4-1 Chemical Reactions and
Chemical Equations
Petrucci • Harwood • Herring • Madura
GENERAL
CHEMISTRY
Ninth
Edition
As reactants are converted to products (new
compounds) we observe:
Principles and Modern Applications
Chapter 4: Chemical Reactions
Yrd. Doç. Dr. Burak Esat
Fatih University
–
–
–
–
Color change
Precipitate formation
Gas evolution
Heat absorption or evolution
Chemical evidence may be necessary.
Chemical Equations
Qualitative Information:
Reactants
Chemical Reaction
Nitrogen monoxide + oxygen ? nitrogen dioxide
Products
Step 1: Write the reaction using chemical symbols.
Step 2: Balance the chemical equation.
States of Matter: (s) solid
(l) liquid
(g) gaseous
(aq) aqueous: dissolved in water
2 H2 (g) + O2 (g)
2 H2O (l)
2 NO + 1 O2 ? 2 NO2
Molecular Representation
Balancing Equations
• Never introduce extraneous atoms to balance.
NO + O2 ?
NO2 + O
• Never change a formula for the purpose of balancing an
equation.
NO + O2 ?
Balancing Equation Strategy
• Balance elements that occur in only one
compound on each side first.
• Balance free elements last.
• Balance unchanged poly-atomics as groups.
• Fractional coefficients are acceptable and
can be cleared at the end by multiplication.
NO3
Example 4-2
Writing and Balancing an Equation: The Combustion of a
Carbon-Hydrogen-Oxygen Compound.
Liquid triethylene glycol, C 6H14O4, is used a a solvent and
plasticizer for vinyl and polyurethane plastics. Write a
balanced chemical equation for its complete combustion.
Interpreting Chemical Equations:
Example 4-2
2 Al (s) + Fe 2O 3 (s)
Chemical Equation:
C 6H14O4 + 15
2 O2 ? 66 CO2 + 7 H2O
1. Balance C.
2 moles
1 formula
unit
1 mole
2. Balance H.
53.964 g
26.982 g
3. Balance O.
2 atoms
4. Multiply by two
2 C 6H14O4 + 15 O2 ? 12 CO2 + 14 H2O
and check all elements.
2 H 2 (g) + O 2 (g)
2 molecules 1 molecule
2 H 2O (l)
2 molecules
2 moles
1 mole
2 moles
4.0318 g
31.9988 g
36.0306 g
∆
Al 2O 3 ( s) + 2 Fe (s
(s )
1 formula
unit
1 mole
2 atoms
159.69 g
101.96 g
111.69 g
79.845 g
50.980 g
55.845 g
2 moles
Chemical equations express the quantitative (Stochiometric)
relationships among the reactants and products in a
chemical reaction.
4-2 Chemical Equations and
Stoichiometry
• Stoichiometry includes all the quantitative
relationships involving:
– atomic and formula masses
– chemical formulas.
• Mole ratio is a central conversion factor.
Example 4-3
Example 4-6
Relating the Numbers of Moles of Reactant and Product.
Additional Conversion Factors ina Stoichiometric
Calculation: Volume, Density, and Percent Composition.
How many moles of H2O are produced by burning 2.72
mol H2 in an excess of O2?
Write the Chemical Equation:
Balance the Chemical Equation:
2 H2 + O2 ?
2 H2O
An alloy used in aircraft structures consists of 93.7% Al
and 6.3% Cu by mass. The alloy has a density of 2.85
g/cm 3. A 0.691 cm 3 piece of the alloy reacts with an
excess of HCl(aq). If we assume that all the Al but none
of the Cu reacts with HCl(aq), what is the mass of H2
obtained?
Use the stoichiometricfactor or mole ratio in an equation:
nH2 O = 2.72 mol H2 × 2 mol H2O = 2.72 mol H2O
2 mol H2
Example 4-6
Example 4-6
Write the Chemical Equation:
Balance the Chemical Equation:
2 Al + 6 HCl ?
2Al + 6 HCl ? 2 AlCl3 + 3 H2
cm 3 alloy ? g alloy ? g Al ? mole Al ? mol H2 ? g H2
We need 5 conversion factors!
