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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!