theoretical yield: which reactant is limiting? 1) calculate moles (or mass) of product formed by...

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.

An Ice Cream Sundae Analogy for Limiting Reactions

Fig. 3.10

Add: 14 mol 20 mol

Limiting Reactant Example 2

4NH3 + 5O2 → 4NO + 6H2O

Could make14 mol NO

Could make16 mol NO

NH3 is the limiting reagent.(Use this as basis for all

further calculations)

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

1.What is the % yield of HCN in this reaction?

2.How many grams of NH3 remain?

Limiting Reactant Example 3

Mass to moles66.6 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.

% yield = actual yield theoretical yield

x 100

% yield = 36.4 g HCN 37.5 g HCN

x 100 = 97.1%

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

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 NH3 initially 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

• 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?

Molar massesNH3 1(14.01) + 3(1.008) = 17.02 gCO2 1(12.01) + 2(16.00) = 44.01 gCH4N2O 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.

ONCHg13.6

ONCHmol1

ONCHg60.06

COmol1

ONCHmol1

COg44.01

COmol1COg10.0

24

24

24

2

24

2

22

ONCHg17.6

ONCHmol1

ONCHg60.06

NHmol2

ONCHmol1

NHg17.024

NHmol1NHg10.0

24

24

24

3

24

3

33

10.0 at start-7.73 reacted 2.27 g remains

2.3 g NH3 is left unreacted.(1 decimal place)

reactedNHg7.73 3

3

3

2

3

2

22 NHmol1

NHg17.02

COmol1

NHmol2

COg44.01

COmol1COg10.0

3NHg17.73460577

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

Now subtract the amount reacted from the starting amount:

• 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 gActual yield = 9.3 g

= 68% yield(2 significant figures)

100%x g 13.6

g 9.3

Acid-Base

Two Definitions of Acids & Bases

1) Arrhenius: When dissolved in water,

• acids produce H+ HCl (aq) → H+ (aq) + Cl (aq)

• bases produce OHNaOH (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)

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

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

(aq)

When gaseous hydrogen iodide dissolves in water, the attraction of theoxygen 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 HIwas dissolved.

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

(aq)

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

(aq)H2O

Figure 4.8B: Red cabbage juice added to solutions in the beakers.

Photo courtesy of James Scherer.

Molecular representation of ammonium hydroxide.

NH3(aq) + H2O(l) NH4+

(aq) + OH-(aq)

Reaction of nitric acid with water.

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

(aq)

Two Types of Acid-Base Reactions

1) A-B Neutralization:

LiOH (aq) + HCN (aq) → LiCN (aq) + H2O (l)

H2SO4 (aq) + Ca(OH)2 (aq) → ?

“salt” water

2) A-B Reactions with Gas Formation:

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

Li2SO3 (aq) + NaOH (aq) → ?

gas

Which salts? carbonates CO2

sulfites SO2

sulfides H2S

Beaker with Na+(aq), C2H302-(aq), and SrS04 (solid).

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

2 Na+(aq) + SO4

-2 (aq)+ Sr+2

(aq) + 2C2H3O2 -(aq)

SrSO4 (s)+ 2 Na+(aq) + 2C2H3O2 -

(aq)

SO4-2

(aq)+ Sr+2(aq) SrSO4 (s)

Figure 4.9: Reaction of a carbonate with an acid.

Photo courtesy of American Color.

Oxidation-Reduction

2Na (s) + Cl2(g) 2NaCl(s)

Figure 4.10: Iron nail and copper ( II) sulfate.Photo courtesy of American Color.

Figure 4.10: Fe reacts with Cu2+(aq) and makes Cu(s).

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)

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)

Rules for Assigning Oxidation Numbers

“Molarity” or “M”

mol of soluteliter of solution

NOT:

mol soluteliter of solvent

0.2500 Lmark

= M

Molar concentration

1.Use molarity to convert volume of solution to moles of solute: Mi × Vi = mol solute

2.Use Mi × Vi = Mf × Vf to calculate concentrations of solutions after dilutionNever use this for reactionsNever use this for reactions (e.g. neutralization)

3.Use Mi × Vi = mol solute & stoichiometry to calculate concentrations of sample solutions in reactions (e.g. titrations).

