Percent Yield When you make a peanut butter sandwich you need 2 slices of bread and 1 tablespoon of peanut butter The reactant that runs out first is called the limiting reactant Learning Goal ID: 600081
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Slide1
7.8 Limiting Reactants, Percent Yield
When you make a peanut butter sandwich, you need 2 slices of bread and 1 tablespoon of peanut butter. The reactant that runs out first is called the limiting reactant.
Learning Goal Identify a limiting reactant and calculate the amount of product formed from the limiting reactant. Given the actual quantity of product, determine the percent yield for a reaction.Slide2
Limiting Reactant
A
limiting reactant in a chemical reaction is the substance that is used up first. limits the amount of product that can
form.
The reactant that does not completely react and is left over at the end of the reaction is called the
excess reactant
.Slide3
Calculating Moles of Product from a Limiting Reactant
In many reactions, there is a limiting reactant that determines the amount of product that can be formed.
Given a chemical reaction, from each reactant we can
calculate the amount of product possible when it is completely consumed.
determine the limiting reactant, the one that runs out first and produces the smaller amount of product.
Core Chemistry Skill
Calculating the Quantity of Product from a Limiting ReactantSlide4
Guide to Calculating Moles of Product from a Limiting ReactantSlide5
Limiting Reactant, Moles of Product
If 3.00 moles of CO and 5.00 moles of H
2 are the initial reactants, how many moles of methanol (CH3OH) can be produced and what is the limiting reactant? CO(
g
) + 2H
2
(
g
)
CH
3
OH(
g
)
STEP 1
State the given and needed moles.
ANALYZE Given Need
THE PROBLEM
3.00 moles of CO moles CH
3
OH produced
5.00 moles
of
H
2
limiting reactant
Equation
CO(
g
) + 2H
2
(
g
)
CH
3
OH(
g
)Slide6
Limiting Reactant, Moles of Product
STEP 2
Write a plan to convert the moles of each reactant to moles of product. moles of CO moles of CH
3
OH
moles of H
2
moles of
CH
3
OH
Mole–mole factor
Mole–mole factorSlide7
Limiting Reactant, Moles of Product
STEP 3
Write the mole–mole factors from the equation. Slide8
Limiting Reactant, Moles of Product
STEP 4
Calculate the number of moles of product from each reactant, and select the smaller number of moles as the amount of product from the limiting reactant.
H
2
is the limiting reactant,
and 2.50 moles of CH
3
OH is produced.
Smaller amount
of product
Limiting
reactantSlide9
Limiting Reactant, Mass of Product
The quantities of reactants can also be given in grams.
The calculations to identify the limiting reactant arefirst, convert the grams of each reactant to moles of reactant using molar mass conversion factors;second, use mole–mole factors to convert moles of reactant to moles of product; and third, use molar mass to convert the moles of product to grams of product.The reactant that produces the least amount of product is the limiting reactant. Slide10
Guide to Calculating the Grams of Product from a Limiting ReactantSlide11
Limiting Reactant, Mass of Product
Silicon carbide,
SiC, is a ceramic material that tolerates extreme temperatures and is used as an abrasive and in the brake discs
of sports cars.
How many grams of CO are
formed from 70.0 g of SiO
2
and
50.0 grams of C?
SiO
2
(
s
) + 3C(
s
) SiC
(
s
) + 2CO(
g
)
A ceramic brake disc in a sports
car withstands
temperatures
of 1400
°
C
.Slide12
Limiting Reactant, Mass of Product
STEP 1
State the given and needed grams.
ANALYZE Given Need
THE PROBLEM
70.0 g of SiO
2
grams of CO produced
50.0 g
of
C
from limiting reactant
Equation
SiO2(s) + 3C(s) SiC
(
s
) + 2CO(
g
)Slide13
Limiting Reactant, Mass of Product
STEP 2
Write a plan to convert the grams of each reactant to grams of product.grams moles moles gramsof SiO
2
of
SiO
2
of CO of CO
grams
moles moles
grams
of H
2
of H2
of CO of CO
Mole–mole factor
Mole–mole factor
Molar
mass
Molar
mass
Molar
mass
Molar
massSlide14
Limiting Reactant, Mass of Product
STEP 3
Write the molar mass and mole–mole factors from the equation. Molar mass factors Slide15
Limiting Reactant, Mass of Product
STEP 4
Calculate the number of grams of product from each reactant, and select the smaller number of grams as the amount of product from the limiting reactant.
