OB: intro to STOICHIOMETRY Using ratios of balanced reactions to manipulate the chemical variables...

OB: intro to STOICHIOMETRY

Using ratios of balanced reactions to manipulate the chemical variables in

these reactions

Calculators, reference tables, notes.Get your thinking caps on!

Hydrogen + oxygen makes waterMeans many different things

1 Skeleton: H2 + O2 H2O2 Balanced: 3 # of atoms: 4 # of molecules: 5 AMU: 6 Grams: 7 # of MOLES:

Hydrogen + oxygen makes water

1 Skeleton: H2 + O2 H2O2 Balanced: 2H2 + O2 2H2O3 # of atoms: four + two makes six4 # of molecules: two + one makes two5 AMU: 4 amu + 32 amu = 36 amu6 Grams: 4 g + 32 g = 36 grams7 # of MOLES: 2 moles + 1 mole = 2 moles

8. For this reaction, the number of MOLES is

2 moles + 1 mole = 2 moles

9. The mole ratio for this reaction is 2:1:2

This is like the “recipe” for the reaction.

For every single chemical reaction that’s balanced, we can determine the “mole ratio”, or ratio of moles to moles to moles for that reaction.

11. What’s the mole ratio for this?

4Al + 3O2 2Al2O3

For every single chemical reaction that’s balanced, we can determine the “mole ratio”, or ratio of moles to moles to moles for that reaction.

11. What’s the mole ratio for this?

4Al + 3O2 2Al2O3

That would be 4:3:2. It takes 4 moles of aluminum to react with three moles of oxygen, which forms two molesof aluminum oxide.

To make brownies you use 1 box mix, 3 eggs, 1 cup water, and ½ cup oil.

The “recipe” ratio is 1:3:1:½

12. How many eggs needed for a double batch?

2H2 + O2 2H2OThis equation has a 2:1:2 mole ratio.

14. If you had 9.0 moles of hydrogen, how many moles of oxygen are required to complete this reaction? (hint, let’s use proportions)

2H2 + O2 2H2O This has a 2:1:2 mole ratio.

14. If you had 9.0 moles of hydrogen, how many moles of oxygen are required to complete this reaction?

MR =H2 2O2 1

9.0x

2x = 9.0

X = 4.5 moles oxygen

THINK:

Could you convert the 4.5 moles oxygen into grams?

Could you covert it into liters of gas (at STP)?

How about particles (if you really had to)?

Using this mole ratio provided by every single balanced chemical reaction, we can use that ratio to “play” with every equation and do lots of conversions to it. This is Stoichiometry, the math manipulation of balanced reactions by conversions.

15. Not long after the mole islands were found, just to the east, a new set of islands were found. Chemists, not being that creative, named them this way:

Mole Island

Volume Island

Particle Island

Mass Island

Mole Island II

Volume Island II

Particle Island II

Mass Island II

Naming was easy, but no money for a new bridge, so they built the MOLE RATIO TUNNEL!

Mole Island

Volume Island

Particle Island

Mass Island

Mole Island II

Volume Island II

Particle Island II

Mass Island II

Simple new names, but not connected together. So they started digging A TUNNEL

Mole Island

Volume Island

Particle Island

Mass Island

Mole Island II

Volume Island II

Particle Island II

Mass Island II

Mole Ratio Tunnel

This bigger map shows all the math conversions for all Stoichiometry. The same “tolls” or equalities apply on both sides of the map.

16. So, how many grams of oxygen are required to completely react with 50.0 grams of hydrogen?

Let’s look over the map to figure this out.

This bigger map shows all the math conversions for all Stoichiometry. The same “tolls” or equalities apply on both sides of the map.

