Transcript Limiting Reactant/Reagent Problems - Dr. Vernon-

Slide 1

Limiting Reactant/Reagent Problems
• Consider the balanced equation for the
production of ammonia:
N2 + 3 H2  2NH3
• If you have 2 moles of nitrogen and 3 moles of
hydrogen, which reactant will be used up?
• How much ammonia will you be able to
make?
• How much extra (and of which chemical) will
you have?

Limiting Reactant/Reagent Problems
• Spend 2 minutes discussing your thought
process for the previous question with your
neighbor.
• How did you decide which reactant would be
used up?
• Write some simple math to explain how you
figured out how much of the excess you would
have left over. (label what your numbers
mean)

Limiting Reactant/Reagent Problems
• A stoichiometry problem where quantities of
reactants are not available in the EXACT molar
ratio.
• This means one reactant is “in excess” and the
other is the “limiting reactant.”
• You will be given the mass/volume/moles of
more than one reactant.
• You find the actual theoretical amount of
product(s) produced by determining which
reactant LIMITS the reaction.

Steps for Identifying Limiting
Reactants:
1. Write the balanced chemical equation.
2. Solve two (or more) stoichiometry problems
– one for each reactant given.
3. Compare the results of step #2 above. The
reactant that would produce the smaller
amount of product is the limiting reactant;
the other reactant is in excess.

OR
1. Write the balanced chemical equation.
2. Find the number of moles of each reactant
given. (convert from grams, volume, or particles
if necessary)
3. Compare the ratio of moles from step #2 above
to the ratio of moles in the balanced equation
from step #1.
4. Use the limiting reactant to do one full
stoichiometry problem to determine the
theoretical yield from this reaction.

Example Problem:
• What is the limiting reactant when 1.5 moles
of hydrogen react with 0.5 moles of oxygen to
produce water?
2H2 + O2  2H2O
• 1.5 mol H2 x 1 mol O2 = 0.75 mol O2
2 mol H2
• This means, to completely react with 1.5 mol
of hydrogen, you would need 0.75 mol of
oxygen.

You don’t have that much oxygen!
• Which means oxygen is the limiting reagent.
• You will use up all of your oxygen and have
hydrogen left over.
• How much water will you be able to make?
• 0.5 mol O2 x 2 mol H2O = 1.0 mol H2O
1 mol O2
You could convert this to grams, particles, etc if
necessary.

Example Problem
• When 2.60 g of hydrogen react with 10.20 g of
oxygen, what mass of water will be produced?
What is the limiting reactant?
• First, find the moles of each reactant.
• 2.60 g H2 x 1 mol H2 = 1.29 mol H2
2.016 g H2
• 10.20 g O2 x 1 mol O2 = 0.3188 mol O2
32.00 g O2

Next, relate them to one another using
the mole ratio
• 1.29 mol H2 x 1 mol O2 = 0.645 mol O2

2 mol H2
• This is how much oxygen you would need to
exactly react with the hydrogen.
• Do you have that much oxygen?
• NO!
• Oxygen is again the limiting reactant.

Now, figure out how much water you
can make.
• Begin with oxygen, because we know we will
use all of that up.
10.20 g O2 x 1 mol O2 x 2 mol H2O x 18.016 g H2O = 11.49 g H2O
32.00 g O2
1 mol O2
1 mol H2O

Slide 2

Limiting Reactant/Reagent Problems
• Consider the balanced equation for the
production of ammonia:
N2 + 3 H2  2NH3
• If you have 2 moles of nitrogen and 3 moles of
hydrogen, which reactant will be used up?
• How much ammonia will you be able to
make?
• How much extra (and of which chemical) will
you have?

Limiting Reactant/Reagent Problems
• Spend 2 minutes discussing your thought
process for the previous question with your
neighbor.
• How did you decide which reactant would be
used up?
• Write some simple math to explain how you
figured out how much of the excess you would
have left over. (label what your numbers
mean)

Limiting Reactant/Reagent Problems
• A stoichiometry problem where quantities of
reactants are not available in the EXACT molar
ratio.
• This means one reactant is “in excess” and the
other is the “limiting reactant.”
• You will be given the mass/volume/moles of
more than one reactant.
• You find the actual theoretical amount of
product(s) produced by determining which
reactant LIMITS the reaction.

