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Theoretical Yield: Which Reactant is Limiting?. 1) calculate moles (or mass) of product formed by complete reaction of each reactant. 2) the reactant that yields the least product is the limiting reactant (or limiting reagent).

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theoretical yield which reactant is limiting
Theoretical Yield: Which Reactant is Limiting?

1) calculate moles (or mass) of product formed by complete reaction of each reactant.

2) the reactant that yields the least product is the limiting reactant (or limiting reagent).

3) the theoretical yield for a reaction is the maximum amount of product that could be generated by complete consumption of the limiting reagent.

slide3

Limiting Reactant Example 2

4NH3 + 5O2→ 4NO + 6H2O

Add: 14 mol 20 mol

Could make

16 mol NO

Could make

14 mol NO

NH3 is the limiting reagent.

(Use this as basis for all further calculations)

slide5

Limiting Reactant Example 3

  • When 66.6 g of O2 gas is mixed with 27.8 g of NH3 gas and 25.1 g of CH4 gas, 36.4 g of HCN gas is produced by the following reaction:
  • 2CH4 + 2NH3 + 3O2→ 2HCN + 6H2O
  • 16.04 17.03 32.00 27.03 g/mol
  • What is the % yield of HCN in this reaction?
  • How many grams of NH3 remain?
slide6

Mass to moles

66.6 g of O2→ 2.08 mol O2

27.8 g of NH3 → 1.63 mol NH3

25.1 g of CH4→ 1.56 mol CH4

Which reactant is limiting?

2.08 mol O2 can yield 1.39 mol (or 37.5 g) HCN

1.63 mol NH3 can yield 1.63 mol (or 44.1 g) HCN

1.56 mol CH4 can yield 1.56 mol (or 42.2 g) HCN

Conclusion?

O2 is the limiting reagent.

slide7
% yield = actual yield

O2 is the limiting reagent. So, the theoretical yield is based on 100% consumption of O2.

2.08 mol O2 can yield 1.39 mol (or 37.5 g) HCN

x 100

theoretical yield

% yield = 36.4 g HCN

x 100 = 97.1%

37.5 g HCN

slide8

2. How many grams of NH3 remain?

36.4 g (or 1.35 mol) of HCN gas is produced

2CH4 + 2NH3 + 3O2→ 2HCN + 6H2O

Since the reaction stoichiometry is 1:1, 1.35 mol of NH3 is consumed:

1.63 mol NH3initially present

– 1.35 mol NH3 consumed

0.28 mol NH3 remaining

0.28 mol NH3 x (17.03 g NH3/mol) = 4.8 g NH3 remain

slide9
Urea, CH4N2O, is used as a nitrogen fertilizer. It is manufactured from ammonia and carbon dioxide at high pressure and high temperature:

2NH3 + CO2(g)  CH4N2O + H2O

In a laboratory experiment, 10.0 g NH3 and 10.0 g CO2 were added to a reaction vessel. What is the maximum quantity (in grams) of urea that can be obtained? How many grams of the excess reactant are left at the end of the reactions?

slide10

Molar masses

NH3 1(14.01) + 3(1.008) = 17.02 g

CO2 1(12.01) + 2(16.00) = 44.01 g

CH4N2O 1(12.01) + 4(1.008) + 2(14.01) + 1(16.00) = 60.06 g

CO2 is the limiting reactant.

13.6 g CH4N2O will be produced.

slide11

To find the excess NH3, we find how much NH3 reacted:

Now subtract the amount reacted from the starting amount:

10.0 at start

-7.73 reacted

2.27 g remains

2.3 g NH3 is left unreacted.

(1 decimal place)

slide12
2NH3 + CO2(g)  CH4N2O + H2O

When 10.0 g NH3 and 10.0 g CO2 are added to a reaction vessel, the limiting reactant is CO2. The theoretical yield is 13.6 of urea. When this reaction was carried out, 9.3 g of urea was obtained. What is the percent yield?

