Unit 9 Stoich Part 2

8/2/2019 Unit 9 Stoich Part 2

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Limiting Reagents

andPercent Yield


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If one of our ingredients gets used upduring our preparation it is called thelimiting reactant (LR)The LR limits the amount of product

we can form

It is equally impossible for a chemistto make a certain amount of a desiredcompound if there isnt enough of one

of the reactants.


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As weve been learning, a balancedchemical rxn is a chemists recipe.Which allows the chemist to predict

the amount of product formed fromthe amounts of ingredients available

Lets look at the reaction equation for

the formation of ammonia:N2(g) + 3H2(g) 2NH3(g)

When 1 mole of N2 reacts with 3 moles of

H2, 2 moles of NH3 are produced. How much NH3 could be made if 2 moles

of N2 were reacted with 3 moles of H2?


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The amount of H2 limits the amount ofNH

3that can be made.

From the amount of N2 available wecan make 4 moles of NH3From the amount of H2 available we

can only make 2 moles of NH3.H2 is our limiting reactant here.It runs out before the N2 is used up.

Therefore, at the end of the reactionthere should be N2 left over.When there is reactant left over it is

said to be in excess.

N2(g) + 3H2(g) 2NH3(g)


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How much N2 will be left over afterthe reaction?In our rxn it takes 1 mol of N2 to react

all of 3 mols of H2, so there must be1 mol of N2 that remains unreacted.

We can use our new stoich calculation

skills to determine 3 possible types ofLR type calculations.1. Determine which of the reactants

will run out first (limiting reactant)

2. Determine amount of product3. Determine how much excess

reactant is wasted


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Limiting Reactant Problems:

Given the following reaction:

2Cu + S Cu2S

What is the limiting reactant when

82.0 g of Cu reacts with 25.0 g S?What is the maximum amount ofCu2S that can be formed?

How much of the other reactant iswasted?


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Our 1st goal is to calculate howmuch S would react if all of the Cu

was reacted.From that we can determine thelimiting reactant (LR).

Then we can use the LimitingReactant to calculate the amountof product formed and the amountof excess reactant left over.

82g Cu mol Cumol S g S


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2Cu + S Cu2S

82.0gCu 1molCu

63.5gCu

1mol S

2molCu

32.1g S

1mol S

=20.7 g SSo if all of our 82.0g of Copper were

reacted completely it would require

only 20.7 grams of Sulfur.Since we initially had 25g of S, we aregoing to run out of the Cu, the limitingreactant) & end up with 4.3 grams of S


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Copper being our Limiting Reactant isthen used to determine how muchproduct is produced.

The amount of Copper we initiallystart with limits the amount of productwe can make.

82.0gCu1molCu

63.5gCu

1molCu2S

2molCu

________159gCu2S

1molCu2S

= 103 g Cu2S


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So the reaction between 82.0g of Cuand 25.0g of S can only produce 103gof Cu2S.

The Cu runs out before the S and wewill end up wasting 4.7 g of the S.

Ex 2: Hydrogen gas can be produced inthe lab by the rxn of Magnesium metalwith HCl according to the following rxn

equation: Mg + 2HCl MgCl2 + H2

What is the LR when 6.0 g HCl reactswith 5.0 g Mg? What is the maximumamount of H2 that can be formed? Andhow much of the other reactant is

wasted?


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5.0g Mg 1molMg24.3gMg

2molHCl1molMg

36.5gHCl

1molHCl

= 15.0g HCl

5.0g Mgmol Mg 2mol HClg HCl

So if 5.0g of Mg were used up it wouldtake 15.0g HCl, but we only had 6.0g

of HCl to begin with.Therefore, the 6.0g of HCl will run outbefore the 5.0g of Mg, so HCl is ourLimiting Reactant.


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6.0g HCl 1molHCl36.5gHCl

1molH2

2molHCl

2.0gH2

1molH2

= 0.164 g H2 produced

6.0g HCl2mol HCl 1mol H2 g H2

6.0g HCl 1molHCl36.5gHCl

1molMg2molHCl

24.3gMg

1molMg

= 1.997 g Mg

6.0g HCl2mol HCl 1mol Mg g Mg

- 5.0 g Mg = 3.01g Mg extra


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Calculating Percent Yield

In theory, when a teacher gives an

exam to the class, every studentshould get a grade of 100%.