Steps for Identifying Limiting
Reactants:
1. Write the balanced chemical equation.
2. Solve two (or more) stoichiometry problems
– one for each reactant given.
3. Compare the results of step #2 above. The
reactant that would produce the smaller
amount of product is the limiting reactant;
the other reactant is in excess.

OR
1. Write the balanced chemical equation.
2. Find the number of moles of each reactant
given. (convert from grams, volume, or particles
if necessary)
3. Compare the ratio of moles from step #2 above
to the ratio of moles in the balanced equation
from step #1.
4. Use the limiting reactant to do one full
stoichiometry problem to determine the
theoretical yield from this reaction.

Example Problem:
• What is the limiting reactant when 1.5 moles
of hydrogen react with 0.5 moles of oxygen to
produce water?
2H2 + O2  2H2O
• 1.5 mol H2 x 1 mol O2 = 0.75 mol O2
2 mol H2
• This means, to completely react with 1.5 mol
of hydrogen, you would need 0.75 mol of
oxygen.

You don’t have that much oxygen!
• Which means oxygen is the limiting reagent.
• You will use up all of your oxygen and have
hydrogen left over.
• How much water will you be able to make?
• 0.5 mol O2 x 2 mol H2O = 1.0 mol H2O
1 mol O2
You could convert this to grams, particles, etc if
necessary.

Example Problem
• When 2.60 g of hydrogen react with 10.20 g of
oxygen, what mass of water will be produced?
What is the limiting reactant?
• First, find the moles of each reactant.
• 2.60 g H2 x 1 mol H2 = 1.29 mol H2
2.016 g H2
• 10.20 g O2 x 1 mol O2 = 0.3188 mol O2
32.00 g O2

Next, relate them to one another using
the mole ratio
• 1.29 mol H2 x 1 mol O2 = 0.645 mol O2

2 mol H2
• This is how much oxygen you would need to
exactly react with the hydrogen.
• Do you have that much oxygen?
• NO!
• Oxygen is again the limiting reactant.

Now, figure out how much water you
can make.
• Begin with oxygen, because we know we will
use all of that up.
10.20 g O2 x 1 mol O2 x 2 mol H2O x 18.016 g H2O = 11.49 g H2O
32.00 g O2
1 mol O2
1 mol H2O

Slide 3

Limiting Reactant/Reagent Problems
• Consider the balanced equation for the
production of ammonia:
N2 + 3 H2  2NH3
• If you have 2 moles of nitrogen and 3 moles of
hydrogen, which reactant will be used up?
• How much ammonia will you be able to
make?
• How much extra (and of which chemical) will
you have?

Limiting Reactant/Reagent Problems
• Spend 2 minutes discussing your thought
process for the previous question with your
neighbor.
• How did you decide which reactant would be
used up?
• Write some simple math to explain how you
figured out how much of the excess you would
have left over. (label what your numbers
mean)

Limiting Reactant/Reagent Problems
• A stoichiometry problem where quantities of
reactants are not available in the EXACT molar
ratio.
• This means one reactant is “in excess” and the
other is the “limiting reactant.”
• You will be given the mass/volume/moles of
more than one reactant.
• You find the actual theoretical amount of
product(s) produced by determining which
reactant LIMITS the reaction.

Steps for Identifying Limiting
Reactants:
1. Write the balanced chemical equation.
2. Solve two (or more) stoichiometry problems
– one for each reactant given.
3. Compare the results of step #2 above. The
reactant that would produce the smaller
amount of product is the limiting reactant;
the other reactant is in excess.

OR
1. Write the balanced chemical equation.
2. Find the number of moles of each reactant
given. (convert from grams, volume, or particles
if necessary)
3. Compare the ratio of moles from step #2 above
to the ratio of moles in the balanced equation
from step #1.
4. Use the limiting reactant to do one full
stoichiometry problem to determine the
theoretical yield from this reaction.

