Transcript Chapter 9 Notes Part III Limiting Reagents & Percent Yield What are limiting reagents? Up until now, we have assumed that all reactants are.

Chapter 9 Notes Part III
Limiting Reagents
& Percent Yield
What are limiting reagents?
Up until now, we have assumed
that all reactants are used up in
a reaction.
In actuality, usually one chemical
runs out first, stopping the
reaction—that chemical is
called the limiting reagent.
For Example:
How many sandwiches can you make from
a full jar of peanut butter, a full jar of jelly
and 4 slices of bread?
For Example:
How many sandwiches can you make from
a full jar of peanut butter, a full jar of jelly
and 4 slices of bread?
For Example:
How many sandwiches can you make from
a full jar of peanut butter, a full jar of jelly
and 4 slices of bread?
+
For Example:
How many sandwiches can you make from
a full jar of peanut butter, a full jar of jelly
and 4 slices of bread?
+
For Example:
How many sandwiches can you make from
a full jar of peanut butter, a full jar of jelly
and 4 slices of bread?
+
+
For Example:
How many sandwiches can you make from
a full jar of peanut butter, a full jar of jelly
and 4 slices of bread?
+
+
For Example:
How many sandwiches can you make from
a full jar of peanut butter, a full jar of jelly
and 4 slices of bread?
+
+
=
For Example:
How many sandwiches can you make from
a full jar of peanut butter, a full jar of jelly
and 4 slices of bread?
+
+
=
The limiting reagent limits or
determines the amount of product
that can be formed in a reaction.
On the flip side, the excess reagent
is the one not completely used up.
How do you find what the
limiting reagent is?
To find the limiting reagent, use
dimensional analysis to determine
how much of the product will be
formed by both reactants.
The smaller of the two results will
indicate which reactant is the
limiting reagent.
Here’s an example:
In the reaction:
Mg + HCl a MgCl2 + H2
you combine 35.5 grams of
magnesium with 28.1 grams of
hydrochloric acid. Which one is
the limiting reagent, and how
much magnesium chloride is
produced?
Mg + 2 HCl a MgCl2 + H2
Mg + 2 HCl a MgCl2 + H2
35.5 g Mg x 1 mole Mg x 1 mole MgCl2 x 95.3 g MgCl2 = 139 g MgCl2
24.3 g Mg
1 mole Mg
1 mole MgCl2
Mg + 2 HCl a MgCl2 + H2
35.5 g Mg x 1 mole Mg x 1 mole MgCl2 x 95.3 g MgCl2 = 139 g MgCl2
24.3 g Mg
1 mole Mg
1 mole MgCl2
28.1 g HCl x 1 mole HCl x 1 mole MgCl2 x 95.3 g MgCl2 = 36.6 g MgCl2
36.5 g HCl
2 moles HCl
1 mole MgCl2
Mg + 2 HCl a MgCl2 + H2
35.5 g Mg x 1 mole Mg x 1 mole MgCl2 x 95.3 g MgCl2 = 139 g MgCl2
24.3 g Mg
1 mole Mg
1 mole MgCl2
28.1 g HCl x 1 mole HCl x 1 mole MgCl2 x 95.3 g MgCl2 = 36.6 g MgCl2
36.5 g HCl
2 moles HCl
1 mole MgCl2
HCl is the limiting reagent and
36.6 g of MgCl2 are produced.
Percent Yield
In the real world, you never form
100% of the products that you think
you should based on theoretical
calculations.
By comparing how much of a
product forms to how much should
form based on stoichiometry, you
can find the percent yield.
Percent Yield
%Yield= actual yield
theoretical yield
x 100
Practice Problem
In the reaction of nitrogen gas
and hydrogen gas producing
ammonia (NH3), 250.2g H2
produces 1095.1g NH3. What
is the % yield of this reaction?
N2 + 3 H2 → 2 NH3
First find out how much NH3 should be produced:
N2 + 3 H2 → 2 NH3
First find out how much NH3 should be produced:
250.2 g H2 x 1 mole H2 x 2 moles NH3 x 17 g NH3 = 2836 g NH3
1 g H2
3 moles H2
1 mole NH3
N2 + 3 H2 → 2 NH3
First find out how much NH3 should be produced:
250.2 g H2 x 1 mole H2 x 2 moles NH3 x 17 g NH3 = 2836 g NH3
1 g H2
3 moles H2
1 mole NH3
Then calculate the percent yield:
N2 + 3 H2 → 2 NH3
First find out how much NH3 should be produced:
250.2 g H2 x 1 mole H2 x 2 moles NH3 x 17 g NH3 = 2836 g NH3
1 g H2
3 moles H2
1 mole NH3
Then calculate the percent yield:
% Yield = Actual x 100
Theor.
