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REMINDER

Due Tonight at Midnight (January 21 at 11:59 pm) via email

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*** Kinesthetic: If you do not know how to use Prezi you may

do a power point otherwise email your Prezi link.

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approximately 10 points averaged with your other grades in

the category.

It can only HELP! Not harm your grade.

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Important things to Remember

The Mole

A counting unit

Similar to a dozen, except instead

of 12, it’s

602,000,000,000,000,000,000,000

6.02 X 1023 (in scientific notation)

This number is named in honor of

Amedeo Avogadro (1776 – 1856)

The Mass of 1 mole (in grams)

Equal to the numerical value of the average

atomic mass (get from periodic table), or add

the atoms together for a molecule

1 mole of C atoms = 12.0 g

1 mole of Mg atoms = 24.3 g

1 mole of O2 molecules = 32.0 g

Diatomic elements are: H, N, O, F, Cl, Br, I

Molar Mass

Molar Mass of Compounds

The molar mass (MM) of a compound is determined the same way, except now you add up all the atomic masses for the molecule (or compound) Ex. Molar mass of CaCl2

Avg. Atomic mass of Calcium = 40.08g

Avg. Atomic mass of Chlorine = 35.45g

Molar Mass of calcium chloride = 40.08 g/mol Ca + (2 X 35.45) g/mol Cl 110.98 g/mol CaCl2

20

Ca 40.08 17

Cl

35.45

http://pearl1.lanl.gov/periodic/elements/20.html

http://pearl1.lanl.gov/periodic/elements/17.html

Flowchart Atoms or

Molecules

Moles

Mass

(grams)

Divide by 6.02 X 1023

Multiply by 6.02 X 1023

Multiply by

atomic/molar mass

from periodic table Divide by

atomic/molar mass

from periodic table

Practice

Calculate the Molar Mass of

calcium phosphate

Formula =

Masses elements: Ca: 3 Ca’s X 40.1 =

P: 2 P’s X 31.0 =

O: 8 O’s X 16.0 =

Molar Mass =

Ca3(PO4)2

120.3 g

62.0 g

128.0 g

120.3g + 62.0g +128.0g 310.3 g/mol

molar mass Avogadro’s number Grams Moles particles

Everything must go through

Moles!!!

Calculations

Atoms/Molecules and Grams

How many atoms of Cu are

present in 35.4 g of Cu?

35.4 g Cu 1 mol Cu 6.02 X 1023 atoms Cu

63.5 g Cu 1 mol Cu

= 3.4 X 1023 atoms Cu

Chocolate Chip Cookies!! 1 cup butter

1/2 cup white sugar

1 cup packed brown sugar

1 teaspoon vanilla extract

2 eggs

2 1/2 cups all-purpose flour

1 teaspoon baking soda

1 teaspoon salt

2 cups semisweet chocolate chips

Makes 3 dozen

How many eggs are needed to make 3 dozen cookies?

How much butter is needed for the amount of chocolate chips used?

How many eggs would we need to make 9 dozen cookies?

How much brown sugar would I need if I had 1 ½ cups white sugar?

Cookies and Chemistry…Huh!?!?

Just like chocolate chip

cookies have recipes,

chemists have recipes as well

Instead of calling them

recipes, we call them reaction

equations

Furthermore, instead of using

cups and teaspoons, we use

moles

Lastly, instead of eggs, butter,

sugar, etc. we use chemical

compounds as ingredients

Chemistry Recipes

Looking at a reaction tells us how much of something you need to react with something else to get a product (like the cookie recipe)

Be sure you have a balanced reaction before you start!

Example: 2 Na + Cl2 2 NaCl

This reaction tells us that by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chloride

What if we wanted 4 moles of NaCl? 10 moles? 50 moles?

Mole Ratios

These mole ratios can be used to calculate

the moles of one chemical from the given

amount of a different chemical

Example: How many moles of chlorine is

needed to react with 5 moles of sodium

(without any sodium left over)?

2 Na + Cl2 2 NaCl

5 moles Na 1 mol Cl2

2 mol Na = 2.5 moles Cl2

Mole-Mole Conversions How many moles of sodium chloride will

be produced if you react 2.6 moles of

chlorine gas with an excess (more than

you need) of sodium metal?

