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LECTURE TWELVECHM 151 ©slg
TOPICS:Overview: Balancing Limiting Reagent Precipitation Reactions Acid/Base Reactions
GROUP WORK, BALANCE
C6H12O6 + O2 --> CO2 + H2O (sugar)
C5H11OH + O2 --> CO2 + H2O (alcohol)
a) Balance Cb) Balance Hc) Balance Od) Check
C6H12O6 + O2 --> CO2 + H2O
C6H12O6 + O2 --> 6 CO2 + H2O
C6H12O6 + O2 --> 6 CO2 + 6 H2O
C6H12O6 + 6 O2 --> 6 CO2 + 6 H2O
6C 12H 6 O + 12 O ---> 6 C 12 O + 12 H 6 O
Solution, Sugar
12 O + 6 O = 18 O
C5H11OH + O2 --> CO2 + H2OC5H11OH + O2 --> CO2 + H2O
C5H11OH + O2 --> 5 CO2 + H2O
C5H11OH + O2 --> 5 CO2 + 6 H2O
C5H11OH + 7.5 or 15/2 O2 --> 5 CO2 + 6 H2O 1 O 15 O 10 O 6 O
#1
#2
#3
C5H11OH + 7.5 or 15/2 O2 --> 5 CO2 + 6 H2O 10 O 6 O
X 2
2 C5H11OH + 15 O2 --> 10 CO2 + 12 H2O
[10C 24H 2 O] + 30 O --> [10C 20 O] + [24H 12 O]
Ch 4. Prob 24: Which is LR, what is theoretical yield,how much of excess reagent remains after reaction is complete?
CO(g) + 2 H2 (g) ----> CH3OH (l) 28.01 g/mol 2.016 g/mol 32.04 g/mol
74.5 g 12.0 g ? g theoretical ? g excess reagent leftover
Solving the question: How much of the excess reagent isleft after reaction is theoretically complete?
CO(g) + 2 H2 (g) ----> CH3OH (l) 28.01 g/mol 2.016 g/mol 32.04 g/mol
74.5 g 12.0 g ? g theoretical
74.5 g CO 1 mol CO 1 mol CH3OH 32.04 g CH3OH 28.01 g CO 1 mol CO 1 mol CH3OH = 85.2 g CH3OH
12.0 g H2 1 mol H2 1 mol CH3OH 32.04 g CH3OH 2.106 g H2 2 mol H2 1 mol CH3OH
= 95.4 g CH3OH TOO MUCH!
CO(g) + 2 H2 (g) ----> CH3OH (l) 28.01 g/mol 2.016 g/mol 32.04 g/mol
74.5 g ( 12.0 g)excess 85.2 g theoretical ? How much leftover
Figure out how much is used up! Calculate from eithervalue, LR or product from LR:
74.5 g CO 1 mol CO 2 mol H2 2.016 g H2
28.01 g CO 1 mol CO 1 mol H2
= 10.7 g H2 required
12.0 g H2 (original) - 10.7 g H2 (required) = 1.3 g H2 (leftover)
CO(g) + 2 H2 (g) ----> CH3OH (l) 28.01 g/mol 2.016 g/mol 32.04 g/mol
74.5 g ( 12.0 g)excess 85.2 g theoretical ? How much leftover
85.2 g CH3OH 1 mol CH3OH 2 mol H2 2.016 g H2
32.04 g CH3OH 1 mol CH3OH 1 mol H2
= 10.7 g H2 required
12.0 g H2 (original) - 10.7 g H2 (required) = 1.3 g H2 (leftover)
Mass of product, grams
12
10
8
6
4
2
1 2 3 4
Mass of Fe, grams
Constant mass, Bromine, variable masses Fe
Ch 4,#60
(a) What mass of Br2 is used when the reaction consumes 2.0 g Fe?
Product 10.8 g; 10.8 - 2.0 = 8.8 g Br
(b) What is the mole ratio of Fe to Br in this reaction?
2.0 g Fe 1 mol Fe = .036 mol Fe .036 = 1 55.85 g Fe .036
8.8 g Br 1 mol Br = .11 mol Br .11 = 3 79.9 g Br .036
(c) What is the empirical formula of the product?
