Second Exam: Friday February 15?

1
Second Exam Friday February 15? Chapters 3 and
4. Please note that there is a class at 1 pm so
you will need to finish by 1255 pm. Electronic
Homework due R by 1130 pm Office hours this
week T 2-3 pm R 1-2 pm SL
130
2
Figure 4.9 Reaction of a carbonate with an
acid.Photo courtesy of American Color.
3
(No Transcript)
4
(No Transcript)
5
Molarity (Concentration of Solutions) M
Moles of Solute Moles Liters of
Solution L
M
solute material dissolved into the solvent In
air , Nitrogen is the solvent and oxygen, carbon
dioxide, etc.

are the solutes. In sea water , Water is the
solvent, and salt, magnesium chloride, etc.

are the solutes. In
brass , Copper is the solvent (90), and Zinc is
the solute(10)
6
Fig. 3.11
7
Preparing a Solution - I

• Prepare a solution of Sodium Phosphate by
  dissolving 3.95g of Sodium Phosphate into water
  and diluting it to 300.0 ml or 0.300 l !
• What is the Molarity of the salt and each of the
  ions?
• Na3PO4 (s) H2O(solvent) 3 Na(aq)
  PO4-3(aq)

8
Preparing a Solution - II

• Mol wt of Na3PO4 163.94 g / mol
• 3.95 g / 163.94 g/mol 0.0241 mol Na3PO4
• dissolve and dilute to 300.0 ml
• M 0.0241 mol Na3PO4 / 0.300 l 0.0803 M
• 
  Na3PO4
• for PO4-3 ions 0.0803 M
• for Na ions 3 x 0.0803 M 0.241 M

9
Dilution of Solutions

• Take 25.00 ml of the 0.0400 M KMnO4
• Dilute the 25.00 ml to 1.000 l - What is the
  resulting Molarity of the diluted solution?
• moles Vol x M
• 0.0250 l x 0.0400 M 0.00100 Moles
• 0.00100 Mol / 1.00 l 0.00100 M

10
Converting a Concentrated Solution to a Dilute
Solution
11
The Dilution Dogma NEVER FORGET IT!
M1V1M2V2
12
How could you make 5.0 L of 0.025 M sucrose from
a solution which is 0.100 M sucrose?
Mix 1.250 L of 0.100 M sucrose with 3.75 L water.
13
Chemical Equation Calculation - III
Mass
Atoms (Molecules)
Molecular Weight
Avogadros Number
g/mol
6.02 x 1023
Molecules
Reactants
Products
Moles
Molarity
moles / liter
Solutions
14
Calculating Mass of Solute from a Given Volume
of Solution
Volume (L) of Solution
Molarity M (mol solute / Liters of solution)
M/L
Moles of Solute
Molar Mass (M) ( mass / mole) g/mol
Mass (g) of Solute
15
Calculating Amounts of Reactants and Products
for a Reaction in Solution
Al(OH)3 (s) 3 HCl (aq)
3 H2O(l) AlCl3 (aq)

Given 10.0 g Al(OH)3, what volume of 1.50 M HCl
is required to neutralize the base?
Mass (g) of Al(OH)3
M (g/mol)
10.0 g Al(OH)3 78.00 g/mol Al(OH)3
Moles of Al(OH)3

molar ratio
Moles of HCl

M ( mol/L)
Volume (L) of HCl
16
Calculating Amounts of Reactants and Products
for a Reaction in Solution
Al(OH)3 (s) 3 HCl (aq)
3 H2O(l) AlCl3 (aq)

Given 10.0 g Al(OH)3, what volume of 1.50 M HCl
is required to neutralize the base?
Mass (g) of Al(OH)3
M (g/mol)
10.0 g Al(OH)3 78.00 g/mol
0.128 mol Al(OH)3
Moles of Al(OH)3
0.128 mol Al(OH)3 x

molar ratio
Moles HCl
Moles of HCl


M ( mol/L)
Volume (L) of HCl
17
Calculating Amounts of Reactants and Products
for a Reaction in Solution
Al(OH)3 (s) 3 HCl (aq)
3 H2O(l) AlCl3 (aq)

Given 10.0 g Al(OH)3, what volume of 1.50 M HCl
is required to neutralize the base?
Mass (g) of Al(OH)3
M (g/mol)
10.0 g Al(OH)3 78.00 g/mol
0.128 mol Al(OH)3
Moles of Al(OH)3
3 moles HCl moles Al(OH)3
0.128 mol Al(OH)3 x

molar ratio
0.385 Moles HCl
Moles of HCl
Vol HCl
M ( mol/L)
Volume (L) of HCl
18
Calculating Amounts of Reactants and Products
for a Reaction in Solution
Al(OH)3 (s) 3 HCl (aq)
3 H2O(l) AlCl3 (aq)

