Good Afternoon! 9/13/2014
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Transcript of Good Afternoon! 9/13/2014
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Good Afternoon! 04/21/23
• Today we will…
• Hand out the Schedules
• Finish the problems from yesterday
• Hand out books
• Discuss tow other areas of Stoichiometry – Limiting Reactants (it is just like it sounds)– Percent yield, theoretical yield
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Equation 2
First you have to balance it, so you First you have to balance it, so you can have the ratios.can have the ratios.
PP ++ O O22 P P22OO554 5 2
Given: 3 moles of P Find: _____ moles of PGiven: 3 moles of P Find: _____ moles of P22OO55
x2 Moles P2O5
4 moles P
Start with what moles of what you
are given.
3 moles P
Multiply by the Mole Ratio
= 1.5 moles P2O5
Finally an answer pops
out
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Equation 2 Part II
How many Moles of Phosphorus do you How many Moles of Phosphorus do you need to produce 2.5 moles of Pneed to produce 2.5 moles of P22OO55??
PP ++ O O22 P P22OO554 5 2
Given: 2.5 moles of PGiven: 2.5 moles of P22OO55 Find ____ moles P Find ____ moles P
x2 Moles P
1 moles P2O5
Start with what moles of what you
are given.
= 5 moles P
Multiply by the Mole Ratio
2.5 moles P2O5
Finally an answer pops
out
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Equation 2 Part II Step 2
How many grams of Phosphorus would How many grams of Phosphorus would that be?that be?
Given: 5 moles of P Find ______ grams PGiven: 5 moles of P Find ______ grams P
x30.97 grams P
1 mole P
Start with what moles of what you
are given.
= 154.85 grams P
Multiply by the Molar mass
5 moles P
Finally an answer pops out
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One more practice, only this one will be a big one.
• Use the following reaction to answer the question below.
• Li3N + H2O NH3 + LiOH
• Use the steps, balance, convert to moles, use the ratio, then go to what you are asked.
• If you have 32.9 grams of Li3N how many grams of water do you need?
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Here is the answer.
Li3N + 3 H2O NH3 + 3 LiOH
32.9 g Li3N x
1 Moles Li3N
34.83 g Li3N
x
3 moles H2O
x1 Moles
Li3N
18.02 g H2O
1 Moles H2O
= 51.06 g H2O
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Practice Problem 2
First you have to balance it, so you First you have to balance it, so you can have the ratios.can have the ratios.
AlAl ++ Cl Cl22 AlCl AlCl332 3 2
Given: 5.67 grams of Al Given: 5.67 grams of Al Find: _____ liters of ClFind: _____ liters of Cl22
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5.67 g Al x
1 Moles Al
26.98 g Al
x
3 moles Cl2 x
2 Moles Al
22.4 L Cl2
1 Moles Cl2
= 7.06 L Cl2
AlAl ++ Cl Cl22 AlCl AlCl332 3 2
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Limiting ReactantThe limiting reactant is the reactant that is
consumed first, and then limits the amount of product formed. What is the limiting reactant in the reaction below?
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Limiting Reagents - Combustion
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If you have the following amounts for the following reaction, which
one is the limiting Reactant?
• 2 H2 + O2 2 H2O
6.4 moles H2
3.4 moles O2
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So what you have to do is just pick one, doesn’t matter which just pick.
Find out how many moles are needed to consume that one and
then compare.
= 3.2 moles O2
needed6.4 Moles H2 x
Since we have more O2 than we need H2 is limiting reactant
1 Moles O2
2 Moles H2
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Percent Yield!• So here’s the thing, if you take the reaction
that we looked at before, only this time we’ll see how much NH3 we can make.
• Li3N + 3 H2O NH3 + 3 LiOH
32.9 g Li3N x
1 Moles Li3N
34.83 g Li3N
x
1 moles NH3
x1 Moles
Li3N
17.03 g NH3
1 Moles NH3
= 16.09 g NH3
• If I send you to the lab with 32.9 g Li3N, do you think you’ll get 16.09 g NH3?
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Here are a few Vocabulary words you’ll need.
• Theoretical Yield: this is the amount that you are supposed to get (assuming not one atom is lost) You have to calculate this using the process.
• Actual Yield: This is the amount that you get when you do the reaction in the lab.
• Percent Yield: This is what % the Actual Yield is of the Theoretical yield.
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Theoretical yield
Let’s say you do the following reaction in the presence of excess (more than
enough) Iron and 15.0 grams of Sb2S3
Sb2S3 + 3 Fe 2 Sb + 3 FeS
15.0 g Sb2S3
x
2 moles Sb
1 Moles Sb2S3
1 moles Sb2S3
x 339.68 g Sb2S3
x121.75 g Sb
1 moles Sb
= 10.75 g Sb
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Actual Yield
• Let’s say you do the experiment and end up with 9.84 grams of Sb, what is your percent yield?
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Percent Yield
10.75 g Sb is our theoretical yield & 9.84 g Sb is our Actual yield, so…
Percent Yield =Actual Yield
x 100 =Theoretical Yield
Percent Yield =9.84 g Sb
x 100 = 91.5% yield10.75 g Sb