1. Unit 6 –Stoichiometry and Reaction Types
Cartoon courtesy of NearingZero.net
Unit 6 –Stoichiometry and Reaction Types

2. I. Types of Chemical Reactions

3. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
A. Combustion
The burning of any substance in O2 to produce heat
A + O2  B
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)

4. A. Combustion 4 2 Na(s)+ O2(g)  Na2O(s) C3H8(g)+ O2(g)  5 3 4
Products:
contain oxygen
hydrocarbons form CO2 + H2O
Na(s)+ O2(g) 
Na2O(s)
C3H8(g)+ O2(g) 
CO2(g)+ H2O(g)

5. B. Synthesis
The combination of 2 or more substances to form a compound
Only one product
A + B  AB

6. B. Synthesis
H2(g) + Cl2(g)  2 HCl(g)

7. B. Synthesis Al(s)+ Cl2(g)  2 3 2 AlCl3(s) Products:
ionic - cancel charges
covalent - hard to tell
Al(s)+ Cl2(g) 
AlCl3(s)

8. AB  A + B C. Decomposition
a compound breaks down into 2 or more simpler substances
only one reactant
AB  A + B

9. C. Decomposition
2 H2O(l)  2 H2(g) + O2(g)

10. C. Decomposition 2 2 K(s) + Br2(l) KBr(l)  NI3(s)  Products:
binary - break into elements
others - hard to tell
K(s) + Br2(l)
KBr(l) 
NI3(s) 

11. A + BC  B + AC D. Single Replacement
one element replaces another in a compound
metal replaces metal (+)
nonmetal replaces nonmetal (-)
A + BC  B + AC

12. Cu(s) + 2AgNO3(aq)  Cu(NO3)2(aq) + 2Ag(s)
D. Single Replacement
Cu(s) + 2AgNO3(aq)  Cu(NO3)2(aq) + 2Ag(s)

13. D. Single Replacement Fe(s)+ CuSO4(aq)  Cu(s)+ FeSO4(aq)
Products:
metal  metal (+)
nonmetal  nonmetal (-)
free element must be more active (check activity series)
Fe(s)+ CuSO4(aq) 
Cu(s)+ FeSO4(aq)
Br2(l)+ NaCl(aq) 
N.R.

14. Aluminum + Ferric oxide 
Reaction Type: _____________________________

15. AB + CD  AD + CB E. Double Replacement
ions in two compounds “change partners”
cation of one compound combines with anion of the other
AB + CD  AD + CB

16. Pb(NO3)2(aq) + K2CrO4(aq)  PbCrO4(s) + 2KNO3(aq)
E. Double Replacement
Pb(NO3)2(aq) + K2CrO4(aq)  PbCrO4(s) + 2KNO3(aq)

17. E. Double Replacement 2 2 PbI2(s)+ KNO3(aq) Pb(NO3)2(aq)+ KI(aq) 
Products:
switch negative ions
Pb(NO3)2(aq)+ KI(aq) 
PbI2(s)+ KNO3(aq)

18. Plumbous nitrate + Sodium iodide 
Reaction Type: _____________________________
Cobaltous chloride + Sodium hydroxide

19. IQ 1 #1: Calculate the number of atoms in 47.5 g carbon.
#2: Calculate the mass of 5.1 x 1023 molecules of water.
47.5 g C
1 mol C
6.022 x 1023 atoms C
12.01 g C
1 mol C
= x atoms C
5.1 x 1023 molec H2O
1 mol H2O
18.02 g H2O
1 mol H2O
6.022 x 1023 molec H2O
= 15 g H2O or 1.5 x 101 g H2O

20. Stoichiometry
“In solving a problem of this sort, the grand thing is to be able to reason backward. This is a very useful accomplishment, and a very easy one, but people do not practice it much.”
Sherlock Holmes, in Sir Arthur Conan Doyle’s A Study in Scarlet
Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions.

21. Flowchart Atoms or Molecules Moles Mass (grams) Divide by 6.02 X 1023
Multiply by 6.02 X 1023
Moles
Multiply by atomic/molar mass from periodic table
Divide by atomic/molar mass from periodic table
Mass (grams)

22. Everything must go through Moles!!!
Calculations
molar mass Avogadro’s number Grams Moles particles
Everything must go through Moles!!!