2 Na + Cl2 2 NaCl

Mole-Mass Conversions

Most of the time in chemistry, the amounts are

given in grams instead of moles

We still go through moles and use the mole ratio,

but now we also use molar mass to get to grams

Example: How many grams of chlorine are required

to react completely with 5.00 moles of sodium to

produce sodium chloride?

2 Na + Cl2 2 NaCl

5.00 moles Na 1 mol Cl2 70.90g Cl2

2 mol Na 1 mol Cl2 = 177g Cl2

Practice

Calculate the mass in grams of Iodine

required to react completely with 0.50

moles of aluminum.

2 Al + 3 I2 2 AlI3

Mass-Mole

We can also start with mass and convert to

moles of product or another reactant

We use molar mass and the mole ratio to get

to moles of the compound of interest

Calculate the number of moles of ethane (C2H6)

needed to produce 10.0 g of water

2 C2H6 + 7 O2 4 CO2 + 6 H20

10.0 g H2O 1 mol H2O 2 mol C2H6

18.0 g H2O 6 mol H20

= 0.185

mol C2H6

Practice Calculate how many moles of oxygen are

required to make 10.0 g of aluminum oxide

4 Al + 3 O2 2 Al2O3

Practice

Write the balanced reaction for hydrogen gas reacting with oxygen gas.

H2 + O2 H2O

How many moles of each reactants are needed?

What if we wanted 4 moles of water how many moles of each reactant would you need?

What if we had 3 moles of oxygen, how much hydrogen would we need to react and how much water would we

get?

What if we had 50 moles of hydrogen, how much oxygen would we need and how much water produced?

Practice

Write the balanced reaction for hydrogen gas

reacting with oxygen gas.

2 H2 + O2 2 H2O

How many moles of reactants are needed?

What if we wanted 4 moles of water?

What if we had 3 moles of oxygen, how much hydrogen

would we need to react and how much water would we

get?

What if we had 50 moles of hydrogen, how much oxygen

would we need and how much water produced?

2 mol H2

1 mol O2

4 mol H2

2 mol O2

6 mol H2, 6 mol H2O

25 mol O2, 50 mol H2O

Practice

A2 + 2B 2AB

What is the mole ratio of substance A to substance AB?

What is the mole ratio of substance A to substance B?

What is the mole ratio of substance B to substance AB?

Show the work for the problems below:

If you have 4 moles of substance A, how many moles of substance AB can you produce?

If you have 10 moles of substance B and an excess of substance A, how many moles of substance AB can you

produce?

Mass-Mass Conversions

Most often we are given a starting mass and

want to find out the mass of a product we will get

(called theoretical yield)

Or how much of another reactant we need to

completely react with it (no leftover ingredients!)

Now we must go from grams to moles, mole

ratio, and back to grams of compound we are

interested in

gA mol A mol B gB

Mass-Mass Conversion Ex. Calculate how many grams of

ammonia are produced when you react

2.00g of nitrogen with excess hydrogen.

N2 + H2 NH3

1. Make sure you have a balanced equation.

2. Convert grams of nitrogen to moles of nitrogen

3. Convert moles of nitrogen to moles of ammonia

4. Then, convert moles of ammonia to grams of

ammonia

With a partner discuss which values you will need for

each step above. Then, work the problem.

Hint: gA mol A mol B gB

Mass-Mass Conversion

Ex. Calculate how many grams of

ammonia are produced when you react

2.00g of nitrogen with excess hydrogen.

N2 + 3 H2 2 NH3

gA mol A mol B gB

Practice

How many grams of calcium nitride are

produced when 2.00 g of calcium reacts

with an excess of nitrogen?

Ca + N2 Ca3N2

General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.

Theoretical, Actual, and Percent

Yield

Theoretical yield:

the maximum amount of product, which is calculated

using the balanced equation.

Actual yield:

the amount of product obtained when the reaction

takes place

Percent yield:

the ratio of actual yield to theoretical yield

Percent yield = actual yield (g) x 100

theoretical yield (g)

26


Percent Yield

27


Calculating Percent Yield

Suppose you have prepared cookie dough to make 5

dozen cookies. The phone rings and you answer. While

you talk, a sheet of 12 cookies burns, and you have to

throw them out. The rest of the cookies you make are

okay. What is the percent yield of edible cookies?