Tips for Molarity-Based Calculations

L solutionmol solute

Molarity

Molarity (Concentration of Solutions)= M

M = = Moles of Solute MolesLiters of Solution L

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%)

Fig. 3.11

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

• 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

Stoichiometry & Ion Dissociation

For a 0.27 M aq. solution of sodium carbonate:

1. Write the dissociation reaction & identify solute(s).

2. Find the molarity of Na+ (aq) & CO32- (aq).

3. If you had 185 mL of this solution, how many moles of Na+ (aq) & CO3

2- (aq) would be present?

4. If you added excess MgBr2 (aq), would you expect a rxn.? If so, how many moles of solid would form?

Concept Check

Converting a Concentrated Solution to a Dilute Solution

Mi x Vi = mol solute = Mf x Vf

Add more solvent(“Dilution”)

Mass of solute does not change!

mol solute

liter of solutionx liter of sample = mol solute

i = initial f = final

Constant!

M1V1=M2V2

The Dilution Dogma: NEVER FORGET IT!

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

How much 0.20 M HCl is needed to make50 mL of 10 mM HCl solution?

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

Mi × Vi = Mf × Vf

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

0.20 M HCl

Dilution CalculationConcept 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)?

H2SO4(aq) + CaCl2(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.

Concept Check

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.

Two Types of Acid-Base Reactions

1) A-B Neutralization:

LiOH (aq) + HCN (aq) → LiCN (aq) + H2O (l)

H2SO4 (aq) + Ca(OH)2 (aq) → ?

“salt” water

2) A-B Reactions with Gas Formation:

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

Li2SO3 (aq) + NaOH (aq) → ?

gas

Which salts? carbonates CO2

sulfites SO2

sulfides H2S

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

or…..

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

acid + base → “salt” + water

Neutralization Calculation

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

N.I.E.: H+ (aq) + OH

(aq) → H2O (l)

Concept Check

1) __ HNO3 (aq) + __ KOH (aq) →

Neutralization CalculationsDetermine 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

2) + __ KOH (aq) →

3) + __ KOH (aq) →

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

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

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?

Stoichiometry not 1:1 !!!

Starting

pointAt equivalence

point

∆V = 4.49 mL

Too far!

0.2500 M NaOH

25.00 mL[H2SO4] = ?

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

xL of NaOH (aq)

required to getto endpoint

Measuredwith buret

=mol NaOHconsumed in neutralization

Calculated

mol NaOH

liter of solution

Knownmoleratio

mol H2SO425.00 mLsample

[H2SO4] (M)

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

0.2500 mol/L x 0.00449 L = 0.00112 mol

xL of NaOH (aq)

required to getto endpoint

=mol NaOHconsumed in neutralization

mol NaOH

liter of solution

0.00112 mol NaOH consumed by neutralization

0.000560 molH2SO4 present in the initial sample

=1 mol H2SO4

2 mol NaOH(from chemical

equation)

x

mol H2SO4 present in initial sample

volume initial sample=

[H2SO4] (M)in the

initial sample

0.02500 L (known)

0.000560 (calculated)

[H2SO4 (aq)] = 0.0224 M

=mol

liter

• 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?

• Molar mass of ZnS = 97.47 g

• = 0.157 L = 157 mL HCl solution

HClmol0.0512

solutionL1 x

ZnSmol1

HClmol2 x

ZnSg97.47

ZnSmol1 x ZnSg0.392

Chemical Equation Calculation - III

Reactants ProductsMolecules

Moles

MassMolecularWeight g/mol

Atoms (Molecules)

Avogadro’sNumber

6.02 x 1023

Solutions

Molaritymoles / liter

Calculating Mass of Solute from a Given Volume of Solution

Volume (L) of Solution

Moles of Solute

Mass (g) of Solute

Molarity M = (mol solute / Liters of solution) = M/L

Molar Mass (M) = ( mass / mole) = g/mol

CaCO3(s) + 2 HCl(aq) CaCl2(aq) + CO2(g) + H2O(l)

2 g 10 mL0.75 M

Which is limiting?

