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

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

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

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

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?

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

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

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

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?

Limiting Reagents - Combustion

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

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

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?

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.

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

Actual Yield

• Let’s say you do the experiment and end up with 9.84 grams of Sb, what is your percent yield?

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