Theoretical yield = 13.6 g

Actual yield = 9.3 g

= 68% yield

(2 significant figures)

slide15

Two Definitions of Acids & Bases

  • 1) Arrhenius: When dissolved in water,
  • acids produce H+ HCl(aq)→ H+(aq) + Cl-(aq)
  • bases produce OH- NaOH(aq)→ Na+(aq) + OH-(aq)
  • 2) Brønsted-Lowry: Proton transfer
  • acids are proton donors
  • bases are proton acceptors

NaOH(aq) + HCl(aq)→?

Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq)→ ?

NaOH(aq) + HCl(aq)→ H2O(l) + NaCl(aq)

slide16

Acids - A Group of Covalent Molecules Which Lose Hydrogen Ions to Water Molecules in Solution

When gaseous hydrogen iodide dissolves in water, the attraction of the

oxygen atom of the water molecule for the hydrogen atom in HI is

greater that the attraction of the of the iodide ion for the hydrogen atom,

and it is lost to the water molecule to form an hydronium ion and an

iodide ion in solution. We can write the hydrogen atom in solution as

either H+(aq) or as H3O+(aq) they mean the same thing in solution. The

presence of a hydrogen atom that is easily lost in solution is an “Acid”

and is called an “acidic” solution. The water (H2O) could also be written

above the arrow indicating that the solvent was water in which the HI

was dissolved.

HI(g) + H2O(L) H+(aq) + I -(aq)

HI(g) + H2O(L) H3O+(aq) + I -(aq)

H2O

HI(g) H+(aq) + I -(aq)

figure 4 8b red cabbage juice added to solutions in the beakers photo courtesy of james scherer
Figure 4.8B: Red cabbage juice added to solutions in the beakers.Photo courtesy of James Scherer.
molecular representation of ammonium hydroxide
Molecular representation of ammonium hydroxide.

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

reaction of nitric acid with water
Reaction of nitric acid with water.

HNO3(aq)+ H2O(l)  NO3-(aq) + H3O+(aq)

slide22

“salt”

water

Two Types of Acid-Base Reactions

1) A-B Neutralization:

LiOH(aq) + HCN(aq)→ LiCN(aq) + H2O(l)

H2SO4(aq) + Ca(OH)2(aq)→?

gas

2) A-B Reactions with Gas Formation:

Na2CO3(aq) + 2HBr(aq)→ 2NaBr(aq) + H2O(l) + CO2(g)

Li2SO3(aq) + NaOH(aq)→?

Which salts?carbonates  CO2

sulfites  SO2

sulfides  H2S

beaker with na aq c 2 h 3 0 2 aq and srs0 4 solid
Beaker with Na+(aq), C2H302-(aq), and SrS04 (solid).

Na2SO4 (aq)+ Sr(C2H3O2)2(aq)  SrSO4 (s)+ NaC2H3O2 (aq)

SO4-2 (aq)+ Sr+2(aq) SrSO4 (s)

2 Na+(aq) + SO4-2 (aq)+ Sr+2(aq) + 2C2H3O2-(aq)

 SrSO4 (s)+ 2 Na+(aq) + 2C2H3O2-(aq)

oxidation reduction
Oxidation-Reduction

2Na (s) + Cl2(g) 2NaCl(s)

figure 4 10 the copper metal plates out on the nail
Figure 4.10: The copper metal plates out on the nail.

Write a net ionic equation for this reaction!

Cu+2(aq) + Fe(s) Cu(s) + Fe+2(aq)

slide31

Redox: changing oxidation numbers

Oxidation number:

a real charge (ionic compounds) or hypothetical charge (molecular compounds, polyatomic ions) associated with an individual atom in a compound.

