93 93 Learning Targets Identify the limiting reactant in a chemical equation Identify the excess reactant and calculate the amount remaining after the reaction is complete Calculate the mass of a product when the amount of more than one reactant is given ID: 788230
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Slide1
Limiting Reactants and Percent Yield
9-3
Slide29-3 Learning Targets
Identify the limiting reactant in a chemical equation
Identify the excess reactant and calculate the amount remaining after the reaction is complete
Calculate the mass of a product when the amount of more than one reactant is given
Calculate the theoretical yield of a chemical reaction from data
Determine the percent yield for a chemical reaction
Slide3Limiting Reactants
Limiting reactant
(reagent)- the reactant that limits how much product is formed
When reactant is gone reaction stops
CH
4
+ H
2
O
→
3H
2
+ CO
249 g CH
4
takes 279 g of H
2
O to react
If you have 300g of H
2
O and 249 g of CH
4
, what is the limiting reactant?
CH
4
, it will run out first
Slide4Can not have 2 reactants be limiting
Excess reactant-
the substance that is not used up completely in a reaction
Often the least expensive reactant
Why use excess?
Reaction can be driven to continue until all of limiting reactant is used up
Speed up reaction
Increases efficiency of reaction
Slide5Stiochiometry
with Limiting Reactants
1-Balance equation
2- Mass to moles (have)
3- Determine limiting reactant (need)
4- Limiting reactant and mole ratio to get product
5- Moles to mass
Slide6Limiting reactant problem
N
2
+ H
2
→
NH
3
25kg 5 kg
How much ammonia is produced (grams) when the reaction is run to completion?
1- balance equation
N
2
+ 3H
2
→
2NH
3
Slide7N
2
+ 3H
2
→
2NH
3
2- mass to moles (have)
25000g N
2
X 1
mol
/ 28 g = 892
mol
N
2
5000 g H
2
x 1
mol
/ 2.016 g = 2480
mol
H
2
3- determine limiting reactant (need)
892
mol
N
2
x
3
mol
H
2
= 2680
mol
H
2
1
mol
N
2
2480
mol
H
2
x
1mol
N
2
= 827
mol
N
2
3
mol
H
2
H
2
is limiting reactant because don’t have enough
Slide8N
2
+ 3H
2
→
2NH
3
4- get product
2480
mol
H
2
x
2
mol
NH
3
= 1650
mol
NH
3
3
mol
H
2
5- moles to mass
1650
mol
NH
3
x
17.03 g
= 28100 g NH
3
1
mol
Slide9Limiting reactant problem
Li + N
2
→
Li
3
N
You have 56.0 g lithium and 56.0 g nitrogen, how much product will form?
1- balance equation
6Li + N
2
→ 2
Li
3
N
2- Have
56.0 g Li x
1
mol
= 8.07
mol
Li
6.94 g
56.0 g x
1
mol
N
2
= 2.00
mol
N
2
28 g
Slide106Li + N
2
→ 2
Li
3
N
3- Need
8.07
mol
Li x
1
mol
N
2
= 1.34
mol
N
2
6
mol
Li
2.00
mol
N
2
x
6
mol
Li
= 12.0
mol
Li
1
mol
N
2
Limiting reactant is Li
Slide116Li + N
2
→ 2
Li
3
N
4- get product
8.07
mol
Li x
2
mol
Li
3
N
= 2.69
mol
Li
3
N
6
mol
Li
5- moles to mass
2.69
mol
Li
3
N x
34.82 g
= 93.7 g Li
3
N
1
mol
Slide12Percent Yield
Theoretical Yield-
amount predicted to form from reaction
Maximum amount
Don’t always get because of side reactions, never have prefect conditions
Determined by limiting reactants
Solve for on paper
Slide13Actual yield-
amount of product obtained in lab reaction
Percent yield-
product of the ratio or the actual yield to the theoretical yield expressed as a percent
Percent yield =
Actual yield
x 100
Theoretical yield
Slide14Percent yield example
H
2
+ CO
→
CH
3
OH
8600g 68500 g
You get an actual yield of 3.57 x 10
4
g of CH
3
OH, what is the percent yield of CH
3
OH?
Need to find theoretical yield of limiting reactant.
1- 2H
2
+ CO
→
CH
3
OH
Slide152H
2
+ CO
→
CH
3
OH
2- Have
8600 g H
2
x
1
mol
= 4266
mol
H
2
2.016 g H
2
68500 g CO x
1
mol
= 2446
mol
CO
28.01 g CO
Slide162H
2
+ CO
→
CH
3
OH
3 –Need
4266
mol
H
2
x
1
mol
CO
= 2133
mol
CO
2
mol
H
2
2446
mol
CO x
2
mol
H
2
= 4892
mol
H
2
1
mol
CO
H
2
is limiting reactant
Slide172H
2
+ CO
→
CH
3
OH
4- moles to mass
4266
mol
H
2
x
1
mol
CH
3
OH
x
32.04 g
2
mol
H
2
1mol CH
3
OH
= 68341.3 g CH
3
OH
5- percent yield of
CH
3
OH
35700g
x 100% = 52.2 %
68341g
Slide18During a reaction 38.8 g of chorobenzene
is formed, the theoretical yield is found to be 53.0 g. What is the percent yield?
Percent yield =
Actual yield
x 100
Theoretical yield
(38.8 g/ 53.0 g) x 100 %= 73.2%
Slide192FePO
4
+ 3Na
2
SO
4
→
Fe
2
(SO
4
)
3
+ 2Na
3
PO
4
Determine the theoretical yield of iron (III) sulfate if 25.0 g of iron (III) phosphate is used. (if only one amount is given, it has to be the limiting reactant)
25.0 g FePO
4
x
1
mol
FePO
4
x
1 molFe
2
(SO
4
)
3
X
399.70gFe
2
(SO
4
)
3
150.80g FePO
4
2 molFePO
4
1 molFe
2
(SO
4
)
3
=
= 33.1 g Fe
2
(SO
4
)
3
Slide20If 18. 5 g of iron(III) sulfate is actually made, what is the percent yield?
Percent yield =
Actual yield
x 100
Theoretical yield
Percent yield =
18.5g
x 100
33.1g
= 55.9 %
Slide21Percent yield in the marketplace
Important in calculation or overall cost effectiveness in industrial process
Find most cost effective method to bring price of item down and profit up