C is the limiting reactant;
65.3 grams CO is produced.
Smaller
amount of
product
Limiting
reactant
Four SF
Exact
Four SF
Exact
Exact
Exact
Three SF
Three SF
Three SF
Four SF
Exact
Exact
Exact
Four SF
Three SF
×
×
×
×
×
×Slide16
Study Check
Given the following reaction,
N2(g) + 3H2
(
g
)
2NH
3
(
g
)
calculate the amount of ammonia,
NH
3
, that can be formed when 2.50 grams of nitrogen gas, N
2, reacts with 2.00 grams of hydrogen gas, H2.Slide17
Solution
STEP 1
State the given and needed grams.
ANALYZE Given Need
THE PROBLEM
2.50 g of N
2
grams of NH
3
produced
2.00 g
of
H
2
from limiting reactant
Equation N2(g) + 3H
2
(
g
)
2NH
3
(
g
)Slide18
Solution
STEP 2
Write a plan to convert the grams of each reactant to grams of product.grams moles moles gramsof N
2
of N
2
of NH
3
of NH
3
grams moles moles grams
of H
2
of H
2 of
NH
3
of
NH
3
Mole–mole factor
Mole–mole factor
Molar
mass
Molar
mass
Molar
mass
Molar
massSlide19
Solution
STEP 3
Write the molar mass and mole–mole factors from the equation. Molar mass factors Slide20
Solution
STEP 3
Write the molar mass and mole–mole factors from the equation. Mole–mole factors Slide21
Solution
STEP 4
Calculate the number of grams of product from each reactant, and select the smaller number of grams as the amount of product from the limiting reactant.
N
2
is the limiting reactant;
3.04 grams of NH
3
is produced.
Smaller amount of product
Limiting
reactant
Four SF
Four SF
Exact
Exact
Exact
Exact
Three SF
Three SF
Three SF
Exact
Exact
Exact
Four SF
Three SF
Exact
Three SF
×
×
×
×
×
×Slide22
Actual, Theoretical, and Percent Yield
When the reaction does not go to completion, or some of the reactant or product is lost, the amount of product produced may be less.
Theoretical yield is the maximum amount of product, which is calculated using the balanced equation.Actual yield is the amount of product actually obtained. Percent yield is the ratio of actual yield to theoretical yield.
Core Chemistry Skill
Calculating Percent YieldSlide23
Guide to Calculations for Percent
YieldSlide24
Study Check
On a space shuttle,
LiOH is used to absorb exhaled CO2 from breathing air to form LiHCO3. What is the percent yield of the reaction if 50.0 g of LiOH
gives 72.8 g of LiHCO
3
?
LiOH
(
s
) + CO
2
(
g
)
LiHCO
3(s)Slide25
Solution
On a space shuttle,
LiOH is used to absorb exhaled CO2 from breathing air to form LiHCO3. What is the percent yield of the reaction if 50.0 g of LiOH
gives 72.8 g of LiHCO
3
?
LiOH
(
s
) + Co
2
(
g
)
→
LiHCO3(s)STEP 1
State the given and needed quantities.
ANALYZE Given Need
THE PROBLEM
50.0 g of
LiOH
Percent yield LiHCO
3
72.8 g of LiHCO
3
Equation
LiOH
(s) + CO2(g) LiHCO3(s)Slide26
Solution
STEP 2
Write a plan to calculate theoretical yield and percent yield.grams
moles
moles grams of of of of
LiOH
LiOH
LiHCO
3
LiHCO
3
Mole–mole factor
Molar
mass
Molar
mass Slide27
Solution
STEP 3
Write the molar mass and mole–mole factors from the equation. Molar mass factors Slide28
Solution
STEP 3
Write the molar mass and mole–mole factors from the equation. Mole-mole factors Slide29
Solution
STEP 4
Calculate the percent yield by dividing the actual yield (given) by the theoretical yield and multiplying the result by 100%.
Calculation of percent yield:
Three SF
Four SF
Four SF
Exact
Exact
Exact
Exact
Three SF
Three SF
Three SF
×
×
×
×
×