Theoretical yield: 60 cookies possible

Actual yield: 48 cookies to eat

Percent yield: 48 cookies x 100% = 80.% yield

60 cookies

28


Learning Check

With a limited amount of oxygen, the reaction of

carbon and oxygen produces carbon monoxide.

2C(g) + O2(g) 2CO(g)

What is the percent yield if 40.0 g of CO are

produced when 30.0 g of O2 are used?

1) 25.0%

2) 75.0%

3) 76.2%

29


Solution

STEP 1 Given: 40.0 g of CO produced (actual)

30.0 g of O2 used

Need: percent yield of CO

STEP 2 Write a plan to calculate % yield of CO:

g of O2 moles of moles of g of CO

O2 CO (theoretical)

Percent yield of CO = g of CO (actual) x 100%

g of CO (theoretical)

30


Solution (continued)

STEP 3 Write conversion factors:

1 mole of O2 = 32.0 g of O2

1 mole O2 and 32.0 g O2

32.0 g O2 1 mole O2

1 mole of O2 = 2 moles of CO

1 mole O2 and 2 moles CO

2 moles CO 1 mole O2

1 mole of CO = 28.0 g of CO

1 mole CO and 28.0 g CO

28.0 g CO 1 mole CO 31



STEP 4 Setup to calculate theoretical yield in g of O2:

30.0 g O2 x 1 mole O2 x 2 moles CO x 28.0 g CO

32.0 g O2 1 mole O2 1 mole CO

= 52.5 g of CO (theoretical)

Setup to calculate percent yield:

40.0 g CO (actual) x 100 = 76.2% yield (3)

52.5 g CO (theoretical)

32


Learning Check

When N2 and 5.00 g of H2 are mixed, the reaction

produces 16.0 g of NH3. What is the percent yield

for the reaction?

N2(g) + 3H2(g) 2NH3(g)

1) 31.3% of NH3

2) 56.9% of NH3

3) 80.0% of NH3

33


Solution

2) 56.9%

STEP 1 Given: 16.0 g of NH3 produced (actual)

5.00 g of H2 used

Need: percent yield of NH3

STEP 2 Write a plan to calculate % yield of NH3:

g of H2 moles of moles of g of NH3

H2 NH3 (theoretical)

Percent yield of NH3 = g of NH3 (actual) x 100%

g of NH3 (theoretical)

34



STEP 3 Write conversion factors:

1 mole of H2 = 2.02 g of H2

1 mole H2 and 2.02 g H2

2.02 g H2 1 mole H2

1 mole of H2 = 2 moles of NH3

1 mole H2 and 2 moles NH3

2 moles NH3 1 mole H2

1 mole of NH3 = 17.0 g of NH3

1 mole NH3 and 17.0 g NH3

17.0 g NH3 1 mole NH3



STEP 4 Setup to calculate theoretical yield of g of NH3:

5.00 g H2 x 1 mole H2 x 2 moles NH3 x 17.0 g NH3

2.02 g H2 3 moles H2 1 mole NH3

= 28.1 g of NH3 (theoretical)

Setup to calculate percent yield:

Percent yield = 16.0 g NH3 x 100 = 56.9% yield (2)

28.1 g NH3

36


Guide to Calculations for

Percent Yield

37


Limiting Reactant

A limiting reactant in a chemical reaction is the

substance that

is used up

limits the amount of product that can form and stops

the reaction

38


Reacting Amounts

39

In a table setting, there is 1plate,

1 fork, 1 knife, and 1 spoon.

How many table settings are

possible from 5 plates, 6 forks,

4 spoons, and 7 knives?

What is the limiting item?


Reacting Amounts (continued)

Only 4 place settings are possible.

Initially Used Left over

Plates 5 4 1

Forks 6 4 2

Spoons 4 4 0

Knives 7 4 3

The limiting item is the spoon.

40


Example 1 of an Everyday

Limiting Reactant

How many peanut butter sandwiches could be made

from 8 slices of bread and 1 jar of peanut butter?

With 8 slices of bread, only 4 sandwiches could be

made.

The bread is the limiting item.

41



Limiting Reactant


from 8 slices bread and 1 tablespoon of peanut butter?

