Chemistry 11Chemistry 11Chapter 6Chapter 6

STOICHIOMETRY OF EXCESS QUANTITIESSTOICHIOMETRY OF EXCESS QUANTITIES

Introduction: So far ...Introduction: So far ...

we have we have assumed that a assumed that a given reactant is given reactant is completelycompletely used used up during the up during the reactionreaction

In reality...In reality...

reactions are often reactions are often carried out in such carried out in such a way that one or a way that one or more of the second more of the second reactants actually reactants actually are present in are present in EXCESS amounts.EXCESS amounts.

DefinitionsDefinitions

EXCESS REACTANT = the reactant in excessEXCESS REACTANT = the reactant in excess LIMITING REACTANT = the reactant that LIMITING REACTANT = the reactant that

completely reacts completely reacts THE LIMITING REACTANT determines the yield of THE LIMITING REACTANT determines the yield of

the product (how much product(s) will form)the product (how much product(s) will form)

A Simple AnalogyA Simple Analogy

Imagine you work Imagine you work at McDonalds™ …at McDonalds™ …

You have 10 You have 10 hamburger buns hamburger buns and 5 beef pattiesand 5 beef patties

How many regular How many regular hamburgers can hamburgers can you make?you make?

Da Answer ...Da Answer ...

Indeed, you would Indeed, you would get 5 regular get 5 regular hamburgers!hamburgers!

And what was left And what was left over?over?

Connecting the Lingo …Connecting the Lingo …

There would be 5 There would be 5 hamburger buns in hamburger buns in EXCESS!EXCESS!

Therefore, the beef Therefore, the beef patty is known as the patty is known as the LIMITING ingredient LIMITING ingredient since it “limits” or since it “limits” or determines how many determines how many regular buns can be regular buns can be made!made!

Note thatNote that

we do not predict based on the we do not predict based on the number of hamburger bunsnumber of hamburger buns

Example 1Example 1

If 20.0 g of If 20.0 g of hydrogen gas react hydrogen gas react with 100.0 g of with 100.0 g of oxygen, which oxygen, which reactant is present reactant is present in excess and by in excess and by how many grams?how many grams?

Step 1 …Step 1 …

2H2(g) + O2(g) 2H2O(l)

The balanced equation: The balanced equation:

Step 2 …Step 2 …

First PREDICT which First PREDICT which reactant is limiting reactant is limiting (it’s ok if you predict (it’s ok if you predict wrong)wrong)

USUALLY the reactant USUALLY the reactant with the least number with the least number of moles is limiting of moles is limiting (but not always)(but not always)

Convert masses to molesConvert masses to moles

Number of moles of HNumber of moles of H2 2 present present

= = 20.0 g x 20.0 g x 1 mol H1 mol H22 = 10 mol H10 mol H22

2.0 g H2.0 g H22

Number of moles of ONumber of moles of O22 present present

= 100.0 g x = 100.0 g x 1 mol O1 mol O22 = = 3.1253.125 mol O2

32.0 g O32.0 g O22

Let’s make a prediction ...

Prediction: OPrediction: O22 is limiting is limiting

Mass of HMass of H22 that reacts with 100.0 g O that reacts with 100.0 g O22

= 100.0 g O= 100.0 g O22 x x 1 mol O1 mol O22 x x 2 mol H2 mol H22 x x 2.0 g H2.0 g H22

32.0 g O 32.0 g O22 1 mol O1 mol O22 1 mol H 1 mol H22

= 12.5 g H= 12.5 g H22

Analysing the numbers …Analysing the numbers …

What we What we havehave: 100.0 g O: 100.0 g O22 and 20.0 g H and 20.0 g H22

We predict OWe predict O22 is limiting (i.e. all 100.0 g reacted)

We calculated that we would need 12.5 g H2

Is the prediction correct ?Is the prediction correct ?

Da Answer (again!)Da Answer (again!)

Yes!! Prediction is correct

only 12.5 g H2 is required, so we have an excess of 7.5 g H2 (20.0 g - 12.5 g)

so H2 is in EXCESS of 7.5 g.

The other side of the coinThe other side of the coin

So what if we predicted that HSo what if we predicted that H2 2 was limiting? was limiting?