2 AlCl 3 + 3 H2
Plan the strategy:
Write the Equation and Calculate:
mH2 = 0.691 cm3 alloy × 2.85 g alloy
× 97.3 g Al ×
1 cm3
100 g alloy
1 mol Al × 3 mol H2 × 2.016 g H2 = 0.207 g H
2
26.98 g Al 2 mol Al
1 mol H2
An Ice Cream Sundae Analogy
for Limiting Reactions
4-4 Determining Limiting Reagent
• The reactant that is completely consumed
determines the quantities of the products
formed.
Limiting Reactant (Limiting Reagent):
Reagent): The reactant in
a chemical reaction that is completely used up.
Excess Reactant (Excess Reagent):
Reagent): The reactant that is
left over following a chemical reaction.
Fig. 3.10
Limiting Reactant Problems
aA + bB + cC
Limiting Reactant Problem
dD + eE + f F
Steps to solve
1) Identify it as a limiting Reactant problem - Information on the:
mass, number of moles, number of molecules, volume and
molarity of a solution is given for more than one reactant!
2) Calculate moles of each reactant!
3) Divide the moles of each reactant by the coefficient (a,b,c etc....)!
4) Which ever is smallest, that reactant is the limiting reactant!
5) Use the limiting reactant to calculate the moles of product
desired then convert to the units needed (moles, mass, volume,
number of atoms etc....)!
Problem: A fuel mixture used in the early days of rocketry is composed
of two liquids, hydrazine (N2H4) and dinitrogen tetraoxide (N2O4). They
ignite on contact ( hypergolic) to form nitrogen gas and water vapor.
How many grams of nitrogen gas form when exactly 1.00 x 102 g N2H4
and 2.00 x 102 g N2O4 are mixed?
Plan: First write the balanced equation.
Since amounts of both reactants are given, it is a limiting reactant problem.
Calculate the moles of each reactant, and then divide by the equation coefficient to find
which is limiting and use that one to calculate the moles of nitrogen gas
Calculate mass using the molecular weight of nitrogen gas.
Solution:
2 N2H4 (l) + N2O4 (l)
3 N2 (g) + 4 H2O (g) + Energy
Limiting Reactant Problem - cont.
Acid - Metal Limiting Reactant - I
molar mass N2H4 = ( 2 x 14.01 + 4 x 1.008 ) = 32.05 g/mol
molar mass H2O4 = ( 2 x 14.01 + 4 x 16.00 ) = 92.02 g/mol
2
• 2Al(s) + 6HCl(g)
Moles N2H4 = 1.00 x 10 g
32.05 g/mol
= 3.12 moles N2H4
2.00 x 102 g
Moles N2O4 =
92.02 g/mol
= 2.17 moles N2O4
• Given 30.0g Al and 20.0g HCl, how many moles
of Aluminum Chloride will be formed?
dividing by coefficients 3.12 mol / 2 = 1.56 mol N2H4
2.17 mol / 1 = 2.17 mol N2O4
Nitrogen yielded = 3.12 mol N2H4 x 3 mol N2
2 mol N2H4
Limiting !
= 4.68 moles N2
Mass of Nitrogen = 4.68 moles N2 x 28.02 g N2 / mol = 131 g N2
Acid - Metal Limiting Reactant - II
• since 6 moles of HCl yield 2 moles of AlCl3
• 0.548 moles of HCl will yield:
• 0.548 mol HCl / 6 mol HCl x 2 moles of
•
AlCl3 = 0.183 mol of AlCl 3
2AlCl3(s) + 3H2(g)
• 30.0g Al / 26.98g Al/mol Al = 1.11 mol Al
•
1.11 mol Al / 2 = 0.555
• 20.0g HCl / 36.5gHCl/mol HCl = 0.548 mol HCl
•
O.548 mol HCl / 6 = 0.0913
• HCl is smaller therefore the Limiting reactant!