16. How many grams of oxygen are required to completely react with 50.0 grams of hydrogen?

step 1

50.0 g H2

1x = 25.0 moles H2

1 mole H2 2g H2

16. How many grams of oxygen are required to completely react with 50.0 grams of hydrogen? Step 2 in red

MR H2 O2

2 25.0 1 x

2x = 25.0 moles x = 12.5 moles oxygen

50.0 g H2

1x = 25.0 moles H2

1 mole H2 2g H2

16. How many grams of oxygen are required to completely react with 50.0 grams of hydrogen? Step 3 in black

MRH2 O2

2 25.0 1 x


12.5 moles O2

1

x 32 g O2

1 moles O2

= 400. grams O2

Stoichiometry in three easy steps…

50.0 g H2

1x = 25.0 moles H2

1 mole H2 2g H2

17. Using the same reaction, how many liters of oxygen are required to completely react with 155 grams of hydrogen gas?

Use the map, decide where you need to start, then where you want to go…


Step 1: convert these grams of hydrogen into moles (or get eaten by the shark!)

155 g H2

1x = 77.5 moles H2

1 mole H2 2g H2


155 g H2

1x = 77.5 moles H2

1 mole H2 2g H2

MR H2

O2

2 77.5

1 x


38.8 moles O2

1

x 22.4 L O2

1 moles O2

= 869 liters oxygen

OB: practice problems for Stoich, class #2

18.

45.0 grams of aluminum metal dissolves into sufficient hydrochloric acid to completely react.

How many grams of hydrogen gas form?

Write a balanced chemical reaction, get out your periodic table and your Stoichiometry map. Mind the shark. Show all work.

WRITE BIG. This takes 3 steps

2Al(S) + 6HCl(AQ) 2AlCl3(AQ) +3H2(G)

45.0 grams of Al metal dissolves into sufficient HCl(AQ) to completely react. How many grams of H2(G) form?

2Al(S) + 6HCl(AQ) 2AlCl3(AQ) +3H2(G)

45.0 grams of Al metal dissolves into sufficient HCl(AQ) to completely react. How many grams of H2(G) form?

45.0 g Al1

X = 1.67 moles Al1 mole Al27 g Al

MR Al 2 1.67H2 3 x

2x = 5.01

x = 2.51 moles H2

2.51 moles H2

1 X = 5.02 grams H2

2 g H2 1 mole H2

#18

2Al + 6HCl 2AlCl3 + 3H2

19. Using the same reaction… if you want to use up 23.1 moles of HCl, how many formula units of aluminum chloride form? 2 steps.

MR HCl 6 AlCl3 2

Change this to 3:1 ?



MR HCl 6 AlCl3 2

Never change this to 3:1,

It’s not worth it, you’ll possibly make a boo boo, and the math works fine 6:2

Do all the work on the next slide…



MR HCl 6 AlCl3 2


19. Using the same reaction… if you want to use up 23.1 moles of HCl, how many formula units of aluminum chloride form?

MR HCl 6 23.1 AlCl3 2 x

6x = 46.2

x = 7.70 moles AlCl3

7.70 moles AlCl3

1X = 7.70 x 6.02 x 1023 FU

6.02 x 1023 FU AlCl3

1 mole AlCl3

46.354 x 1023 FU = 46.4 x 1023 = 4.64 x 1024 FU AlCl3

20. 371.5 grams of candle wax (C21H44) combust. How many liters of CO2 gas form, assume STP for the gas. 3 steps.

Start with balancing this combustion reaction.

20. 371.5 grams of candle wax (C21H44) combust. How many liters of CO2 gas form, assume STP for the gas.

C21H44 + 32O2 21CO2 + 22H2O

Mole ratio 1 : 32 : 21 : 22


C21H44 + 32O2 21CO2 + 22H2O

Mole ratio 1 : 32 : 21 : 22

371.5 g wax1

X = 1.255 moles wax

C21H44

C 21 x 12 = 252 gH 44 x 1 = 44 g

296 g/mole

1 mole wax296 g wax


C21H44 + 32O2 21CO2 + 22H2O

371.5 g wax1

X = 1.255 moles wax1 mole wax296 g wax

MR wax 1 1.255CO2 21 x

x = 26.36 moles CO2


C21H44 + 32O2 21CO2 + 22H2O

371.5 g wax1

X = 1.255 moles wax1 mole wax296 g wax

MR wax 1 1.255CO2 21 x

x = 26.36 moles CO2

26.36 moles CO2

1x 22.4 L CO2

1 mole CO2

= 590.5 Liters CO2

21. Using the same wax combustion reaction, if you consume 23.9 moles of oxygen, how many moles of water form? One step. (sometimes it’s easy)