The oxidation number allows for electron accounting

N2(g) + 3H2(g)→2NH3(g)

CH4(g) + 2O2(g)→CO2(g)+ 2H2O(g)

slide33

Molar concentration

“Molarity” or “M”

mol of solute

liter of solution

NOT:

mol solute

liter of solvent

= M

0.2500 L

mark

slide34

Tips for Molarity-Based Calculations

  • Use molarity to convert volume of solution to moles of solute: Mi × Vi = mol solute
  • Use Mi × Vi = Mf × Vfto calculate concentrations of solutions after dilution
    • Never use this for reactions(e.g. neutralization)
  • Use Mi × Vi =mol solute& stoichiometry to calculate concentrations of sample solutions in reactions (e.g. titrations).
slide35

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%)

preparing a solution i
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)

preparing a solution ii
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

slide39

Concept Check

Stoichiometry & Ion Dissociation

For a 0.27 M aq. solution of sodium carbonate:

  • Write the dissociation reaction & identify solute(s).
  • Find the molarity of Na+(aq) & CO32-(aq).
  • If you had 185 mL of this solution, how many moles of Na+(aq) & CO32-(aq) would be present?
  • If you added excess MgBr2(aq), would you expect a rxn.? If so, how many moles of solid would form?
slide41

mol solute

liter of solution

x

=

liter of sample

mol solute

MixVi = mol solute=MfxVf

i = initial

f = final

Add more solvent

(“Dilution”)

Mass of solute does not change!

Constant!

slide42

The Dilution Dogma:

NEVER FORGET IT!

M1V1=M2V2

dilution of solutions
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

slide44

Concept Check

Dilution Calculation

How much 0.20 M HCl is needed to make

50 mL of 10 mM HCl solution?

Mi×Vi = Mf×Vf

(0.20 M HCl)×(L) = (0.010 M HCl)×(0.050 L)

 (L) = (0.010 M HCl)×(0.050 L) = 2.5×10-3L

0.20 M HCl

slide45

Concept Check

Dilution & Solution Examples

A) We have a 3.0 M aqueous solution of H2SO4.

How do you make 100. mL of 1.4 M H2SO4(aq)?

B) Determine how you would make 250. mL of

0.56 M CaCl2(aq)? What is [Ca2+]? [Cl-]?

(Reagent is solid is CaCl2·2H20; 147.01 g/mol)

C) Predict what would happen in you mixed

solutions A and B together.

H2SO4(aq) + CaCl2(aq) →?

slide46

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.

slide47

“salt”

water

Two Types of Acid-Base Reactions

1) A-B Neutralization:

LiOH(aq) + HCN(aq)→ LiCN(aq) + H2O(l)

H2SO4(aq) + Ca(OH)2(aq)→?

gas

2) A-B Reactions with Gas Formation:

Na2CO3(aq) + 2HBr(aq)→ 2NaBr(aq) + H2O(l) + CO2(g)

Li2SO3(aq) + NaOH(aq)→?

Which salts?carbonates  CO2

sulfites  SO2

sulfides  H2S

slide48

Concept Check

Neutralization Calculation

  • How much 2.0 M HCl is needed to “neutralize” 2.3 liters of 0.15 M NaOH?

HCl(aq) + NaOH(aq)→ NaCl(aq) + H2O(l)

acid + base → “salt” + water

or…..

How much 2.0 M HCl should be added such that the mol of H+(aq)=mol OH–(aq)?

N.I.E.: H+(aq) + OH-(aq)→ H2O(l)

slide49

Neutralization Calculations

Determine the volume of 0.10 M KOH(aq)solution required to neutralize 25.00 mL samples of three different acids, all at 0.20 M.

Acids: 1) Nitric acid 2) Carbonic acid 3) Phosphoric acid

  • 1) __ HNO3 (aq) + __ KOH (aq)→
  • 2) + __ KOH (aq)→
  • 3) + __ KOH (aq)→
slide50

Titration Calculation

4.49 mL of 0.2500 M NaOH is required to titrate a 25.00 mL sample of H2SO4 to the endpoint. What is the molar concentration of H2SO4 in the sample?

NaOH(aq) + H2SO4(aq) Na2SO4(aq) + H2O(l)

2NaOH(aq) + 1H2SO4(aq) Na2SO4(aq) + 2H2O(l)

Stoichiometry not 1:1 !!!

slide51

0.2500 M NaOH

∆V = 4.49 mL

25.00 mL

[H2SO4] = ?

