles N2
moles N2
grams N2
6 mol Li = 2 mol Li3N 1 mol N2 = 2 mol Li3N
Limiting Reactant Problem
NLimol
NLig
Limol
NLimolLimol
3
33
1
82.34
6
207.8
NLi g (x) N g 56.0 Li g 56.0
N(s)2Li(g)N (s) Li6
32
32
56.0 g Li 1 mol Li6.941 g Li
56.0 g N2 1 mol N 2
28.02 g N2
93.67 g Li3N
139.3 g Li3N
Limiting reactant
8.07 mol Li
2.00 mol N2
NLimol
NLig
Nmol
NLimolNmol
3
3
2
32 1
82.34
1
200.2
Limiting Reactant Problem
• Lithium is the limiting reactant. We calculated the number of grams of lithium nitride which is formed in the reaction based on the limiting reactant
• This is the calculated amount of lithium nitride formed if the reaction proceeds completely as described by its balanced chemical equation
93.7 g Li3NNLimol
NLigNLimol
3
33 1
83.3469.2
Theoretical yield
Percent Yield• The calculated amount of product that
should be obtained is called the theoretical yield
• Assumes all reactants are converted to product based on the mole-mole ratios of reactant to product
• Rarely do you get the maximum amount of product– Side reactions– Loss during transfer– Accidental spills
Percent Yield• Theoretical Yield
– The calculated amount of product• Actual Yield
– The actual amount of product– Something less than the theoretical
• Percent Yield– The fraction of the theoretical yield actually
obtained is expressed as a percent
% Yield Actual Yield
Theoretical Yield100%
Percent Yield Example• Suppose, in the previous limiting
reactant problem, you actually produced 90.8 g of Li3N. What is the percent yield of this reaction?
%1007.93
8.90
g
g96.9 % yield %100
ltheoretica
actual
Enthalpy• Chemical reactions are associated with an
absorption or evolution of heat
– A change in energy occurs as bonds are broken (reactants) and new ones form (products)
– Nearly all chemical reactions absorb or produce heat
– Measured by the heat of reaction or enthalpy
• Enthalpy change is the amount of heat produced or consumed in a process (∆H )
Sign of ∆Hrxn
• Endothermic reactions absorb heat as they occur
– If (∆H ) is positive, then heat is added to the reaction
– If heat supply is removed, the reaction stops
Sign of ∆Hrxn
• Exothermic reactions produce heat as they occur
– If (∆H ) is negative, then heat is evolved by the reaction
Sign of ∆Hrxn
Enthalpy of Reaction
• Photosynthesis reaction– Carbon dioxide reacts with water to produce glucose
and oxygen
• Cell metabolism– Glucose reacts with oxygen to produce carbon dioxide
and water
(g)O6(s)OHC )(OH6(g)CO6 2612622
∆H = +2801 H = +2801 kJkJ
)(OH6(g)CO6 (g)O6(s)OHC 2226126
∆H = -2801 H = -2801 kJkJ
Stoichiometry of ∆Hrxn
• The coefficients in a given chemical reaction represent the number of moles of reactants and products that produce the given heat of reaction (enthalpy change)
• The combustion of methane gas:
• This information gives a quantitative relationship between the heat evolved per mole of methane and oxygen gas
(g)OH 2(g)CO (g)O2(g)CH 2224 ∆HHrxnrxn = -890 = -890 kJkJ
1 mol CH4 = -890 kJ 2 mol O2 = -890 kJ
Stoichiometry of ∆Hrxn
• If the combustion of 1 mol of CH4 with 2 mol O2 releases 890 kJ of heat, the combustion of 2 mol of CH4 with 4 mol of O2 produces twice as much heat
• The equivalence statements can be used to make conversion factors between the amounts of reactants or products and the amount of heat absorbed or emitted in a given reaction
• Calculate the amount of heat emitted when a certain amount in grams undergoes combustion
∆HHrxnrxn = -1780 = -1780 kJkJ
(g)OH 4(g)CO 2 (g)O4(g)CH2 2224
Stoichiometry Involving ∆HH
• Calculate the heat associated with the complete combustion of 4.50 g of methane gas
(g)OH 2(g)CO (g)O2(g)CH 2224 ∆H H rxnrxn= -890 = -890 kJkJ
Given: 4.50 g CH4 Find: kJ
Conversion Factors: 1 mol CH4= -890 kJ 1 mol CH4 = 16.00 g
Solution Map: g CH4 mol CH4 kJ CH4
4
4
00.16
1
CHg
CHmol
41
890
CHmol
kJ
Solution: 450.4 CHg -2.50 × 10-2.50 × 1022 kJ kJ4
4
00.16
1
CHg
CHmol
41
890
CHmol
kJ
Stoichiometry Involving ∆HH
• The combustion of sulfur dioxide • It reacts with oxygen to produce sulfur
trioxide
• Calculate the heat produced when 75.2 g of sulfur trioxide is produced
(g)SO2 (g)O(g)SO2 322 ∆H H rxnrxn= -99.1 = -99.1 kJkJ
Given 75.2 g SOGiven 75.2 g SO33
Heat in kJ produced Heat in kJ produced when SOwhen SO3 3 is formedis formed
Find:
Stoichiometry Involving ∆HH
Solution Map: Relation between g of SOSolution Map: Relation between g of SO33 and heat and heat released released Grams Grams of SOof SO33
MolesMoles of SOof SO33
kjkj
Write the necessary conversion factors:Write the necessary conversion factors:
3
3
SOmol2
kJ99.1
kJ99.1
SOmol2and
3
3
3
3
SOmol1
SO g80.07
SOg 80.07
SOmol1and
Set up the problem:Set up the problem:
33
33 SOmol2
kJ99.1
SOg80.07
SOmol1SOg75.2 46.5 kJ46.5 kJ
Heat Heat of rxnof rxn
MolarMolarmassmass
33 SOg80.07SOmol1 kJ99.1SOmol2 3
• end