lecture twelve chm 151 ©slg topics: overview: balancing limiting reagent precipitation reactions...

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?


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!


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)


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)


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)


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)


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)

lecture twelve chm 151 ©slg topics: overview: balancing limiting reagent precipitation reactions...

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