23. 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?

24. 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

25. 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?

26. 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?

27. Solving a Stoichiometry Problem
Balance the equation.
Convert masses to moles.
Determine which reactant is limiting.
Use moles of limiting reactant and mole ratios to find moles of desired product.
Convert from moles to grams.

28. Working a Stoichiometry Problem
6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed.
1. Identify reactants and products and write the balanced equation. 4 Al + 3 O2 2
Al2O3
a. Every reaction needs a yield sign!
b. What are the reactants?
c. What are the products?
d. What are the balanced coefficients?

29. 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

30. Stoichiometry Problems
How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas?
2KClO3  2KCl + 3O2
? mol
9 mol
9 mol O2
2 mol KClO3
3 mol O2
= 6 mol
KClO3

31. 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?

32. 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 Cl g Cl2
2 mol Na 1 mol Cl2
= 177g Cl2

33. Practice
Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum.

34. 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
= mol C2H6

35. Practice
Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide

36. 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

37. 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
2.00g N mol N mol NH g NH3
28.02g N2 1 mol N mol NH3
= 2.4 g NH3

38. IQ2
Determine the mass of Iron III oxide formed when 25.1 g of iron reacts with oxygen.
Balanced equations: ______________________________
2. What mass of oxygen is needed to react with 6.8 x 1023 atoms of iron? (Use the above equations).
4 Fe + 3 O2 → 2 Fe2O3
25.1 g Fe
2 mol Fe2O3
g Fe2O3
1 mol Fe
55.85 g Fe
4 mol Fe
1 mol Fe2O3
= 35.9 g Fe2O3
6.8 x 1023 atoms Fe
1 mol Fe
3 mol O2
32.00 g O2
6.022 x 1023 atoms Fe
4 mol Fe
1 mol O2
= 27 g O2

39. Example 2
6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed?
4 Al O2  2Al2O3
6.50 g Al
1 mol Al
2 mol Al2O3
g Al2O3 =
? g Al2O3
26.98 g Al
4 mol Al
1 mol Al2O3
6.50 x 2 x ÷ ÷ 4 =
12.3 g Al2O3

40. Example 3 Cu + 2AgNO3  2Ag + Cu(NO3)2 12.0 g Cu 1 mol Cu 63.55 g Cu
How many grams of silver will be formed from 12.0 g copper?
Cu + 2AgNO3  2Ag + Cu(NO3)2
12.0 g
? g
12.0
g Cu
1 mol Cu
63.55
g Cu
2 mol Ag
1 mol Cu
107.87
g Ag
1 mol Ag
= 40.7 g Ag

41. Practice
How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen?

42. Limiting Reactant: 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
If we had the specified amount of all ingredients listed, could we make 4 dozen cookies?
What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies?
What if we only had one egg, could we make 3 dozen cookies?

43. Limiting Reactant
Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant.
That reactant is said to be in excess (there is too much).
The other reactant limits how much product we get. Once it runs out, the reaction s. This is called the limiting reactant.

44. Limiting Reactant
To find the correct answer, we have to try all of the reactants. We have to calculate how much of a product we can get from each of the reactants to determine which reactant is the limiting one.
The lower amount of a product is the correct answer.
The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it!
Be sure to pick a product! You can’t compare to see which is greater and which is lower unless the product is the same!

45. 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 mol AlCl g AlCl3
27.0 g Al mol Al mol AlCl3
= 49.4g AlCl3
35.0g Cl mol Cl mol AlCl g AlCl3
71.0 g Cl mol Cl mol AlCl3
= 43.9g AlCl3

46. 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 stop .

47. 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.

48. 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?

49. 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 I mol I mol K g K
254 g I mol I mol K
= 4.62 g K USED!
15.0 g K – 4.62 g K = g K EXCESS
Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!
Given amount of excess reactant
Amount of excess reactant actually used

50. Limiting Reactant: Recap
You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT.
Convert ALL of the reactants to the SAME product (pick any product you choose.)
The lowest answer is the correct answer.
The reactant that gave you the lowest answer is the LIMITING REACTANT.
The other reactant(s) are in EXCESS.
To find the amount of excess, subtract the amount used from the given amount.
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!