Example Problem:
• What is the limiting reactant when 1.5 moles
of hydrogen react with 0.5 moles of oxygen to
produce water?
2H2 + O2  2H2O
• 1.5 mol H2 x 1 mol O2 = 0.75 mol O2
2 mol H2
• This means, to completely react with 1.5 mol
of hydrogen, you would need 0.75 mol of
oxygen.

You don’t have that much oxygen!
• Which means oxygen is the limiting reagent.
• You will use up all of your oxygen and have
hydrogen left over.
• How much water will you be able to make?
• 0.5 mol O2 x 2 mol H2O = 1.0 mol H2O
1 mol O2
You could convert this to grams, particles, etc if
necessary.

Example Problem
• When 2.60 g of hydrogen react with 10.20 g of
oxygen, what mass of water will be produced?
What is the limiting reactant?
• First, find the moles of each reactant.
• 2.60 g H2 x 1 mol H2 = 1.29 mol H2
2.016 g H2
• 10.20 g O2 x 1 mol O2 = 0.3188 mol O2
32.00 g O2

Next, relate them to one another using
the mole ratio
• 1.29 mol H2 x 1 mol O2 = 0.645 mol O2

2 mol H2
• This is how much oxygen you would need to
exactly react with the hydrogen.
• Do you have that much oxygen?
• NO!
• Oxygen is again the limiting reactant.

Now, figure out how much water you
can make.
• Begin with oxygen, because we know we will
use all of that up.
10.20 g O2 x 1 mol O2 x 2 mol H2O x 18.016 g H2O = 11.49 g H2O
32.00 g O2
1 mol O2
1 mol H2O

Slide 4

Limiting Reactant/Reagent Problems
• Consider the balanced equation for the
production of ammonia:
N2 + 3 H2  2NH3
• If you have 2 moles of nitrogen and 3 moles of
hydrogen, which reactant will be used up?
• How much ammonia will you be able to
make?
• How much extra (and of which chemical) will
you have?

Limiting Reactant/Reagent Problems
• Spend 2 minutes discussing your thought
process for the previous question with your
neighbor.
• How did you decide which reactant would be
used up?
• Write some simple math to explain how you
figured out how much of the excess you would
have left over. (label what your numbers
mean)

Limiting Reactant/Reagent Problems
• A stoichiometry problem where quantities of
reactants are not available in the EXACT molar
ratio.
• This means one reactant is “in excess” and the
other is the “limiting reactant.”
• You will be given the mass/volume/moles of
more than one reactant.
• You find the actual theoretical amount of
product(s) produced by determining which
reactant LIMITS the reaction.

Steps for Identifying Limiting
Reactants:
1. Write the balanced chemical equation.
2. Solve two (or more) stoichiometry problems
– one for each reactant given.
3. Compare the results of step #2 above. The
reactant that would produce the smaller
amount of product is the limiting reactant;
the other reactant is in excess.

OR
1. Write the balanced chemical equation.
2. Find the number of moles of each reactant
given. (convert from grams, volume, or particles
if necessary)
3. Compare the ratio of moles from step #2 above
to the ratio of moles in the balanced equation
from step #1.
4. Use the limiting reactant to do one full
stoichiometry problem to determine the
theoretical yield from this reaction.

Example Problem:
• What is the limiting reactant when 1.5 moles
of hydrogen react with 0.5 moles of oxygen to
produce water?
2H2 + O2  2H2O
• 1.5 mol H2 x 1 mol O2 = 0.75 mol O2
2 mol H2
• This means, to completely react with 1.5 mol
of hydrogen, you would need 0.75 mol of
oxygen.

You don’t have that much oxygen!
• Which means oxygen is the limiting reagent.
• You will use up all of your oxygen and have
hydrogen left over.
• How much water will you be able to make?
• 0.5 mol O2 x 2 mol H2O = 1.0 mol H2O
1 mol O2
You could convert this to grams, particles, etc if
necessary.