FeBr3
(d) Balanced chemical equation:
Fe + 3 Br2 -----> 2 FeBr3
(e) Name? Iron (III) Bromide
Which best describes graph:
1. When 1.00 g Fe is added, Fe is LR
2. When 3.50 g of Fe is added, there is an excess of Br2
3. When 2.50 g of Fe is added to the Br2, both reactants are used up completely
4. When 2.00 g of Fe is added to the Br2, 10.0 g of product is formed. The percent yield must be 20%
The Electrolyte is Usually Water Soluble if:
THE CATION IS:
• Na+
• K+
• NH4+
OR THE ANION IS*:
• Cl-, Br-, I-
• ClO4-, ClO3
-
• NO3-
• SO42-
• C2H3O2-
THE COMPOUND IS PROBABLY INSOLUBLE IF:
• THE CATION IS NOT:
• Na+
• K+
• NH4+
• THE ANION IS:
• CO32-
• PO43-
• S2-
• O2-
• OH-
There are a few notable exceptions to thesolubility guide on the last slide, principally theones noted below, which you should be aware of:
Insoluble in Water:
AgCl, AgBr, AgIPbCl2, PbBr2, PbI2
BaSO4, PbSO4
Let’s consider names and formulas of common acids and bases which we meet in these reactions:
COMMON STRONG ACIDS:
HCl Hydrochloric acid
HBr Hydrobromic acid
HI Hydroiodic Acid
HNO3 Nitric Acid
HClO4 Perchloric Acid
H2SO4 Sulfuric Acid
Notenames:learn!
Add H to anions: go from ide, ite, ate to “ic or “ous” acid”
H2O HCl (g) ---------> HCl(aq) -----> H+ (aq) + Cl - (aq)
“Hydrogen Chloride”
“Hydrochloric Acid”
HBr(g) “Hydrogen bromide”HBr(aq) “Hydrobromic acid”
HI(g) “Hydrogen iodide”HI(aq) “Hydroiodic acid”
Nitrate, NO3- ---> HNO3 Nitric Acid
Nitrite, NO2- ---> HNO2 Nitrous Acid
Sulfate, SO42- -----> H2SO4 Sulfuric Acid
Sulfite, SO32- -----> H2SO3 Sulfurous Acid
Perchlorate, ClO4- -----> HClO4 Perchloric Acid
COMMON WEAK ACIDS:*
H3PO4 Phosphoric AcidH2CO3 Carbonic Acid
H2SO3 Sulfurous AcidCH3CO2H or HC2H3O2 Acetic Acid
* and many, many more.....
Remember: These are not broken up in net ionicequations because they are “mostly molecular” inwater...
Group Work:
Do NIE’s for the following:
Na2S (aq) + CrCl3 (aq) ----> ?
(NH4)3PO4 (aq) + Fe(CH3CO2)2 (aq) -----> ?
1. Write down ions, decide2. Write out formulas of products3. Write out equation and balance4. Do total Ionic5. Do Net Ionic
Group Work: Solutions
Na2S (aq) + CrCl3 (aq) ----> ?
Step 1: Ions Involved:
Na+ (no ppt) S2- (ppt???) Cr3+ (ppt???) Cl- (no ppt)
Step 2: Formulas of Products:
Na+ + Cl- ---> NaCl (aq)
Cr3+ + S2- ----> Cr2S3 (s)
Step 3: Write unbalanced equation:
Na2S (aq) + CrCl3 (aq) ----> NaCl (aq) + Cr2S3 (s)
Step 4: Balance:
3 Na2S (aq) + 2 CrCl3 (aq) ----> 6 NaCl (aq) + Cr2S3 (s)
Step 5: Total Ionic Equation:
6 Na+ (aq) + 3 S2- (aq) + 2 Cr3+ (aq) + 6 Cl- (aq) ----->
6 Na+ (aq) + 6 Cl- (aq) + Cr2S3 (s)
Step 5: Total Ionic Equation:
6 Na+ (aq) + 3 S2- (aq) + 2 Cr3+ (aq) + 6 Cl- (aq) ----->
6 Na+ (aq) + 6 Cl- (aq) + Cr2S3 (s)
Step 6: Net Ionic Equation:
2 Cr3+ (aq) + 3 S2- (aq) -----> Cr2S3 (s)
(NH4)3PO4 (aq) + Fe(CH3CO2)2 (aq) -----> ?