% Yield = 1095.1 g x 100
2836 g
% Yield = 38.6 %
N2 + 3 H2 → 2 NH3
First find out how much NH3 should be produced:
250.2 g H2 x 1 mole H2 x 2 moles NH3 x 17 g NH3 = 2836 g NH3
1 g H2
3 moles H2
1 mole NH3
Then calculate the percent yield:
% Yield = Actual x 100
Theor.
% Yield = 1095.1 g x 100
2836 g
% Yield = 38.6 %
Reaction #1
K2CrO4 + Pb(NO3)2 a KNO3 + PbCrO4
If 0.947g K2CrO4 are
combined with 0.331g
Pb(NO3)2 what will be the
limiting reagent, and how
much lead chromate is
produced?
K2CrO4 + Pb(NO3)2 a 2 KNO3 + PbCrO4
0.947g K2CrO4 x 1 mole K2CrO4 x 1 mole PbCrO4 x 323.2 g PbCrO4 = 1.97 g PbCrO4
155.1 g K2CrO4 1 mole K2CrO4
1 mole PbCrO4
0.331g Pb(NO3)2 x 1 mole Pb(NO3)2 x 1 mole PbCrO4 x 323.2 g PbCrO4 = 0.323 g PbCrO4
331.2 g Pb(NO3)2 1 mole Pb(NO3)2 1 mole PbCrO4
Pb(NO3)2 is the limiting reagent and
0.323 g PbCrO4 is produced.
Reaction #2
C3H8O + O2 a CO2 + H2O
If 3.78 L of O2 are combined with
7.9 g of C3H8O, what is the
limiting reagent, and how much
water is produced?
2 C3H8O + 9 O2 a 6 CO2 + 8 H2O
3.78 L of O2 x 1 mole O2 x 8 moles H2O x 18 g H2O = 2.7 g H2O
22.4 L O2
9 moles O2
1 mole H2O
7.9 g of C3H8O x 1 mole C3H8O x 8 moles H2O x 18 g H2O = 9.48 g H2O
60 g C3H8O 2 moles C3H8O 1 mole H2O
O2 is the limiting reagent and 2.7 g of H2O are produced.
Reaction #3
K2CrO4 + AgNO3 a KNO3 + Ag2CrO4
If 1.17g K2CrO4 are combined with
1.23g AgNO3, what is the limiting
reagent and how much Ag2CrO4 is
produced?
K2CrO4 + 2 AgNO3 a 2 KNO3 + Ag2CrO4
1.17g K2CrO4 x 1 mole K2CrO4 x 1 mole Ag2CrO4 x 331.8 g Ag2CrO4 = 1.99 g Ag2CrO4
194.2 g K2CrO4
1 mole K2CrO4
1 mole Ag2CrO4
1.23g AgNO3 x 1 mole AgNO3 x 1 mole Ag2CrO4 x 331.8 g Ag2CrO4 = 1.20 g Ag2CrO4
169.9 g AgNO3 2 moles AgNO3
1 mole Ag2CrO4
AgNO3 is the limiting reagent and
1.20 g Ag2CrO4 are produced.
Reaction #4
Zn + HCl a ZnCl2 + H2
What is the limiting reagent and
how many liters of hydrogen
are produced if 15.5 grams of
zinc are combined with 26.3
grams of hydrochloric acid?
Zn + 2 HCl a ZnCl2 + H2
15.5 g Zn x 1 mole Zn x 1 mole H2 x 22.4 L H2 = 5.31 L H2
65.4 g Zn
1 mole Zn 1 mole H2
26.3 g HCl x 1 mole HCl x 1 mole H2 x 22.4 L H2 = 8.07 L H2
36.5 g HCl 2 moles HCl 1 mole H2
Zn is the limiting reagent and 5.31 Liters of H2 are produced.
Reaction #5
Na + H2O a NaOH + H2
If 2.1g of sodium is added to 3.8g
of water, what is the limiting
reagent, and how many grams of
sodium hydroxide are produced?
2 Na + 2 H2O a 2 NaOH + H2
2.1 g Na x 1 mole Na x 2 moles NaOH x 40 g NaOH = 3.7 g NaOH
23 g Na
2 moles Na
1 mole NaOH
3.8 g H2O x 1 mole H2O x 2 moles NaOH x 40 g NaOH = 8.4 g NaOH
18 g H2O
2 moles H2O
1 mole NaOH
Na is the limiting reagent and 3.7 g NaOH are produced.
In Review:
What is a limiting reagent?
What is the excess reagent?
How does a limiting reagent control
the amount of product?