C21H44 + 32O2 21CO2 + 22H2O

MR Oxygen wata

21. Using the same wax combustion reaction, if you consume 23.9 moles of oxygen, how many moles of water form?

MR Oxygen 32 23.9 wata 22 x

32x = 525.8

x = 16.4 moles wata

Homework…

Due Friday: Chemical Reactions Lab

Due Wednesday: Stoich HW #1

OB: Stoich3, Plotting problems, then doing them right.First, get out your maps, and your periodic tables, and a calculator.

Let’s look at the map here…

Mole ratio tunnelmoles moles

vol vol

# of

particles# of

particlesmass mass

22. YOU have 4.56 x 1025 atoms of zinc that you want to put into hydrochloric acid to make them fizz away. How many grams of hydrogen gas form? Don’t do this, let’s talk about where to start, where to go next, and balance this equation.


vol vol

# of

particles# of

particlesmass mass

23. YOU want to know how many molecules of water form when 55.3 Liters of methane combust. Balance the equation, show the plan of attack on the map.


vol vol

# of

particles# of

particlesmass mass

24. YOU need (sorry, it’s true) to know how many liters of nitrogen gas are required to combine with 809 liters of hydrogen when ammonia forms. Balance the equation, show the plan on the map, but don’t do this problem.


vol vol

# of

particles# of

particlesmass mass

22. YOU have 4.56 x 1025 atoms of zinc that you want to put into hydrochloric acid to make them fizz away. How many grams of hydrogen gas form?

22A. Do this now. Zn + 2HCl(AQ) ZnCl2 + H2

22. YOU have 4.56 x 1025 atoms of zinc that you wan to put into hydrochloric acid to make them fizz away. How many grams of hydrogen gas form? Zn + 2HCl(AQ) ZnCl2 + H2

1. 4.56 x 1025 atoms 1

x 1 mole Zn6.02 x 1023 atoms Zn =

4.566.02 x

1025

1023 = 75.7 moles Zn

2. MRZnH2

11

75.7X X = 75.7 moles hydrogen

3. 75.7 moles H2

1x

2 grams H2

1 mole H2= 151 grams H2

23.YOU want to know how many molecules of water form when 55.3 Liters of methane combust. Do this now. CH4 + 2O2 CO2 + 2H2O

23. YOU want to know how many molecules of water form when 55.3 Liters of methane combust.

CH4 + 2O2 CO2 + 2H2O

1. 55.3 L CH4

1x

1 mole methane22.4 Liters CH4

= 2.47 moles methane

2. MRMethane

Water12

2.47X

X = 4.94 moles of water

3. 4.94 moles water1

x6.02 x 1023 molecules

1 mole water = 29.7388 x 1023 molecules

2.97 x 1024 moleculesof water

24. YOU need (sorry, it’s true) to know how many liters of nitrogen gas are required to combine with 809 liters of hydrogen when ammonia forms. 3H2 + N2 2NH3

24. YOU need (sorry, it’s true) to know how many liters of nitrogen gas are required to combine with 809 liters of hydrogen when ammonia forms. 3H2 + N2 2NH3

1. 809 L H2

1x 1 mole hydrogen

22.4 L H2

= 36.1 moles H2

2. MRnitrogenhydrogen

13

X36.1

3X = 36.1 moles

3. 12.0 moles N2

1x

22.4 L N2

1 mole N2

= 269 Liters nitrogen gas

X = 12.0 moles nitrogen

For Thursday…

PROBLEM 25

2C2H6(G) + 7O2(G) --> 4CO2(G) + 6H2O(L)

If exactly 15.6 moles of ethane gas combusts, how many moles of oxygen are used?