Ostwald Process Limiting Reactant Problem
• What mass of NO could be formed by the reaction 30.0g of
Ammonia gas and 40.0g of Oxygen gas?
• 4NH3 (g) + 5 O2 (g)
4NO(g) + 6 H2O(g)
• 30.0g NH3 / 17.0g NH3/mol NH3 = 1.76 mol NH3
1.76 mol NH3 / 4 = 0.44 mol NH3
• 40.0g O2 / 32.0g O2 /mol O2 = 1.25 mol O2
1.25 mol O2 / 5 = 0.25 mol O2
• Therefore Oxygen is the Limiting Reagent!
• 1.25 mol O2 x 4 mol NO = 1.00 mol NO
5 mol O2
• mass NO = 1.00 mol NO x 30.0 g NO = 30.0 g NO
1 mol NO
Example 4-12
Example 4-12
Determining the Limiting Reactant in a Reaction.
Phosphorus trichloride , PCl3, is a commercially important
compound used in the manufacture of pesticides, gasoline
additives, and a number of other products. It is made by the
direct combination of phosphorus and chlorine
nCl2 = 323 g Cl2 × 1 mol Cl2 = 4.56 mol Cl 2
70.91 g Cl2
nP4 = 125 g P 4 × 1 mol P 4 = 1.01 mol P 4
123.9 g P4
P 4 (s) + 6 Cl2 (g) ? 4 PCl 3 (l)
What mass of PCl3 forms in the reaction of 125 g P 4 with
323 g Cl 2?
Strategy: Compare the actual mole ratio
to the required mole ratio.
χ=
nCl
nP
χactual
= 4.55 mol Cl 2/mol P 4
4
2
χtheoretical =
6.00 mol Cl2/mol P4
Chlorine gas is the limiting reagent.
4-5 Other Practical Matters in
Reaction Stoichiometry
Theoretical, Actual and Percent Yield
Theoretical yield is the expected yield from given amount of
reactants.
• When actual yield = % theoretical yield the
reaction is said to be quantitative.
• Side reactions reduce the percent yield.
• By-products are formed by side reactions.
Actual yield is the amount of product actually produced.
Percent yield =
Actual yield
× 100%
Theoretical Yield
Percent Yield / Limiting Reactant Problem - I
Problem: Ammonia is produced by the Haber Process using Nitrogen
and Hydrogen Gas. If 85.90g of Nitrogen are reacted with
21.66 g Hydrogen and the reaction yielded 98.67 g of
ammonia what was the percent yield of the reaction.
N2 (g) + 3 H 2 (g)
2 NH 3 (g)
Plan: Since we are given the masses of both reactants, this is a limiting
reactant problem. First determine which is the limiting reagent
then calculate the theoretical yield, and then the percent yield.
Solution: Moles of Nitrogen and Hydrogen:
Divide by coefficient
moles N2 = 85.90 g N2 = 3.066 mol N2 to get limiting:
28.02 g N2
3.066 g N2
= 3.066
1 mole N2
1
21.66
g
H
2 = 10.74 mol H
moles H2 =
2
2.016 g H2
10.74 g H2 = 3.582
1 mole H2
3
Consecutive Reactions,
Simultaneous Reactions and
Overall Reactions
• Multistep synthesis is often unavoidable.
• Reactions carried out in sequence are called
consecutive reactions.
• When substances react independently and at
the same time the reaction is a simultaneous
reaction.
Percent Yield/Limiting Reactant Problem - II
Solution Cont.