With 1 tablespoon of peanut butter, only 1 sandwich

could be made. The peanut butter is the limiting item.

42


Limiting Reactant

When 4.00 moles of H2 is mixed with 2.00 moles of

Cl2, how many moles of HCl can form?

H2(g) + Cl2(g) 2HCl(g)

4.00 moles 2.00 moles ??? Moles

Calculate the moles of product that each reactant, H2

and Cl2, could produce.

The limiting reactant is the one that produces the

smaller number of moles of product.

43


Limiting Reactant (continued)

HCl from H2

4.00 moles H2 x 2 moles HCl = 8.00 moles of HCl

1 moles H2

HCl from Cl2

2.00 moles Cl2 x 2 moles HCl = 4.00 moles of HCl

1 mole Cl2

4.00 moles of HCl is the smaller number of moles

produced. Thus, Cl2 will be used up.

The limiting reactant is Cl2.

44


Limiting Reactants Using Mass

If 4.80 moles Ca are mixed with 2.00 moles N2, which is

the limiting reactant? 3Ca(s) + N2(g) Ca3N2(s)

Moles of Ca3N2 from Ca

4.80 moles Ca x 1 mole Ca3N2 = 1.60 moles of Ca3N2

3 moles Ca (Ca is used up)

Moles of Ca3N2 from N2

2.00 moles N2 x 1 mole Ca3N2 = 2.00 moles of Ca3N2

1 mole N2

Ca is used up. Thus, Ca is the limiting reactant. 45


Learning Check

What is the mass of water that can be produced when

8.00 g of H2 and 24.0 g of O2 react?

2H2(g) + O2(g) 2H2O(l)

1) 8.0 g of H2O

2) 27.0 g of H2O

3) 72 g of H2O

46


Solution

3) 72 g of H2O

Moles of H2O from H2:

8.00 g H2 x 1 mole H2 x 2 moles H2O = 4.0 moles of H2O

2.0 g H2 2 moles H2

Moles of H2O from O2:

24.0 g O2 x 1 mole O2 x 2 moles H2O = 1.50 moles of H2O

32.0 g O2 1 mole O2 Smaller number

of moles of H2O

1.50 moles H2O x 18.0 g H2O = 27.0 g of H2O

1 mole H2O

47


Guide to Calculating Product

from a Limiting Reactant

48


Check Calculations

Equation

Initially

H2

4.00 moles

Cl2

2.00 moles

2HCl

0 mole

Reacted/

Formed

–2.00 moles –2.00 moles +4.00 moles

Left after

reaction

2.00 moles

(4.00 – 2.00)

Excess

0 moles

(2.00 – 2.00)

Limiting

4.00 moles

(0 + 4.00)

Product

possible

49


Limiting Reactant

A limiting reactant in a chemical reaction is the

substance that

is used up

limits the amount of product that can form and stops

the reaction

50


Reacting Amounts

51

In a table setting, there is 1plate,

1 fork, 1 knife, and 1 spoon.

How many table settings are

possible from 5 plates, 6 forks,

4 spoons, and 7 knives?

What is the limiting item?


Reacting Amounts (continued)

Only 4 place settings are possible.

Initially Used Left over

Plates 5 4 1

Forks 6 4 2

Spoons 4 4 0

Knives 7 4 3

The limiting item is the spoon.

52



Limiting Reactant


from 8 slices of bread and 1 jar of peanut butter?

With 8 slices of bread, only 4 sandwiches could be

made.

The bread is the limiting item.

53



Limiting Reactant


from 8 slices bread and 1 tablespoon of peanut butter?

With 1 tablespoon of peanut butter, only 1 sandwich

could be made. The peanut butter is the limiting item.

54


Limiting Reactant

When 4.00 moles of H2 is mixed with 2.00 moles of

Cl2, how many moles of HCl can form?

H2(g) + Cl2(g) 2HCl(g)

4.00 moles 2.00 moles ??? Moles

Calculate the moles of product that each reactant, H2

and Cl2, could produce.

The limiting reactant is the one that produces the

smaller number of moles of product.

55


Limiting Reactant (continued)

HCl from H2

4.00 moles H2 x 2 moles HCl = 8.00 moles of HCl

1 moles H2

HCl from Cl2

2.00 moles Cl2 x 2 moles HCl = 4.00 moles of HCl

1 mole Cl2

4.00 moles of HCl is the smaller number of moles

produced. Thus, Cl2 will be used up.