Example Problem
• When 2.60 g of hydrogen react with 10.20 g of
oxygen, what mass of water will be produced?
What is the limiting reactant?
• First, find the moles of each reactant.
• 2.60 g H2 x 1 mol H2 = 1.29 mol H2
2.016 g H2
• 10.20 g O2 x 1 mol O2 = 0.3188 mol O2
32.00 g O2

Next, relate them to one another using
the mole ratio
• 1.29 mol H2 x 1 mol O2 = 0.645 mol O2

2 mol H2
• This is how much oxygen you would need to
exactly react with the hydrogen.
• Do you have that much oxygen?
• NO!
• Oxygen is again the limiting reactant.

Now, figure out how much water you
can make.
• Begin with oxygen, because we know we will
use all of that up.
10.20 g O2 x 1 mol O2 x 2 mol H2O x 18.016 g H2O = 11.49 g H2O
32.00 g O2
1 mol O2
1 mol H2O

Slide 5

Limiting Reactant/Reagent Problems
• Consider the balanced equation for the
production of ammonia:
N2 + 3 H2  2NH3
• If you have 2 moles of nitrogen and 3 moles of
hydrogen, which reactant will be used up?
• How much ammonia will you be able to
make?
• How much extra (and of which chemical) will
you have?

Limiting Reactant/Reagent Problems
• Spend 2 minutes discussing your thought
process for the previous question with your
neighbor.
• How did you decide which reactant would be
used up?
• Write some simple math to explain how you
figured out how much of the excess you would
have left over. (label what your numbers
mean)

Limiting Reactant/Reagent Problems
• A stoichiometry problem where quantities of
reactants are not available in the EXACT molar
ratio.
• This means one reactant is “in excess” and the
other is the “limiting reactant.”
• You will be given the mass/volume/moles of
more than one reactant.
• You find the actual theoretical amount of
product(s) produced by determining which
reactant LIMITS the reaction.

Steps for Identifying Limiting
Reactants:
1. Write the balanced chemical equation.
2. Solve two (or more) stoichiometry problems
– one for each reactant given.
3. Compare the results of step #2 above. The
reactant that would produce the smaller
amount of product is the limiting reactant;
the other reactant is in excess.

OR
1. Write the balanced chemical equation.
2. Find the number of moles of each reactant
given. (convert from grams, volume, or particles
if necessary)
3. Compare the ratio of moles from step #2 above
to the ratio of moles in the balanced equation
from step #1.
4. Use the limiting reactant to do one full
stoichiometry problem to determine the
theoretical yield from this reaction.

Example Problem:
• What is the limiting reactant when 1.5 moles
of hydrogen react with 0.5 moles of oxygen to
produce water?
2H2 + O2  2H2O
• 1.5 mol H2 x 1 mol O2 = 0.75 mol O2
2 mol H2
• This means, to completely react with 1.5 mol
of hydrogen, you would need 0.75 mol of
oxygen.

You don’t have that much oxygen!
• Which means oxygen is the limiting reagent.
• You will use up all of your oxygen and have
hydrogen left over.
• How much water will you be able to make?
• 0.5 mol O2 x 2 mol H2O = 1.0 mol H2O
1 mol O2
You could convert this to grams, particles, etc if
necessary.

Example Problem
• When 2.60 g of hydrogen react with 10.20 g of
oxygen, what mass of water will be produced?
What is the limiting reactant?
• First, find the moles of each reactant.
• 2.60 g H2 x 1 mol H2 = 1.29 mol H2
2.016 g H2
• 10.20 g O2 x 1 mol O2 = 0.3188 mol O2
32.00 g O2

Next, relate them to one another using
the mole ratio
• 1.29 mol H2 x 1 mol O2 = 0.645 mol O2

2 mol H2
• This is how much oxygen you would need to
exactly react with the hydrogen.
• Do you have that much oxygen?
• NO!
• Oxygen is again the limiting reactant.

