Second Exam: Friday February 15

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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
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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)
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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
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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
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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)


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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


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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


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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
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Figure 4.21A Gravimetric analysis for barium
solution is poured.Photo courtesy of James
Scherer.
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Figure 4.21B Gravimetric analysis for barium
solution filtered.
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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
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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
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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!
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Percent Yield of a reaction Actual Yield x
100 Theortetical Yield
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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
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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?
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Fig. 3.14
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Figure 4.22A Titration of an unknown amount of
HCl with NaOH (1). Photo courtesy of American
Color.
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Figure 4.22B Titration of an unknown amount of
HCl with NaOH (2). Photo courtesy of American
Color.
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Figure 4.22C Titration of an unknown amount of
HCl with NaOH (3). Photo courtesy of American
Color.
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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

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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

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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