Mass of OMass of O22 that reacts with 20.0 g H that reacts with 20.0 g H22

= 20.0 g H= 20.0 g H22 x x 1 mol H1 mol H22 x x 1 mol O1 mol O22 x x 32.0 g O32.0 g O22

2.0 g H2.0 g H22 2 mol H 2 mol H22 1 mol O 1 mol O22

= 160.0 g O= 160.0 g O22

Therefore …Therefore …

If ALL 20.0 g of HIf ALL 20.0 g of H22 were to completely react we were to completely react we

would need 160.0 g of Owould need 160.0 g of O22

BUT we only have 100.0 g of O2

So the prediction that H2 limiting is INCORRECT!

Example 2.

If 79.1 g of Zn reacts

with 1.05 L of 2.00 M

HCl,

a) Which reactant is in excess and by how much?

b) What is the mass of each product?

a) which reactant is excess?a) which reactant is excess?

The balanced equation:

Zn + 2HCl ZnCl2 + H2

79.1 g 1.05 L, 2.00 M x g y g

1.21 molmol 2.10 mol

(what we HAVE)

Prediction: Zn is limiting

Moles of HCl requiredMoles of HCl required

== 1.21 mol Zn x 1.21 mol Zn x 2 mol HCl = 2.43 mol HCl

1 mol Zn

Therefore 2.42 mol HCl would be required to react

with 1.21 mol Zn.

We ONLY have 2.10 mol HCl

So is our prediction correct?

Uh Oh! You’re wrong!Uh Oh! You’re wrong!

We would need more HCl (2.42 mol) than what we have (2.10 mol) if all the Zn were to react

Thus: Zn is in excess, and HCl is limiting!

To find how much in excess:

We must find how many

moles of Zn is required

to react with 2.10 molHCl

Mol of Zn

= 2.10 mol HCl x 1 molZn

2 molHCl

= 1.05 mol Zn

Excess Zn = 1.21 - 1.05

= 0.16 mol Zn

b) mass of products?

Since HCl is limiting we MUST use this amount toSince HCl is limiting we MUST use this amount to

calculate the mass of productscalculate the mass of products

x g ZnClx g ZnCl22 = 2.10 mol HCl x = 2.10 mol HCl x 1 mol ZnCl1 mol ZnCl22 x x 136.4 g136.4 g

2 mol HCl 1 mol ZnCl2

= 143 g ZnCl2

y g H2 = 2.10 mol HCl x 1 mol H2 x 2.0 g

2 mol HCl 1 mol H2

= 2.1 g H2

Example 3:Example 3:

3.00 L of 0.1 M NaCl reacts with 2.50 L of 0.125 M 3.00 L of 0.1 M NaCl reacts with 2.50 L of 0.125 M AgNOAgNO33. Calculate the yield of solid AgCl (in . Calculate the yield of solid AgCl (in

grams) that will be produced.grams) that will be produced.

This problem requires us to determine how much This problem requires us to determine how much product (AgCl) will form, so we will need to first product (AgCl) will form, so we will need to first determine which reactant is limiting. determine which reactant is limiting.

The balanced equation:The balanced equation:

NaClNaCl(aq)(aq) + AgNO + AgNO3(aq)3(aq) NaNO3(aq) + AgCl(s)

3.00 L3.00 L 2.50 L 2.50 L ? g? g

0.1M 0.1M 0.125 M 0.125 M

0.300 mol0.300 mol 0.325 mol 0.325 mol

LimitingLimiting Excess Excess (since 1:1 ratio)(since 1:1 ratio)

NaCl limiting ...NaCl limiting ...

Therefore: mol NaCl = mol AgCl = 0.300 molTherefore: mol NaCl = mol AgCl = 0.300 mol

(also 1:1 ratio)(also 1:1 ratio)

Mass of AgCl = 0.300 mol AgCl x Mass of AgCl = 0.300 mol AgCl x 143.5 g AgCl143.5 g AgCl

1 mol AgCl

= 43.1 g AgCl

Percent YieldPercent Yield

Often 100% of the Often 100% of the expected amount of expected amount of product cannot be product cannot be obtained from a reactionobtained from a reaction

The term “Percent The term “Percent Yield” is used to Yield” is used to describe the amount of describe the amount of product actually product actually obtained as a obtained as a percentage of the percentage of the expected amountexpected amount

Reasons for reduced yieldsReasons for reduced yields

A) the reactants may not all A) the reactants may not all react because:react because:i) not all of the pure materiali) not all of the pure material

actually reactsactually reactsii) the reactants may be ii) the reactants may be

impureimpure

B) B) Some of the products are Some of the products are lost during procedures lost during procedures such as solvent extraction, such as solvent extraction, filtration etcfiltration etc

The equation:The equation:

Percent Yield = Percent Yield = ACTUAL YIELDACTUAL YIELD x 100%x 100%

THEORETICAL YIELDTHEORETICAL YIELD

Actual yield = amount of product Actual yield = amount of product obtainedobtained (determined experimentally)(determined experimentally)

Theoretical yield = amount of product Theoretical yield = amount of product expectedexpected (determined from calculations based on the (determined from calculations based on the stoichiometry of the reaction)stoichiometry of the reaction)

The amounts may be expressed in g, mol, The amounts may be expressed in g, mol, moleculesmolecules

Types of calculationsTypes of calculations

A) Find the percentage yield, given the mass of A) Find the percentage yield, given the mass of reactant used and mass of product formedreactant used and mass of product formed

B) Find the mass of product formed, given the mass B) Find the mass of product formed, given the mass of reactant used and the percentage yieldof reactant used and the percentage yield

C) Find the mass of reactant used, given the mass C) Find the mass of reactant used, given the mass of product formed and percentage yieldof product formed and percentage yield

Note that the percentage yield must be less than Note that the percentage yield must be less than 100%100%

But when calculating the theoretical yield But when calculating the theoretical yield assumeassume

a 100% yield a 100% yield

Example 1Example 1

When 15.0 g of CHWhen 15.0 g of CH44 is is

reacted with an excess ofreacted with an excess of

ClCl22 according to the according to the

reaction:reaction:

CHCH44 + Cl + Cl22 CH CH33Cl + HClCl + HCl

a total of 29.7 g of CHa total of 29.7 g of CH33Cl Cl

is formed. Calculate theis formed. Calculate the

percentage yield.percentage yield.

The solution ...The solution ...

The actual yield of CHThe actual yield of CH33Cl = 29.7 g Cl = 29.7 g

To find the theoretical yield of CHTo find the theoretical yield of CH33Cl: (assuming aCl: (assuming a

100% yield) 100% yield)

g of CHg of CH33Cl = 15.0 g CHCl = 15.0 g CH44 x x 1 mol CH1 mol CH44 x 1 mol CH3Cl x 50.5 g

16.0g CH16.0g CH44 1 mol CH 1 mol CH44 1 mol CH 1 mol CH33ClCl

= 47.34 g = 47.34 g

Then:Then:

Percentage yield = Percentage yield = actual yieldactual yield x 100% x 100%

theoretical yield theoretical yield

= = 29.7 g29.7 g x 100% x 100% = 62.7 % = 62.7 %

47.34 g47.34 g

Example deux!Example deux!

What mass of KWhat mass of K22COCO33 is produced when 1.50 g of KO is produced when 1.50 g of KO22 is is

reacted with an excess of COreacted with an excess of CO22 if the reaction has a if the reaction has a

76.0% yield? The reaction is:76.0% yield? The reaction is:

4KO4KO2(s) 2(s) + 2 CO + 2 CO2(g)2(g) 2K 2K22COCO3(s)3(s) + 3O + 3O2(g)2(g)

The solution:The solution:

We are looking for the actual yield (some idiot forgot to

weigh and record the mass of product!)

First calculate the mass of K2CO3 produced (assuminga 100% yield) i.e. the theoretical yield

g of K2CO3 = 1.50 g KO2 x 1 mol KO2 x 2 mol K2CO3 x 138.2 g

71.1 g KO2 4 mol KO2 1 mol K2CO3

= 1.458 g

actual yield = 76.0 % x 1.458 g = 0.760 x 1.458 = 1.11 g K2CO3

Last (but not least) Last (but not least) example ...example ...

What mass of CuO is required to make 10.0 g of Cu What mass of CuO is required to make 10.0 g of Cu

according to the reactionaccording to the reaction

2NH2NH33 + 3CuO + 3CuO N N22 + 3Cu + 3Cu + 3H2O

if the reaction has 58.0 % yield?if the reaction has 58.0 % yield?

Here we go again …Here we go again …

Actual yield = 10.0 g Cu

From the percentage yield equation, calculate the

theoretical yield of Cu.

Theoretical yield of Cu = Actual yield x 100%

Percentage yield

= 10.0 g x 100 %

58.0 %

= 17.24 g

Now find the mass of CuO:Now find the mass of CuO:

Use this theoretical yield and find the mass of CuO that

would be needed:

g CuO = 17.24 g Cu x 1 mol Cu x 3 mol CuO x 79.5g

63.5 g Cu 3 mol Cu 1 mol CuO

= 21.6 g CuO