Now, figure out how much water you
can make.
• Begin with oxygen, because we know we will
use all of that up.
10.20 g O2 x 1 mol O2 x 2 mol H2O x 18.016 g H2O = 11.49 g H2O
32.00 g O2
1 mol O2
1 mol H2O

Slide 6

Limiting Reactant/Reagent Problems
• Consider the balanced equation for the
production of ammonia:
N2 + 3 H2  2NH3
• If you have 2 moles of nitrogen and 3 moles of
hydrogen, which reactant will be used up?
• How much ammonia will you be able to
make?
• How much extra (and of which chemical) will
you have?

Limiting Reactant/Reagent Problems
• Spend 2 minutes discussing your thought
process for the previous question with your
neighbor.
• How did you decide which reactant would be
used up?
• Write some simple math to explain how you
figured out how much of the excess you would
have left over. (label what your numbers
mean)

Limiting Reactant/Reagent Problems
• A stoichiometry problem where quantities of
reactants are not available in the EXACT molar
ratio.
• This means one reactant is “in excess” and the
other is the “limiting reactant.”
• You will be given the mass/volume/moles of
more than one reactant.
• You find the actual theoretical amount of
product(s) produced by determining which
reactant LIMITS the reaction.

Steps for Identifying Limiting
Reactants:
1. Write the balanced chemical equation.
2. Solve two (or more) stoichiometry problems
– one for each reactant given.
3. Compare the results of step #2 above. The
reactant that would produce the smaller
amount of product is the limiting reactant;
the other reactant is in excess.

OR
1. Write the balanced chemical equation.
2. Find the number of moles of each reactant
given. (convert from grams, volume, or particles
if necessary)
3. Compare the ratio of moles from step #2 above
to the ratio of moles in the balanced equation
from step #1.
4. Use the limiting reactant to do one full
stoichiometry problem to determine the
theoretical yield from this reaction.

Example Problem:
• What is the limiting reactant when 1.5 moles
of hydrogen react with 0.5 moles of oxygen to
produce water?
2H2 + O2  2H2O
• 1.5 mol H2 x 1 mol O2 = 0.75 mol O2
2 mol H2
• This means, to completely react with 1.5 mol
of hydrogen, you would need 0.75 mol of
oxygen.

You don’t have that much oxygen!
• Which means oxygen is the limiting reagent.
• You will use up all of your oxygen and have
hydrogen left over.
• How much water will you be able to make?
• 0.5 mol O2 x 2 mol H2O = 1.0 mol H2O
1 mol O2
You could convert this to grams, particles, etc if
necessary.

Example Problem
• When 2.60 g of hydrogen react with 10.20 g of
oxygen, what mass of water will be produced?
What is the limiting reactant?
• First, find the moles of each reactant.
• 2.60 g H2 x 1 mol H2 = 1.29 mol H2
2.016 g H2
• 10.20 g O2 x 1 mol O2 = 0.3188 mol O2
32.00 g O2

Next, relate them to one another using
the mole ratio
• 1.29 mol H2 x 1 mol O2 = 0.645 mol O2

2 mol H2
• This is how much oxygen you would need to
exactly react with the hydrogen.
• Do you have that much oxygen?
• NO!
• Oxygen is again the limiting reactant.

Now, figure out how much water you
can make.
• Begin with oxygen, because we know we will
use all of that up.
10.20 g O2 x 1 mol O2 x 2 mol H2O x 18.016 g H2O = 11.49 g H2O
32.00 g O2
1 mol O2
1 mol H2O

Slide 7

Limiting Reactant/Reagent Problems
• Consider the balanced equation for the
production of ammonia:
N2 + 3 H2  2NH3
• If you have 2 moles of nitrogen and 3 moles of
hydrogen, which reactant will be used up?
• How much ammonia will you be able to
make?
• How much extra (and of which chemical) will
you have?

Limiting Reactant/Reagent Problems
• Spend 2 minutes discussing your thought
process for the previous question with your
neighbor.
• How did you decide which reactant would be
used up?
• Write some simple math to explain how you
figured out how much of the excess you would
have left over. (label what your numbers
mean)

Limiting Reactant/Reagent Problems
• A stoichiometry problem where quantities of
reactants are not available in the EXACT molar
ratio.
• This means one reactant is “in excess” and the
other is the “limiting reactant.”
• You will be given the mass/volume/moles of
more than one reactant.
• You find the actual theoretical amount of
product(s) produced by determining which
reactant LIMITS the reaction.