ANSWER PROBLEM 25

2C2H6(G) + 7O2(G) --> 4CO2(G) + 6H2O(L)

If exactly 15.6 moles of ethane gas combusts, how many moles of oxygen are used?

MR Ethaneoxygen

2 15.67 x

2X = 109.2

X = 54.6 moles O2

That was easy, one step stoich!

PROBLEM 26

4NH3(G) + 5O2(G) --> 4NO(G) + 6H2O(L)

If exactly 649.6 L of NO(G) form, how many liters of O2 are used?

ANSWER PROBLEM 26

4NH3(G) + 5O2(G) --> 4NO(G) + 6H2O(L)

If exactly 649.6 L of NO(G) form, how many liters of O2 are used?

649.6 L NO

1X 1 mole NO

22.4 L NO= 29.00 mole NO(G) (4 SF)

MR NOO2

4 29.005 x

4X = 145.0

X = 36.25 moles O2

36.25 mole O2

1X 22.4 L O2

1 mole O2

= 812.0 Liters O2 (4 SF)

PROBLEM 27.

2C8H18(L) + 25O2(G) --> 16CO2(G)

+18H2O(G)

During this combustion, 125 g of oxygen are used up. How many g of H2O are produced?

ANSWER PROBLEM 27

2C8H18(L) + 25O2(G) --> 16CO2(G)

+18H2O(G)

During this combustion, 125 g of oxygen are used up. How many g of H2O are produced?

125 g O2

1X 1 mole O2

32 g O2

= 3.91 mole O2(G) (3 SF)

MR H2OO2

18 X25 3.91

25X = 70.38

X = 2.82 moles water

2.82 moles water1

X 18 g water1 mole H2O

= 50.8 g water 3 SF

PROBLEM 28.

2N2(G) + 5O2(G) --> 2N2O5(G)

105 g of N2 react with oxygen to form dinitrogen pentoxide. How many molecules of N2O5 actually form in

this reaction?

ANSWER PROBLEM 28

2N2(G) + 5O2(G) --> 2N2O5(G)

105 g of N2 react with oxygen to form dinitrogen pentoxide. How many molecules of N2O5 actually form in

this reaction? 105 g N2

1X 1 mole N2

28 g N2

= 3.75 mole N2(G) (3 SF)

MR N2

N2O5

2 3.752 X

X = 3.75 moles N2O5

3.75 moles N2O5

1X 6.02 x 1023 molecules

1 mole= 22.575 x 1023

= 22.6 x 1023

= 2.26 x 1024 molecules

PROBLEM 29.

4Fe(S) + 3O2(G) --> 2Fe2O3(S)

In synthesis between an iron truck + the air, 12,525 moles of oxygen will convert how many moles of iron into

rust?

ANSWER PROBLEM 29

4Fe(S) + 3O2(G) --> 2Fe2O3(S)

In synthesis between an iron truck + the air, 12,525 moles of oxygen will convert how many moles of iron into

rust?

This is a one step – MOLES to MOLES problem (no problem)

MR O2

Fe2O3

3 12,5252 X

3X = 25,050

X = 8350 moles of rust 3 SF

PROBLEM 30

2NO2(G) + 7H2(G) --> 2NH3(G) + 4H2O(L)

In a test tube reaction, 0.135 moles of H2 will react to form how many grams of NH3?

ANSWER PROBLEM 30

2NO2(G) + 7H2(G) --> 2NH3(G) + 4H2O(L)

In a test tube reaction, 0.135 moles of H2 will react to form how many grams of NH3?

Moles to Grams - this is a two step stoich problem

MR H2

NH3

7 0.135 2 X

7X = 0.270X = 0.0386 moles of NH3

0.0386 moles NH3

1X 17 g NH3

1 mole NH3

= 0.656 g NH3

There are 2 moles of NH3 in this product, but the MOLAR MASS is the mass of one mole. If the equation had 1 mole or 3 moles of ammonia, the molar mass is a CONSTANT.

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OB: intro to STOICHIOMETRY Using ratios of balanced reactions to manipulate the chemical variables...

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