N2 (g) + 3 H 2 (g)
2 NH 3 (g)
We have 3.066 moles of Nitrogen, and it is limiting, therefore the
theoretical yield of ammonia is:
2 mol NH3 = 6.132 mol NH
3.066 mol N2 x
3
1 mol N2
(Theoretical Yield)
6.132 mol NH3 x 17.03 g NH3
1 mol NH3
Percent Yield =
Percent Yield =
= 104.427 g NH3
(Theoretical Yield)
Actual Yield
x 100%
Theoretical Yield
98.67 g NH3
104.427 g NH3
x 100% = 94.49 %
Overall Reactions and Intermediates
• The Overall Reaction is a chemical equation
that expresses all the reactions occurring in
a single overall equation.
• An intermediate is a substance produced in
one step and consumed in another during a
multi-step synthesis.
Calculating the Amounts of Reactants and
Products in a Reaction Sequence - I
Problem: Calcium Phosphate could be prepared in the following
reaction sequence:
4 P4 (s) + 10 KClO 3 (s)
4 P4O10 (s) + 10 KCl (s)
P4O10 (s) + 6 H2O (l)
2 H 3PO4 (aq) + 3 Ca(OH)2 (aq)
4 H3PO4 (aq)
6 H2O(aq) + Ca3(PO4)2 (s)
Calculating the Amounts of Reactants and
Products in a Reaction Sequence - II
Solution:
1 mole P 4
moles of Phosphorous = 15.50 g P4 x 123.88 g P4 = 0.1251 mol P4
For Reaction #1 [ 4 P 4 (s) + 10 KClO4 (s)
4 P 4O10 (s) + 10 KCl (s) ]
For Reaction #2 [ 1 P 4O10 (s) + 6 H2O (l)
4 H3PO4 (aq) ]
For Reaction #3 [ 2 H3PO4 + 3 Ca(OH)2
Given: 15.5 g P 4 and sufficient KClO3 , H2O and Ca(OH)2. What mass
of Calcium Phosphate could be formed?
Plan: (1) Calculate moles of P4.
(2) Use molar ratios to get moles of Ca3(PO4)2.
(3) Convert the moles of product back into mass by using the
molar mass of Calcium Phosphate.
Calculating the Amounts of Reactants and
Products in a Reaction Sequence - III
Molar mass of Ca3(PO4)2 = 310.18 g mole
mass of product = 0.2502 moles Ca3(PO4)2 x 310.18 g Ca3(PO4)2 =
1 mole Ca3(PO4)2
= 77.61 g Ca3(PO4)2
1Ca3(PO4)2 + 6 H2O]
0.1251 moles P4 x 4 moles P 4O10 x4 moles H3PO4 x1 mole Ca3(PO4)2
4 moles P 4
1 mole P 4O10
2 moles H3PO4
= 0.2502 moles Ca3(PO4)2
4-3 Chemical Reactions in Solution
• Close contact between atoms, ions and molecules
necessary for a reaction to occur.
• Solvent: The component with the highest quantity
in the mixture
– We will usually use aqueous (aq) solution.
• Solute
– A material dissolved by the solvent.
Molarity
Molarity (M) =
Amount of solute (mol solute)
Volume of solution (L)
Preparation of a Solution
Weigh the solid sample.
Dissolve it in a volumetric flask partially filled with solvent.
Carefully fill to the mark.
If 0.444 mol of urea is dissolved in enough water to make
1.000 L of solution the concentration is:
curea = 0.444 mol urea = 0.444 M CO(NH2)2
1.000 L
Solution
Dilution
M ×V
Example 4-6
i
i
M=
n
V
M f × Vf
Calculating the mass of Solute in a solution of Known
Molarity.
We want to prepare exactly 0.2500 L (250 mL) of an 0.250 M
K2CrO4 solution in water. What mass of K2CrO4 should we
use?
Plan strategy:
Volume ? moles ? mass
We need 2 conversion factors!
Mi × Vi = ni = nf = Mf × Vf
Write equation and calculate:
194.02 g = 12.1 g
mK2CrO4 = 0.2500 L × 0.250 mol× 1.00
mol
1.00 L
Mf =
Mi × Vi
Vf
= Mi
Vi
Vf
The number of moles of the solute in the initial solution and in
the final solution is the same (# of moles does not change)
Calculating Amounts of Reactants and
Products for a Reaction in Solution
Example 4-10
Preparing a solution by dilution.