Steps for Identifying Limiting
Reactants:
1. Write the balanced chemical equation.
2. Solve two (or more) stoichiometry problems
– one for each reactant given.
3. Compare the results of step #2 above. The
reactant that would produce the smaller
amount of product is the limiting reactant;
the other reactant is in excess.

OR
1. Write the balanced chemical equation.
2. Find the number of moles of each reactant
given. (convert from grams, volume, or particles
if necessary)
3. Compare the ratio of moles from step #2 above
to the ratio of moles in the balanced equation
from step #1.
4. Use the limiting reactant to do one full
stoichiometry problem to determine the
theoretical yield from this reaction.

Example Problem:
• What is the limiting reactant when 1.5 moles
of hydrogen react with 0.5 moles of oxygen to
produce water?
2H2 + O2  2H2O
• 1.5 mol H2 x 1 mol O2 = 0.75 mol O2
2 mol H2
• This means, to completely react with 1.5 mol
of hydrogen, you would need 0.75 mol of
oxygen.

You don’t have that much oxygen!
• Which means oxygen is the limiting reagent.
• You will use up all of your oxygen and have
hydrogen left over.
• How much water will you be able to make?
• 0.5 mol O2 x 2 mol H2O = 1.0 mol H2O
1 mol O2
You could convert this to grams, particles, etc if
necessary.

Example Problem
• When 2.60 g of hydrogen react with 10.20 g of
oxygen, what mass of water will be produced?
What is the limiting reactant?
• First, find the moles of each reactant.
• 2.60 g H2 x 1 mol H2 = 1.29 mol H2
2.016 g H2
• 10.20 g O2 x 1 mol O2 = 0.3188 mol O2
32.00 g O2

Next, relate them to one another using
the mole ratio
• 1.29 mol H2 x 1 mol O2 = 0.645 mol O2

2 mol H2
• This is how much oxygen you would need to
exactly react with the hydrogen.
• Do you have that much oxygen?
• NO!
• Oxygen is again the limiting reactant.

Now, figure out how much water you
can make.
• Begin with oxygen, because we know we will
use all of that up.
10.20 g O2 x 1 mol O2 x 2 mol H2O x 18.016 g H2O = 11.49 g H2O
32.00 g O2
1 mol O2
1 mol H2O

Slide 8

Limiting Reactant/Reagent Problems
• Consider the balanced equation for the
production of ammonia:
N2 + 3 H2  2NH3
• If you have 2 moles of nitrogen and 3 moles of
hydrogen, which reactant will be used up?
• How much ammonia will you be able to
make?
• How much extra (and of which chemical) will
you have?

Limiting Reactant/Reagent Problems
• Spend 2 minutes discussing your thought
process for the previous question with your
neighbor.
• How did you decide which reactant would be
used up?
• Write some simple math to explain how you
figured out how much of the excess you would
have left over. (label what your numbers
mean)

Limiting Reactant/Reagent Problems
• A stoichiometry problem where quantities of
reactants are not available in the EXACT molar
ratio.
• This means one reactant is “in excess” and the
other is the “limiting reactant.”
• You will be given the mass/volume/moles of
more than one reactant.
• You find the actual theoretical amount of
product(s) produced by determining which
reactant LIMITS the reaction.

Steps for Identifying Limiting
Reactants:
1. Write the balanced chemical equation.
2. Solve two (or more) stoichiometry problems
– one for each reactant given.
3. Compare the results of step #2 above. The
reactant that would produce the smaller
amount of product is the limiting reactant;
the other reactant is in excess.

OR
1. Write the balanced chemical equation.
2. Find the number of moles of each reactant
given. (convert from grams, volume, or particles
if necessary)
3. Compare the ratio of moles from step #2 above
to the ratio of moles in the balanced equation
from step #1.
4. Use the limiting reactant to do one full
stoichiometry problem to determine the
theoretical yield from this reaction.

