Limiting Reagents

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Limiting Reagents
Note it would be greatly to your benefit to
have gone through my Stoichiometry tutorial
before attempting this one (bfrench.ppt)
Easy
Not so easy

• Bobby French
• pd4

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• You need help with limiting reagents, or you
  wouldnt be here. Try this tutorial it might
  help! Ok, the problem is
• Ca HCl CaCl2 H2
• Your job find the limiting reagent and how much
  calcium chloride will be made from 80.2g Ca and
  73.0g HCl
• Whats the first step?

Start with 80.2g Ca and go from there
Balance the equation
Start with 73.0g HCl and go from there
3
Correct! Ok, the problem is Ca 2HCl
CaCl2 H2 Your job find the limiting reagent
and how much calcium chloride will be made from
80.2g Ca and 73.0g HCl Whats the next step?
Go from 80.2g Ca to g of HCl and see if that
number Is larger
Find out how much product will be made
from 80.2g Ca or from 73.0g HCl
Convert both to moles of H2
4
Correct! Ok, the problem is Ca 2HCl
CaCl2 H2 Your job find the limiting reagent
and how much calcium chloride will be made from
80.2g Ca and 73.0g HCl How many moles of CaCl2
will 80.2g Ca make?
2.00
1.53
1.0
5
Correct! Ok, the problem is Ca 2HCl
CaCl2 H2 Your job find the limiting reagent
and how much calcium chloride will be made from
80.2g Ca and 73.0g HCl How many moles of CaCl2
will 73.0g HCl make?
2.0
3.00
1.00
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Correct! Ok, the problem is Ca 2HCl
CaCl2 H2 Your job find the limiting reagent
and how much calcium chloride will be made from
80.2g Ca and 73.0g HCl So now that we know that,
which one is the limiter?
HCl
H2
Ca
7
Correct! Ok, the problem is Ca 2HCl
CaCl2 H2 Your job find the limiting reagent
and how much calcium chloride will be made from
80.2g Ca and 73.0g HCl So then, how many moles of
product are produced?
1.00
2.00
1.530
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Correct! Ok, the problem is Ca 2HCl
CaCl2 H2 Your job find the limiting reagent
and how much calcium chloride will be made from
80.2g Ca and 73.0g HCl So then, how many grams of
product are produced?
110
111.1
111
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Correct! Ok, the problem is Ca 2HCl
CaCl2 H2 Your job find the limiting reagent
and how much calcium chloride will be made from
80.2g Ca and 73.0g HCl Summarize what the answers
are.
HCl, 1.53 mol, 111g
HCl, 1.00mol, 111g
HCl, 1.00mol, 110g
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Good job! You finished the easy limiting reagent
tutorial! You should try the other one now.
Not so easy?
End show?
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Good!

• If you know that already, dont bother with
  the easy stuff. Good job!

Back to the easy anyways?
Not so easy one now?
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Youve been Lego-ified!

• Go back to your work and get it right!

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Confident, are you?

• Good. The next problem
• Al(OH)3 HCl AlCl3 H2O
• Find out which reactant is the limiting reagent
  and find how much AlCl3 is produced. You will
  also find out how much of the excess reagent is
  left over. You will do this by going from one
  reactant to the other. You are given 117.0g
  Al(OH)3 and 182.5g HCl. What is your first move?

Go from either reactant to AlCl3
Use Stoichiometry to go from Al(OH)3 to HCl
Balance the chemical equation
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Correct! The problem Al(OH)3 3HCl AlCl3
3H2O Find out which reactant is the limiting
reagent, how much excess reagent is left over,
and find out how much AlCl3 is produced. You
will do this by going from one reactant to the
other. You are given 117.0g Al(OH)3 and 182.5g
HCl. Whats your next move?
Go from Al(OH)3 to HCl
Go from either reactant to AlCl3
Re-balance the chemical equation
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Correct! The problem Al(OH)3 3HCl AlCl3
3H2O Find out which reactant is the limiting
reagent, how much excess reagent is left over,
and find out how much AlCl3 is produced. You
will do this by going from one reactant to the
other. You are given 117.0g Al(OH)3 and 182.5g
HCl. You get 164.3g HCl when you go one way, so
what now?
Try the other way
Say Al(OH)3 is the excess reagent
Say HCl is the excess reagent
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Excellent! That is correct, but I have to go
slowly for some people. Since you are smarter
than that, you get to go ahead. Yay!
Go on!
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Correct! The problem Al(OH)3 3HCl AlCl3
3H2O Find out which reactant is the limiting
reagent, how much excess reagent is left over,
and find out how much AlCl3 is produced. You
will do this by going from one reactant to the
other. You are given 117.0g Al(OH)3 and 182.5g
HCl. You get 164.3g HCl when you go one way, and
130.0g Al(OH)3 the other way, so what now?
Say that neither one is the excess reagent
Say Al(OH)3 is the excess reagent
Say HCl is the excess reagent
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Correct! The problem Al(OH)3 3HCl AlCl3
3H2O Find out which reactant is the limiting
reagent, how much excess reagent is left over,
and find out how much AlCl3 is produced. You
will do this by going from one reactant to the
other. You are given 117.0g Al(OH)3 and 182.5g
HCl. You get 164.3g HCl when you go one way, and
130.0g Al(OH)3 the other way, so you know that
HCl is the excess reagent. How much of it is
left over?
0.5000mol
1.500mol
0.500mol
Explanation
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Correct! The problem Al(OH)3 3HCl AlCl3
3H2O Find out which reactant is the limiting
reagent, how much excess reagent is left over,
and find out how much AlCl3 is produced. You
will do this by going from one reactant to the
other. You are given 117.0g Al(OH)3 and 182.5g
HCl. You get 164.3g HCl when you go one way, and
130.0g Al(OH)3 the other way, so you know that
HCl is the excess reagent. You have 0.5000mol
HCl left over how much AlCl3 is produced?
0.5000mol
1.5000mol
1.500mol
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Correct! The problem Al(OH)3 3HCl AlCl3
3H2O Find out which reactant is the limiting
reagent, how much excess reagent is left over,
and find out how much AlCl3 is produced. You
will do this by going from one reactant to the
other. You are given 117.0g Al(OH)3 and 182.5g
HCl. You get 164.3g HCl when you go one way, and
130.0g Al(OH)3 the other way. Summarize
the answers
HCl, 1.500mol HCl, and 1.500mol AlCl3
Al(OH)3, 0.5000mol Al(OH)3, and 0.5000mol AlCl3
HCl, 0.5000 mol HCl, and 1.500mol AlCl3
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Excellent! You finished the Not so easy
tutorial! You get no prize. Still, you might have
learned something, and thats worth it. Good for
you.
End show
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You thought that you could get through this one
that easily, eh? Go back to the easy one! (You
got Lego-ified again!)
Fine. back to your mistake
Yes, get me to easy!
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Ok, you want an explanation. We started with
117.0g Al(OH)3, and 182.5g HCl, and when we went
from Al(OH)3 to HCl, we got 164.3g HCl. When
we went from HCl to Al(OH)3, we got 130.0g
Al(OH)3. So, that means that if we just had
182.5g HCl, we would need 130.0g Al(OH)3 to
react fully with the HCl. Since that is more than
we started with, we cannot have a reaction that
uses up all of the HCl with 182.5g HCl and
117.0g Al(OH)3. Also, since we got 164.3g HCl
when we started with Al(OH)3, that means that a
full reaction (going by Al(OH)3) would only
need 164.3g HCl, so we would have some HCl left
over. That means that HCl is the limiting
reagent. Once you figure that out, its just
straight stoichiometry from there.
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