Too far!

Starting

point

At equivalence

point

slide52

L of NaOH(aq)

required to get

to endpoint

mol NaOH

consumed in

neutralization

mol NaOH

liter of solution

x

=

Measured

with buret

Calculated

Known

mole

ratio

[H2SO4] (M)

mol H2SO4

25.00 mL

sample

2NaOH(aq) + H2SO4(aq)→ Na2SO4(aq)+ 2H2O(l)

slide53

L of NaOH(aq)

required to get

to endpoint

mol NaOH

consumed in

neutralization

mol NaOH

liter of solution

x

=

1 mol H2SO4

2 mol NaOH

0.000560 mol

H2SO4 present in

the initial sample

x

=

(from chemical equation)

2NaOH(aq) + H2SO4(aq)→ Na2SO4(aq)+ 2H2O(l)

0.2500 mol/L x 0.00449 L = 0.00112 mol

0.00112 mol

NaOHconsumed by neutralization

slide54

molH2SO4 present in initial sample

mol

liter

=

=

volume initial sample

0.000560 (calculated)

[H2SO4] (M)

in the

initial sample

0.02500 L (known)

[H2SO4(aq)] = 0.0224 M

slide56
Zinc sulfide reacts with hydrochloric acid to produce hydrogen sulfide gas:

ZnS(s) + 2HCl(aq) ZnCl2(aq) + H2S(g)

How many milliliters of 0.0512 M HCl are required to react with 0.392 g ZnS?

slide57
Molar mass of ZnS = 97.47 g

= 0.157 L = 157 mL HCl solution

chemical equation calculation iii
Chemical Equation Calculation - III

Mass

Atoms (Molecules)

Molecular

Weight

Avogadro’s

Number

g/mol

6.02 x 1023

Molecules

Reactants

Products

Moles

Molarity

moles / liter

Solutions

slide60

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

slide61

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?

slide62

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

slide63

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

slide64

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

slide65

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

slide66

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)

slide67

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

slide68

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

Divide by

equation

coefficient

Amount (mol)

of Na2S

Divide by

equation

coefficient

Smallest

Molar Ratio

Amount (mol)

of PbS

Mass (g) of PbS

slide69

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) =

slide70

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:

slide71

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:

1 mol PbS

1 mol Pb(NO3)2

Moles PbS = 0.012065 Mol Pb+2x = 0.012065 Mol Pb+2

slide72

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:

1 mol PbS

1 mol Pb(NO3)2

Moles PbS = 0.012065 Mol Pb+2x = 0.012065 Mol Pb+2

0.012065 Mol Pb+2 = 0.012065 Mol PbS

0.012065 Mol PbS x = 2.89 g PbS

239.3 g PbS

1 Mol PbS

slide73
Quantitative Analysis

The determination of the amount of a substance or species present in a material

slide74
Gravimetric Analysis

A type of quantitative analysis in which the amount of a species in a material is determined by converting the species to a product that can be isolated completely and weighed

slide75
The figure on the right shows the reaction of Ba(NO3)2 with K2CrO4 forming the yellow BaCrO4 precipitate.
slide76
The BaCrO4 precipitate is being filtered in the figure on the right. It can then be dried and weighed.
slide77
A soluble silver compound was analyzed for the percentage of silver by adding sodium chloride solution to precipitate the silver ion as silver chloride. If 1.583 g of silver compound gave 1.788 g of silver chloride, what is the mass percent of silver in the compound?
slide78

Molar mass of silver chloride (AgCl) = 143.32 g

= 1.346 g Ag in the compound

= 85.03% Ag

slide79

http://www.learnerstv.com/animation/chemistry/ruther14.swf

http://www.dartmouth.edu/~chemlab/info/resources/mashel/MASHEL.html

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/thermochem/solutionSalt.swf

other resources
Other Resources

Visit the student website at college.hmco.com/pic/ebbing9e

other resources81
Other Resources

Visit the student website at college.hmco.com/pic/ebbing9e