Example Problem:
• What is the limiting reactant when 1.5 moles
of hydrogen react with 0.5 moles of oxygen to
produce water?
2H2 + O2  2H2O
• 1.5 mol H2 x 1 mol O2 = 0.75 mol O2
2 mol H2
• This means, to completely react with 1.5 mol
of hydrogen, you would need 0.75 mol of
oxygen.

You don’t have that much oxygen!
• Which means oxygen is the limiting reagent.
• You will use up all of your oxygen and have
hydrogen left over.
• How much water will you be able to make?
• 0.5 mol O2 x 2 mol H2O = 1.0 mol H2O
1 mol O2
You could convert this to grams, particles, etc if
necessary.

Example Problem
• When 2.60 g of hydrogen react with 10.20 g of
oxygen, what mass of water will be produced?
What is the limiting reactant?
• First, find the moles of each reactant.
• 2.60 g H2 x 1 mol H2 = 1.29 mol H2
2.016 g H2
• 10.20 g O2 x 1 mol O2 = 0.3188 mol O2
32.00 g O2

Next, relate them to one another using
the mole ratio
• 1.29 mol H2 x 1 mol O2 = 0.645 mol O2

2 mol H2
• This is how much oxygen you would need to
exactly react with the hydrogen.
• Do you have that much oxygen?
• NO!
• Oxygen is again the limiting reactant.

Now, figure out how much water you
can make.
• Begin with oxygen, because we know we will
use all of that up.
10.20 g O2 x 1 mol O2 x 2 mol H2O x 18.016 g H2O = 11.49 g H2O
32.00 g O2
1 mol O2
1 mol H2O

Slide 9

Limiting Reactant/Reagent Problems
• Consider the balanced equation for the
production of ammonia:
N2 + 3 H2  2NH3
• If you have 2 moles of nitrogen and 3 moles of
hydrogen, which reactant will be used up?
• How much ammonia will you be able to
make?
• How much extra (and of which chemical) will
you have?

Limiting Reactant/Reagent Problems
• Spend 2 minutes discussing your thought
process for the previous question with your
neighbor.
• How did you decide which reactant would be
used up?
• Write some simple math to explain how you
figured out how much of the excess you would
have left over. (label what your numbers
mean)

Limiting Reactant/Reagent Problems
• A stoichiometry problem where quantities of
reactants are not available in the EXACT molar
ratio.
• This means one reactant is “in excess” and the
other is the “limiting reactant.”
• You will be given the mass/volume/moles of
more than one reactant.
• You find the actual theoretical amount of
product(s) produced by determining which
reactant LIMITS the reaction.

Steps for Identifying Limiting
Reactants:
1. Write the balanced chemical equation.
2. Solve two (or more) stoichiometry problems
– one for each reactant given.
3. Compare the results of step #2 above. The
reactant that would produce the smaller
amount of product is the limiting reactant;
the other reactant is in excess.

OR
1. Write the balanced chemical equation.
2. Find the number of moles of each reactant
given. (convert from grams, volume, or particles
if necessary)
3. Compare the ratio of moles from step #2 above
to the ratio of moles in the balanced equation
from step #1.
4. Use the limiting reactant to do one full
stoichiometry problem to determine the
theoretical yield from this reaction.

Example Problem:
• What is the limiting reactant when 1.5 moles
of hydrogen react with 0.5 moles of oxygen to
produce water?
2H2 + O2  2H2O
• 1.5 mol H2 x 1 mol O2 = 0.75 mol O2
2 mol H2
• This means, to completely react with 1.5 mol
of hydrogen, you would need 0.75 mol of
oxygen.

You don’t have that much oxygen!
• Which means oxygen is the limiting reagent.
• You will use up all of your oxygen and have
hydrogen left over.
• How much water will you be able to make?
• 0.5 mol O2 x 2 mol H2O = 1.0 mol H2O
1 mol O2
You could convert this to grams, particles, etc if
necessary.