2 g CaCO3 x 1 mol CaCl2 = 0.01 mol CaCl2100 g CaCO3 1 mol CaCO3

0.01 L HCl x 0.75 mol HCl x 1 mol CaCl2 =

L HCl 2 mol HCl0.004 mol CaCl2

How many g of CaCO3 remain?What is the [Cl-] after the reaction?

Calculating Amounts of Reactants and Products for a Reaction in Solution

Al(OH)3 (s) + 3 HCl (aq) 3 H2O(l) + AlCl3 (aq)

Mass (g) of Al(OH)3

Moles of Al(OH)3

Moles of HCl

Volume (L) of HCl

M (g/mol)

molar ratio

M ( mol/L)

Given 10.0 g Al(OH)3, what volume of1.50 M HCl is required to neutralize thebase?

10.0 g Al(OH)3

78.00 g/mol Al(OH)3



Mass (g) of Al(OH)3

Moles of Al(OH)3

Moles of HCl

Volume (L) of HCl

M (g/mol)

molar ratio

M ( mol/L)


10.0 g Al(OH)3

78.00 g/mol= 0.128 mol Al(OH)3

0.128 mol Al(OH)3 x =

Moles HCl



Mass (g) of Al(OH)3

Moles of Al(OH)3

Moles of HCl

Volume (L) of HCl

M (g/mol)

molar ratio

M ( mol/L)


10.0 g Al(OH)3

78.00 g/mol= 0.128 mol Al(OH)3

0.128 mol Al(OH)3 x 3 moles HClmoles Al(OH)3

=

0.385 Moles HCl

Vol HCl



Mass (g) of Al(OH)3

Moles of Al(OH)3

Moles of HCl

Volume (L) of HCl

M (g/mol)

molar ratio

M ( mol/L)


10.0 g Al(OH)3

78.00 g/mol= 0.128 mol Al(OH)3

0.128 mol Al(OH)3 x 3 moles HClmoles Al(OH)3

=

0.385 Moles HCl

Vol HCl = x 0.385 Moles HCl

Vol HCl = 0.256 L = 256 ml

1.00 L HCl1.50 Moles HCl

Solving Limiting Reactant Problems in Solution - Precipitation Problem - I

Problem: Lead has been used as a glaze for pottery for years, and can bea 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)

Volume (L) of Pb(NO3)2

solution

Mass (g) of PbS

Amount (mol) of Pb(NO3)2

Volume (L) of Na2S solution

Amount (mol) of Na2S

Amount (mol) of PbS

Amount (mol) of PbS

Multiply by M (mol/L)


Molar Ratio Molar Ratio

Choose the lower numberof PbS and multiply by M (g/mol)

Volume (L) of Pb(NO3)2

solution

Volume (L) of Na2S solution

Amount (mol) of Pb(NO3)2

Amount (mol) of Na2S

Amount (mol) of PbS

Mass (g) of PbS



Molar Ratio

Divide byequationcoefficient

Divide byequationcoefficientSmallest

Solving Limiting Reactant Problems inSolution - 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) =


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

Calculation of product yield:

Therefore Lead Nitrate is the Limiting Reactant!






Moles PbS = 0.012065 Mol Pb+2x = 0.012065 Mol Pb+21 mol PbS1 mol Pb(NO3)2






Moles PbS = 0.012065 Mol Pb+2x = 0.012065 Mol Pb+21 mol PbS1 mol Pb(NO3)2

0.012065 Mol Pb+2 = 0.012065 Mol PbS

0.012065 Mol PbS x = 2.89 g PbS239.3 g PbS1 Mol PbS

• Quantitative Analysis

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

• 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

• The figure on the right shows the reaction of Ba(NO3)2 with K2CrO4 forming the yellow BaCrO4 precipitate.

• The BaCrO4 precipitate is being filtered in the figure on the right. It can then be dried and weighed.

• 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?

Agmol1

Agg107.9 x

AgClmol1

Agmol1 x

AgClg143.32

AgClmol1 x AgClg1.788

= 1.346 g Ag in the compound

100%compoundsilverg1.583

Agg1.346

= 85.03% Ag

Molar mass of silver chloride (AgCl) = 143.32 g

• 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

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




Other Resources

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

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theoretical yield: which reactant is limiting? 1) calculate moles (or mass) of product formed by...

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