A particular analytical chemistry procedure requires 0.0100 M
K2CrO4. What volume of 0.250 M K2CrO4 should we use to
prepare 0.250 L of 0.0100 M K2CrO4?
Plan strategy:
Mf = Mi
Vi
Vf
Vi = Vf
Mf
Mi
Calculate:
Mass (g) of Al(OH)3
M (g/mol)
3 H 2O(l) + AlCl3 (aq)
Given 10.0 g Al(OH)3, what volume of
1.50 M HCl is required to neutralize the
base?
Moles of Al(OH)3
molar ratio
VK2CrO4 = 0.2500 L × 0.0100 mol × 1.000 L = 0.0100 L
0.250 mol
1.00 L
Solving Limiting Reactant Problems in
Solution - Precipitation Problem - I
Problem: Lead has been used as a glaze for pottery for years, and can be
a problem if not fired properly in an oven, and is leachable from the
pottery. Vinegar is used in leaching tests, followed by Lead precipitated
as a sulfide. If 257.8 ml of a 0.0468 M solution of Lead nitrate is added
to 156.00 ml of a 0.095 M solution of Sodium sulfide, what mass of
solid Lead Sulfide will be formed?
Plan: It is a limiting-reactant problem because the amounts of two
reactants are given. After writing the balanced equation, determine the
limiting reactant, then calculate the moles of product. Convert moles of
product to mass of the product using the molar mass.
Solution: Writing the balanced equation:
Pb(NO3)2 (aq) + Na2S (aq)
Al(OH)3 (s) + 3 HCl (aq)
2 NaNO3 (aq) + PbS (s)
Moles of HCl
10.0 g Al(OH)3 = 0.128 mol Al(OH)
3
78.00 g/mol
3
moles
HCl
0.128 mol Al(OH)3 x
=
moles Al(OH)3
0.385 Moles HCl
1.00 L HCl
M ( mol/L) Vol HCl = 1.50 Moles HCl x 0.385 Moles HCl
Volume (L) of HCl
Vol HCl = 0.256 L = 256 ml
Volume (L) of
Pb(NO3)2
solution
Volume (L)
of Na2S
solution
Multiply by
M (mol/L)
Amount (mol)
of Pb(NO 3)2
Multiply by
M (mol/L)
Amount (mol)
of Na2S
Molar Ratio
Amount (mol)
of PbS
Molar Ratio
Amount (mol)
of PbS
Choose the lower number
of PbS and multiply by
M (g/mol)
Mass (g) of PbS
Volume (L) of
Pb(NO3)2
solution
Volume (L)
of Na2S
solution
Multiply by
M (mol/L)
Multiply by
M (mol/L)
Divide by Amount (mol)
Amount (mol) Divide by
of Pb(NO 3)2
equation
of Na2S
equation
coefficient
Smallest
coefficient
Molar Ratio
Amount (mol)
of PbS
Solving Limiting Reactant Problems in
Solution - Precipitation Problem - II
Moles Pb(NO3)2 = V x M = 0.2578 L x (0.0468 Mol/L) =
= 0.012065 Mol Pb +2
Moles Na2S = V x M = 0.156 L x (0.095 Mol/L) = 0.01482 mol S -2
Therefore Lead Nitrate is the Limiting Reactant!
Calculation of product yield:
Moles PbS = 0.012065 Mol Pb+2x 1 mol PbS
= 0.012065 Mol Pb +2
1 mol Pb(NO3)2
0.012065 Mol Pb +2 = 0.012065 Mol PbS
Mass (g) of PbS
Focus on Industrial Chemistry
0.012065 Mol PbS x 239.3 g PbS
1 Mol PbS
= 2.89 g PbS
Chapter 4 Questions