my lecture ppt for unit 8 spring 2015

1. Nickel III dichromate
reacts with calcium
Write and balance
When you are done: Read IN
YOUR LIFE on Pg 279 then see
#1

2. Stoichiometery
Unit 6 Ch 9
Read 278-316
Mole-mole Mole-gram Gram-gram gram –Liters of gas @STP
% yield limiters Heat
Day 1

3. What is stoichiometry?
• This is the process of using balanced
equations to calculate the mole ratios and
calculating the amounts used, created, or
left over, in a chemical experiment.
• If we are given a reaction and one piece
of information, we can find out ALL the
rest!

4. What is stoichiometry?
• We use the balanced equation to do a mole-
mole conversion. This way we can predict
what quantities we can get when working in
the lab. We call this the theoretical yield.

5. 12.1
Mass Conservation in
Chemical Reactions
– Mass and atoms are conserved in every
chemical reaction.
– Moles don’t have to be conserved, but MASS
does! (see next slide)

6. Relationships given from a
rxn.
• 6HCl (aq) + 2Al (S) 3H2 (g) + 2AlCl3 (Aq)
6 molecules HCl + 2 Al atoms  3 molecules H2 + 2 Form. Units of AlCl3
6 Moles HCl + 2 moles Al 3 moles H2 + 2 mole of AlCl3
218.76 g HCl + 53.96 g Al  6.06 g H2 + 266.66 g of AlCl3

7. In the reaction
•2H202  2H20 + 1O2
• We can write several mole relationship
that can be used as conversion facts.
2moleH202
1mole02
1mole02
2moleH2 02
1mole02
2moleH20
2moleH20
1mole02
2moleH202
2moleH20
2moleH20
2moleH202

8. For example in this equation,
• 2H20 + 2Na  2NaOH + H2(g)
• Can you describe this reaction to your partner?
• 2 moles of water will combine with 2
moles of sodium to make 2 moles of
sodium hydroxide and 1 mole of
hydrogen gas.

9. Sample class problems
(see pg 2 of packet)
2Li + FeS  Li2S + Fe
•Write relationships

10. Sample class problems
2Li + FeS  Li2S + Fe
•If you have 4.5 moles of Li, how many
moles of Lithium sulfide can be
produced?

11. Sample class problems
2Li + FeS  Li2S + Fe
•If you have 4.5 moles of Li, how many
moles of Iron can also be produced?

12. Sample class problems
2Li + FeS  Li2S + Fe
If you used 35.5 grams of Lithium how
many moles of FeS would be required to
make it all react perfectly?

13. Another example: calculate
• 6HCl (aq) + 2Al (S) 3H2 (g) + 2AlCl3 (Aq)
• If you had 0.450 moles of Al, how
many moles of HCl would you need
to completely react with it?
0.450 mol? mol

14. Another example: calculate
• 6HCl (aq) + 2Al (S) 3H2 (g) + 2AlCl3 (Aq)
• If you had 0.450 moles of Al, how
many moles of each product would
you produce.
0.450 mol ? mol ? mol

15. TERMS (fill in your vocab)
• Mole ratio (we just did some)

16. Mole Ratio
• The ratio of moles of one substance to
another from a balanced equation
(coefficients)

17. Limiting reagent
• Any reactant that is used up
first in a chemical reaction; it
determines the amount of
product that can be formed in
the reaction.

18. Theoretical yield
• The amount of product that could
form during a reaction calculated
from a balanced chemical
equation; it represents the
maximum amount of product that
could be formed from a given
amount of reactant

19. Actual yield
• The amount of product that forms
when a reaction is carried out in
the laboratory

20. Percent yield
• The ratio of actual yield to the
theoretical yield for a chemical
reaction expressed as a percentage;
a measure of the efficiency of a
reaction

21. Excess reagent
• A reagent present in a quantity that is
more than sufficient to react with the
limiting reagent; any reactant that
remains after the limiting reagent is
used up in a chemical reaction

22. Calorimeter
• an apparatus for measuring the amount
of heat involved in a chemical reaction

23. Sample problems
mole to mole
• I react 0.1 mol of KI with excess
Pb(NO3)2, how many moles of each
product will I get?
• COPY DOWN MY “ROAD MAP”

24. Roadmap
Gram given 1 mole given mole desired gram desired =
gram given mole given 1 mole desired
We might be
given moles,
mass, volume or
# of particles.
Use mole
ratio from the
balanced
equation.