Given 10.0 g Al(OH)3, what volume of 1.50 M HCl
is required to neutralize the base?
Mass (g) of Al(OH)3
M (g/mol)
10.0 g Al(OH)3 78.00 g/mol
0.128 mol Al(OH)3
Moles of Al(OH)3
3 moles HCl moles Al(OH)3
0.128 mol Al(OH)3 x

molar ratio
0.385 Moles HCl
Moles of HCl
1.00 L HCl 1.50 Moles HCl
Vol HCl x 0.385
Moles HCl Vol HCl 0.256 L
256 ml
M ( mol/L)
Volume (L) of HCl
19
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)
2 NaNO3 (aq) PbS (s)
20
Volume (L) of Pb(NO3)2 solution
Volume (L) of Na2S solution
Multiply by M (mol/L)
Multiply by M (mol/L)
Amount (mol) of Pb(NO3)2
Amount (mol) of Na2S
Molar Ratio
Molar Ratio
Amount (mol) of PbS
Amount (mol) of PbS
Choose the lower number of PbS and multiply by M
(g/mol)
Mass (g) of PbS
21
Volume (L) of Pb(NO3)2 solution
Volume (L) of Na2S solution
Multiply by M (mol/L)
Multiply by M (mol/L)
Amount (mol) of Pb(NO3)2
Amount (mol) of Na2S
Divide by equation coefficient
Divide by equation coefficient
Smallest
Molar Ratio
Amount (mol) of PbS
Mass (g) of PbS
22
Solving Limiting Reactant Problems in Solution -
Precipitation Problem - II
Moles Pb(NO3)2 V x M 0.2578 L x (0.0468
Mol/L)
Moles
Na2S V x M 0.156 L x (0.095 Mol/L)


23
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 Pb2 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


24
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 Pb2 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
1 mol PbS 1 mol Pb(NO3)2
Moles PbS 0.012065 Mol Pb2x
0.012065 Mol Pb2


25
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 Pb2 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
1 mol PbS 1 mol Pb(NO3)2
Moles PbS 0.012065 Mol Pb2x
0.012065 Mol Pb2
0.012065 Mol Pb2 0.012065 Mol PbS 0.012065
Mol PbS x 2.89 g PbS
239.3 g PbS 1 Mol PbS
26
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 H2 (g)
2 NH3 (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
85.90 g N2 28.02 g N2 1 mole N2
moles N2
21.66 g H2 2.016 g H2 1 mole H2
moles H2
27
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 H2 (g)
2 NH3 (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
85.90 g N2 28.02 g N2 1 mole N2
moles N2 3.066 mol N2
21.66 g H2 2.016 g H2 1 mole H2
moles H2 10.74 mol H2
28
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 H2 (g)
2 NH3 (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
2 mol NH3 1 mol N2
6.132 mol NH3
3.066 mol N2
7.16 mol NH3
2 mol NH3 3 mol H2
10.74 mol H
N2 is Limiting!
29
Percent Yield of a reaction Actual Yield x
100 Theortetical Yield
30
Percent Yield/Limiting Reactant Problem - II
N2 (g) 3 H2 (g)
2 NH3 (g)
Solution Cont.
Since Nitrogen is limiting, the theoretical
yield of ammonia is
6.132 mol NH3 x
104.427 g NH3

(Theoretical Yield)
17.03 g NH3 1 mol NH3
Actual Yield Theoretical Yield
Percent Yield x
100
98.67 g NH3 104.427 g NH3
Percent Yield
x 100 94.49
31
CaCO3(s) 2 HCl(aq) ? CaCl2(aq) CO2(g) H2O(l)
2 g 10 mL 0.75 M
Which is limiting?
2 g CaCO3 x 1 mol CaCl2 0.01 mol CaCl2 100 g
CaCO3 1 mol CaCO3
0.01 L HCl x 0.75 mol HCl x 1 mol CaCl2 L
HCl 2 mol HCl 0.004 mol CaCl2
What is the Cl- after the reaction?
How many g of CaCO3 remain?
32
Fig. 3.14
33
Figure 4.22A Titration of an unknown amount of
HCl with NaOH (1). Photo courtesy of American
Color.
34
Figure 4.22B Titration of an unknown amount of
HCl with NaOH (2). Photo courtesy of American
Color.
35
Figure 4.22C Titration of an unknown amount of
HCl with NaOH (3). Photo courtesy of American
Color.
36
(No Transcript)
37
Make a Solution of Potassium Permanganate
Potassium Permanganate is KMnO4 and has a
molecular mass
of 158.04
g / mole
Problem Prepare a solution by dissolving 1.58
grams of KMnO4 into sufficient
water to make 250.00 ml of solution. What is
the molarity of this diluted solution?
1 mole KMnO4 158.04 g KMnO4
1.58 g KMnO4 x

38
Make a Solution of Potassium Permanganate
Potassium Permanganate is KMnO4 and has a
molecular mass
of 158.04
g / mole
Problem Prepare a solution by dissolving 1.58
grams of KMnO4 into sufficient
water to make 250.00 ml of solution. What is
the molarity of this diluted solution?
1 mole KMnO4 158.04 g KMnO4
1.58 g KMnO4 x
0.0100 moles KMnO4

39
Make a Solution of Potassium Permanganate
Potassium Permanganate is KMnO4 and has a
molecular mass
of 158.04
g / mole
Problem Prepare a solution by dissolving 1.58
grams of KMnO4 into sufficient
water to make 250.00 ml of solution. What is
the molarity of this diluted solution?
1 mole KMnO4 158.04 g KMnO4
1.58 g KMnO4 x
0.0100 moles KMnO4
0.0100 moles KMnO4 0.250 liters
Molarity
0.0400 M
Molarity of K ion K ion MnO4- ion
0.0400 M