The limiting reactant is Cl2.

56


Limiting Reactants Using Mass

If 4.80 moles Ca are mixed with 2.00 moles N2, which is

the limiting reactant? 3Ca(s) + N2(g) Ca3N2(s)

Moles of Ca3N2 from Ca

4.80 moles Ca x 1 mole Ca3N2 = 1.60 moles of Ca3N2

3 moles Ca (Ca is used up)

Moles of Ca3N2 from N2

2.00 moles N2 x 1 mole Ca3N2 = 2.00 moles of Ca3N2

1 mole N2

Ca is used up. Thus, Ca is the limiting reactant. 57


Learning Check

What is the mass of water that can be produced when

8.00 g of H2 and 24.0 g of O2 react?

2H2(g) + O2(g) 2H2O(l)

1) 8.0 g of H2O

2) 27.0 g of H2O

3) 72 g of H2O

58


Solution

3) 72 g of H2O

Moles of H2O from H2:

8.00 g H2 x 1 mole H2 x 2 moles H2O = 4.0 moles of H2O

2.0 g H2 2 moles H2

Moles of H2O from O2:

24.0 g O2 x 1 mole O2 x 2 moles H2O = 1.50 moles of H2O

32.0 g O2 1 mole O2 Smaller number

of moles of H2O

1.50 moles H2O x 18.0 g H2O = 27.0 g of H2O

1 mole H2O

59


Guide to Calculating Product

from a Limiting Reactant

60

Limiting Reactant: Example 10.0g of aluminum reacts with 35.0 grams of

chlorine gas to produce aluminum chloride. Which

reactant is limiting, which is in excess, and how

much product is produced?

2 Al + 3 Cl2 2 AlCl3

Start with Al:

Now Cl2:

10.0 g Al 1 mol Al 2 mol AlCl3 133.5 g AlCl3

27.0 g Al 2 mol Al 1 mol AlCl3 = 49.4g AlCl3

35.0g Cl2 1 mol Cl2 2 mol AlCl3 133.5 g AlCl3

71.0 g Cl2 3 mol Cl2 1 mol AlCl3 = 43.9g AlCl3

Limiting

Reactant

LR Example Continued

We get 49.4g of aluminum chloride from the given

amount of aluminum, but only 43.9g of aluminum

chloride from the given amount of chlorine.

Therefore, chlorine is the limiting reactant. Once

the 35.0g of chlorine is used up, the reaction

comes to a complete .

Limiting Reactant Practice

15.0 g of potassium reacts with 15.0 g of

iodine. Calculate which reactant is limiting

and how much product is made.

Finding the Amount of Excess

By calculating the amount of the excess

reactant needed to completely react with

the limiting reactant, we can subtract that

amount from the given amount to find the

amount of excess.

Can we find the amount of excess

potassium in the previous problem?

Finding Excess Practice 15.0 g of potassium reacts with 15.0 g of iodine.

2 K + I2 2 KI

We found that Iodine is the limiting reactant, and

19.6 g of potassium iodide are produced.

15.0 g I2 1 mol I2 2 mol K 39.1 g K

254 g I2 1 mol I2 1 mol K = 4.62 g K

USED!

15.0 g K – 4.62 g K = 10.38 g K EXCESS

Given amount

of excess

reactant

Amount of

excess

reactant

actually

used

Note that we started with

the limiting reactant! Once

you determine the LR, you

should only start with it!

Limiting Reactant: Recap

1. You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT.

2. Convert ALL of the reactants to the SAME product (pick any product you choose.)

3. The lowest answer is the correct answer.

4. The reactant that gave you the lowest answer is the LIMITING REACTANT.

5. The other reactant(s) are in EXCESS.

6. To find the amount of excess, subtract the amount used from the given amount.

7. If you have to find more than one product, be sure to start with the limiting reactant. You don’t have to determine which is the LR over and over again!

1 gram of Chalk

What is the formula for Chalk?

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EXTRA CREDIT REMINDER - PC\|MACimages.pcmac.org/SiSFiles/Schools/GA/HoustonCounty...What if we had 3 moles of oxygen, how much hydrogen would we need to react and how much water would