Example Problem
• When 2.60 g of hydrogen react with 10.20 g of
oxygen, what mass of water will be produced?
What is the limiting reactant?
• First, find the moles of each reactant.
• 2.60 g H2 x 1 mol H2 = 1.29 mol H2
2.016 g H2
• 10.20 g O2 x 1 mol O2 = 0.3188 mol O2
32.00 g O2

Next, relate them to one another using
the mole ratio
• 1.29 mol H2 x 1 mol O2 = 0.645 mol O2

2 mol H2
• This is how much oxygen you would need to
exactly react with the hydrogen.
• Do you have that much oxygen?
• NO!
• Oxygen is again the limiting reactant.

Now, figure out how much water you
can make.
• Begin with oxygen, because we know we will
use all of that up.
10.20 g O2 x 1 mol O2 x 2 mol H2O x 18.016 g H2O = 11.49 g H2O
32.00 g O2
1 mol O2
1 mol H2O

Slide 10

Limiting Reactant/Reagent Problems
• Consider the balanced equation for the
production of ammonia:
N2 + 3 H2  2NH3
• If you have 2 moles of nitrogen and 3 moles of
hydrogen, which reactant will be used up?
• How much ammonia will you be able to
make?
• How much extra (and of which chemical) will
you have?

Limiting Reactant/Reagent Problems
• Spend 2 minutes discussing your thought
process for the previous question with your
neighbor.
• How did you decide which reactant would be
used up?
• Write some simple math to explain how you
figured out how much of the excess you would
have left over. (label what your numbers
mean)

Limiting Reactant/Reagent Problems
• A stoichiometry problem where quantities of
reactants are not available in the EXACT molar
ratio.
• This means one reactant is “in excess” and the
other is the “limiting reactant.”
• You will be given the mass/volume/moles of
more than one reactant.
• You find the actual theoretical amount of
product(s) produced by determining which
reactant LIMITS the reaction.

Steps for Identifying Limiting
Reactants:
1. Write the balanced chemical equation.
2. Solve two (or more) stoichiometry problems
– one for each reactant given.
3. Compare the results of step #2 above. The
reactant that would produce the smaller
amount of product is the limiting reactant;
the other reactant is in excess.

OR
1. Write the balanced chemical equation.
2. Find the number of moles of each reactant
given. (convert from grams, volume, or particles
if necessary)
3. Compare the ratio of moles from step #2 above
to the ratio of moles in the balanced equation
from step #1.
4. Use the limiting reactant to do one full
stoichiometry problem to determine the
theoretical yield from this reaction.

Example Problem:
• What is the limiting reactant when 1.5 moles
of hydrogen react with 0.5 moles of oxygen to
produce water?
2H2 + O2  2H2O
• 1.5 mol H2 x 1 mol O2 = 0.75 mol O2
2 mol H2
• This means, to completely react with 1.5 mol
of hydrogen, you would need 0.75 mol of
oxygen.

You don’t have that much oxygen!
• Which means oxygen is the limiting reagent.
• You will use up all of your oxygen and have
hydrogen left over.
• How much water will you be able to make?
• 0.5 mol O2 x 2 mol H2O = 1.0 mol H2O
1 mol O2
You could convert this to grams, particles, etc if
necessary.

Example Problem
• When 2.60 g of hydrogen react with 10.20 g of
oxygen, what mass of water will be produced?
What is the limiting reactant?
• First, find the moles of each reactant.
• 2.60 g H2 x 1 mol H2 = 1.29 mol H2
2.016 g H2
• 10.20 g O2 x 1 mol O2 = 0.3188 mol O2
32.00 g O2

Next, relate them to one another using
the mole ratio
• 1.29 mol H2 x 1 mol O2 = 0.645 mol O2

2 mol H2
• This is how much oxygen you would need to
exactly react with the hydrogen.
• Do you have that much oxygen?
• NO!
• Oxygen is again the limiting reactant.

Now, figure out how much water you
can make.
• Begin with oxygen, because we know we will
use all of that up.
10.20 g O2 x 1 mol O2 x 2 mol H2O x 18.016 g H2O = 11.49 g H2O
32.00 g O2
1 mol O2
1 mol H2O