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1
Percent Yield and Limiting Reactants
Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin CummingsEdited by bbg
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Theoretical, Actual, and Percent Yield
Theoretical yieldTheoretical yield The maximum amount of product, which is calculated
using the balanced equation.
Actual yieldActual yield The amount of product obtained when the reaction takes
place.
Percent yieldPercent yield The ratio of actual yield to theoretical yield.
percent yield = actual yield (g) x 100 theoretical yield (g)
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Guide to Calculations for Percent Yield
Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings
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You prepared cookie dough to make 5 dozen cookies.The phone rings and you answer. While you talk, a sheetof 12 cookies burn and you throw them out. The restof the cookies are okay. What is the percent yield ofedible cookies?
Theoretical yieldTheoretical yield 60 cookies possible
Actual yieldActual yield 48 cookies to eat
Percent yieldPercent yield 48 cookies x 100 = 80% yield80% yield 60 cookies
Calculating Percent Yield
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Learning Check
Without proper ventilation and limited oxygen, the reaction of carbon and oxygen produces carbon monoxide.
2C(g) + O2(g) 2CO(g)
What is the percent yield if 40.0 g CO are produced when 30.0 g O2 are used?
1) 25.0%
2) 75.0%
3) 76.2%
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Solution
3) 76.2 % yield3) 76.2 % yield
theoretical yield of COtheoretical yield of CO 30.0 g O2 x 1 mole O2 x 2 moles CO x 28.0 g CO
32.0 g O2 1 mole O2 1 mole CO
= 52.5 g CO (theoretical)
percent yieldpercent yield 40.0 g CO (actual) x 100 = 76.2 % yield 76.2 % yield
52.5 g CO (theoretical)
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Learning Check
When N2 and 5.00 g H2 are mixed, the reaction
produces 16.0 g NH3. What is the percent yield for
the reaction?
N2(g) + 3H2(g) 2NH3(g)
1) 31.3 %
2) 56.5 %
3) 80.0 %
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Solution
2)2) 56.5 % 56.5 %
N2(g) + 3H2(g) 2NH3(g)
5.00 g H2 x 1 mole H2 x 2 moles NH3 x 17.0 g NH3
2.02 g H2 3 moles H2 1 mole NH3
= 28.2 g NH3 (theoretical)
Percent yieldPercent yield = 16.0 g NH3 x 100 = 56.7 %56.7 %
28.2 g NH3
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Limiting Reactant
A limiting reactantlimiting reactant in a chemical reaction is the
substance that
Is used up first.
Limits the amount of product that can form and stops the reaction.
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Reacting Amounts
In a table setting, there is 1plate,
1 fork, 1 knife, and 1 spoon.
How many table settings are
possible from 5 plates, 6 forks,
4 spoons, and 7 knives?
What is the limiting item?What is the limiting item?Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings
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Reacting Amounts
Only 4 place settings are possible.
Initially Used Left over
Plates 5 4 1
Forks 6 4 2
Spoons 4 4 0
Knives 7 4 3
The limiting item is the spoon.
Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings
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Example of An Everyday Limiting ReactantHow many peanut butter sandwiches could be made from 8 slices of bread and 1 jar of peanut butter?
With 8 slices of bread, only 4 sandwiches could be made. The bread is the limiting item.
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Example of An Everyday Limiting Reactant
How many peanut butter sandwiches could be made from 8 slices bread and 1 tablespoon of peanut butter?
With 1 tablespoon of peanut butter, only 1 sandwich could be made. The peanut butterpeanut butter is the limiting item.
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Guide to Calculating Product from a Limiting Reactant
Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings
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Limiting Reactant
When 4.00 moles H2 is mixed with 2.00 moles Cl2,how many moles of HCl can form?
H2(g) + Cl(g) 2HCl (g)
4.00 moles 2.00 moles ??? moles
Calculate the moles of product that each reactant, H2 and Cl2, could produce.
The limiting reactant is the one that produces the smallest amount of product.
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Limiting Reactant
HCl from H2
4.00 moles H2 x 2 moles HCl = 8.00 moles HCl
1 moles H2 (not possible)
HCl from Cl2
2.00 moles Cl2 x 2 moles HCl = 4.00 moles HCl
1 mole Cl2 (smaller number of moles, Cl2 will be used up first)
The limiting reactant is Cl2 because it is used up first. Thus Cl2 produces the smaller number of moles of HCl.
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Check Calculations
Initially H2
4.00 moles
Cl2
2.00 moles
2HCl
0 mole
Reacted/
Formed
-2.00 moles -2.00 moles +4.00 moles
Left after reaction
2.00 moles
Excess
0 mole
LimitingLimiting
4.00 moles
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Limiting Reactants Using Mass
If 4.80 moles Ca are mixed with 2.00 moles N2, which is the limiting reactant?
3Ca(s) + N2(g) Ca3N2(s)moles of Ca3N2 from Ca
4.80 moles Ca x 1 mole Ca3N2 = 1.60 moles Ca3N2 3 moles Ca (Ca is used up)
moles of Ca3N2 from N2
2.00 moles N2 x 1 mole Ca3N2 = 2.00 moles Ca3N2
1 mole N2 (not possible)
CaCa is used up when 1.60 mole Ca3N2 forms. Thus, Ca is the limiting reactant.
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Limiting Reactants Using Mass
Calculate the mass of water produced when 8.00 g H2
and 24.0 g O2 react?
2H2(g) + O2(g) 2H2O(l)
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Limiting Reactants Using Mass
Moles H2O from H2:
8.00 g H2 x 1 mole H2 x 2 moles H2O = 3.97 moles H2O
2.02 g H2 2 moles H2 (not possible)
Moles H2O from O2:
24.0 g O2 x 1 mole O2 x 2 moles H2O = 1.50 moles H2O
32.0 g O2 1 mole O2 O2 is limiting
The maximum amount of product is 1.50 moles H2O,which is converted to grams.
1.50 moles H2O x 18.0 g H2O = 27.0 g H27.0 g H22OO
1 mole H2O