Step 1: Ions Involved:
NH4+ (no ppt) PO4
3- (ppt???) Fe2+ (ppt???) CH3CO2 - (no ppt)
Step 2: Formulas of Products:
NH4+ + CH3CO2 - ---> NH4CH3CO2 (aq)
Fe2+ + PO43- ----> Fe3(PO4)2 (s)
Step 3: Write unbalanced equation:
(NH4)3PO4 (aq) + Fe(CH3CO2)2 (aq) ----->
NH4CH3CO2 (aq) + Fe3(PO4)2 (s)
Step 4: Balance:
2 (NH4)3PO4 (aq) + 3 Fe(CH3CO2)2 (aq) ----->
6 NH4CH3CO2 (aq) + Fe3(PO4)2 (s)
2 (NH4)3PO4 (aq) + 3 Fe(CH3CO2)2 (aq) ----->
6 NH4CH3CO2 (aq) + Fe3(PO4)2 (s)
Step 5: Total Ionic Equation:
6 NH4+ (aq) + 2 PO4
3- (aq) + 3 Fe2+ (aq) + 6 CH3CO2- (aq)
-----> 6 NH4+ (aq) + 6 CH3CO2
- (aq) + Fe3(PO4)2 (s)
Step 6: Net Ionic Equation:
3 Fe2+ (aq) + 2 PO43- (aq) -----> Fe3(PO4)2 (s)
Group work 2:
H2SO4 (aq) + Al(OH)3 (s) -----> ?
H3PO4 (aq) + KOH (aq) ---->
HCH3CO2 (aq) + Mg(OH)2 (s) ----->?
1. Form H2O and use leftover ions to form salt2. Write balanced equation3. Do total ionic (watch out for weak acids!)4. Do net ionic equation
Group Work 2 solutions:
H2SO4 (aq) + Al(OH)3 (s) -----> ? ----> H2O + salt
Step One: Formula of salt:
Al3+ + SO42- -----> Al2(SO4)3
Step Two: Write Equation, Balance:
H2SO4 (aq) + Al(OH)3 (s) -----> H2O + Al2(SO4)3 (aq)
3 H2SO4 (aq) + 2 Al(OH)3 (s) -----> 6 H2O + Al2(SO4)3 (aq)
3 H2SO4 (aq) + 2 Al(OH)3 (s) -----> 6 H2O + Al2(SO4)3 (aq)
Step 3: Total Ionic Equation:
6 H+ (aq) + 3 SO42- (aq) + 2 Al(OH)3 (s) ----->
6 H2O + 2 Al3+ (aq) + 3 SO42- (aq)
Step Four: Net Ionic Equation
6 H+ (aq) + 6 OH- (aq) -----> 6 H2O (l) + 2 Al3+ (aq)
H3PO4 (aq) + KOH (aq) ----> ? -----> H2O + salt
Step One: salt formula
K+ + PO43- -----> K3PO4 (aq)
Step Two: Write Equation; Balance
H3PO4 (aq) + KOH (aq) ----> H2O (l) + K3PO4 (aq)
H3PO4 (aq) + 3 KOH (aq) ---->3 H2O (l) + K3PO4 (aq)
H3PO4 (aq) + 3 KOH (aq) ---->3 H2O (l) + K3PO4 (aq)
Step Three: Total Ionic:
H3PO4 (aq) + 3 K+ (aq) + 3 OH- (aq) -----> 3 H2O (l)
+ 3 K+ (aq) + PO43-(aq)
Step Four: Net Ionic:
H3PO4 (aq) + 3 OH- (aq) -----> 3 H2O (l) + PO43-(aq)
HCH3CO2 (aq) + Mg(OH)2 (s) ----->? -----> H2O + salt
Step One: Salt Formula:
Mg2+ + CH3CO2- -----> Mg(CH3CO2)2 (aq)
Step Two: Write Equation, Balance:
HCH3CO2 (aq) + Mg(OH)2(s) -----> H2O + Mg(CH3CO2)2(aq)
2 HCH3CO2 (aq) + Mg(OH)2(s) -----> 2 H2O + Mg(CH3CO2)2(aq)
2 HCH3CO2 (aq) + Mg(OH)2(s) -----> 2 H2O + Mg(CH3CO2)2(aq)
Step Three: Total Ionic Equation:
2 HCH3CO2 (aq) + Mg(OH)2(s) -----> 2 H2O(l) + Mg2+ (aq)
+2 CH3CO2- (aq)
Step Four: Net Ionic Equation
2 HCH3CO2 (aq) + Mg(OH)2(s) -----> 2 H2O(l) + Mg2+ (aq)
+2 CH3CO2- (aq)