Your exam grade, expressed as a

percent, is a quantity that showshow well you did on the examcompared with how well you could

have done if you had answered allquestions correctly


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This calc is similar to the percent yieldcalc that you do in the lab when theproduct from a chemical rxn is lessthan you expected based on thebalanced eqn.

You might have assumed that if we use

stoich to calculate that our rxn willproduce 5.2 g of product, that we willactually recover 5.2 g of product inthe lab.

This assumption is as faulty asassuming that all students will score100% on an exam.


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When an equation is used to calculatethe amount of product that is possibleduring a rxn, a value representing the

theoretical yield is obtained.The theoretical yield is the maximum

amount of product that could be

formed from given amounts ofreactants.

In contrast, the amount of productthat forms when the rxn is carried outin the lab is called the actual yield.The actual yield is often less than the

theoretical yield.


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The percent yield is the ratio of theactual yield to the theoretical yield asa percent

It measures the efficiency of thereaction

Percent yield=actual yield

theoretical yield

x 100

What causes a percent yield to be lessthan 100%? greater than 100%?

actual is less than theoretical < 100%

actual is more than theoretical > 100%


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Rxns dont always go to completion;when this occurs, less than theexpected amnt of product is formed.Impure reactants and competing side

rxns may cause unwanted products toform.

Actual yield can also be lower thanthe theoretical yield due to a loss ofproduct during filtration ortransferring between containers.

If a wet precipitate is recovered itmight weigh heavy due to incompletedrying, etc.


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Calcium carbonate is synthesized byheating,as shown in the followingequation: CaO + CO2 CaCO3

What is the theoretical yield of CaCO3if 24.8 g of CaO is heated with 43.0 gof CO2?

What is the percent yield if 33.1 g ofCaCO3 is produced?

Determine which reactantis the limiting and then decidewhat the theoretical yield is.


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24.8 g

CaO

1molCaO

56g CaO

1mol CO2

1mol CaO

44 g CO2

1molCO2

= 19.5gCO2

24.8gCaOmolCaOmol CO2gCO2

24.8 gCaO

1mol CaO

56g CaO

1molCaCO3

1mol CaO

100g CaCO3

1molCaCO3

= 44.3 g CaCO3

24.8gCaOmolCaOmol CaCO3gCaCO3

LR


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CaO is our LR, so the reaction shouldtheoretically produce 44.3 g of CaCO3(How efficient were we?)Our percent yield is:

Percent yield=

33.1 g CaCO3

44.3 g CaCO3

_____________ x 100

Percent yield = 74.7%


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Limiting Reactant Practice 15.0 g of potassium reacts with 15.0 g of

iodine. Calculate which reactant is limiting

and how much product is made in grams.

2 K + I2 2KI


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Finding the Amount of Excess

By calculating the amount of the excess

reactant needed to completely react with the

limiting reactant, we can subtract that amount

from the given amount to find the amount of

excess.

Can we find the amount of excess potassium

in the previous problem?


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Finding Excess Practice

15.0 g of potassium reacts with 15.0 g of iodine.

2 K + I2 2 KI We found that Iodine is the limiting reactant, and

19.6 g of potassium iodide are produced.

15.0 g I2 1 mol I2 2 mol K 39.1 g K

254 g I2 1 mol I2 1 mol K= 4.62 g K

USED!

15.0 g K 4.62 g K = 10.38 g K EXCESS

Given amount of

excess reactantAmount of

excess

reactant

actually used

Note that we started with the

limiting reactant! Once you

determine the LR, you should

only start with it!


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Limiting Reactant: Recap

1. You can recognize a limiting reactant problem because thereis MORE THAN ONE GIVEN AMOUNT.

2. Convert ALL of the reactants to the SAME product (pick anyproduct you choose.)

3. The lowest answer is the correct answer.

4. The reactant that gave you the lowest answer is theLIMITING REACTANT.

5. The other reactant(s) are in EXCESS.

6. To find the amount of excess, subtract the amount used from

the given amount.7. If you have to find more than one product, be sure to start

with the limiting reactant. You dont have to determinewhich is the LR over and over again!

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Unit 9 Stoich Part 2