25. Roadmap cont.
Gram given 1 mole given mole desired gram desired =
gram given mole given 1 mole desired
These values are found from the
periodic table. Be sure to put
correct labels on everything. Your
teacher is very picky about this.

26. BACK TO THE PROBLEM
• I react 0.1 mol of KI with excess
Pb(NO3)2, how many moles of each
product will I get?
• WRITE THE EQUATION &
BALANCE IT.

27. Here is the equation
• I react 0.1 mol of KI with excess
Pb(NO3)2, how many moles of each
product will I get?
• Pb(NO3)2 + KI  PbI2 + KNO3

28. Now it is balanced
• I react 0.1 mol of KI with excess
Pb(NO3)2, how many moles of each
product will I get?
• Pb(NO3)2 + 2KI  PbI2 + 2KNO3

29. Put in the given information
• I react 0.1 mol of KI with excess
Pb(NO3)2, how many moles of each
product will I get?
• Pb(NO3)2 + 2KI  PbI2 + 2KNO3
0.1 molexcess ? mol ? mol
Write what you are given and want to find
above the reaction.

30. Solve it using dim. Analysis
SHOW ALL WORK & LABELS
• Pb(NO3)2 + 2KI  PbI2 + 2KNO3
0.1 molexcess ? mol ? mol
0.1 mole KI 1 mol PbI2 =
2 mole KI
0.1 mole KI 2 mol KNO3 =
2 mole KI

31. Sample problem gram to gram
• I react 1.2 grams of zinc with
excess HCl, How many
moles of each product will I
get?
• Let’s do this on the board

32. In class practice
• Spring 2015 packet page 3 (get on
it in class) (adjusted to did # 4 in class, do 1-3 & 2nd
part of #4)
• Then read pgs 279- 287 do pg 287
#1,2,4,5

34. Limiting Reagents
• In a chemical reaction, an insufficient
quantity of any of the reactants will
limit the amount of product that forms.
– The limiting reagent is the reagent that
determines the amount of product that can
be formed by a reaction.

35. Limiting Reactant
• You will be given various amounts and
you need to see which one is in excess.
Then use the limiting reagent to
answer the questions.
• To start with, I suggest picking one
reactant and going to the other one.
Day 2

36. Sample limiting
• Pb(NO3)2 + 2KI  PbI2 + 2KNO3
25 g35.00 g ? grams
35.00 g Pb(NO3)2 1 mol Pb(NO3)2 2 mol KI 166g KI = g KI
331.22 g Pb(NO3)2 1 mol Pb(NO3)2 1 mol KI
Our answer would be the required amount of KI
needed. Do we have enough?

37. Solve the problem using the “limiter”
• Pb(NO3)2 + 2KI  PbI2 + 2KNO3
25 g35.00 g ? grams
25 g KI 1 mol KI 1 mol PbI2 461 g PbI2 = g PbI2
166g KI 2 mol KI 1 mol PbI2

38. Limiter Trick
Mass of Reactant
(F.W)(coefficient)

39. Go back to this
• Pb(NO3)2 + 2KI  PbI2 + 2KNO3
25 g35.00 g ? grams
35.00g
(331.22g / mol)(1mole)
25g
(166g / mol)(2mole)

40. Read pg 296-304
• Let’s look at this example.

41. Sample
Worksheet Limiters (answers available)
•#56 pg 315
•WS

43. Limiter WS Solutions

44. Lim ans

45. Lim ans

47. Percent Yield
• The theoretical yield is the maximum
amount of product that could be formed
from given amounts of reactants.
• In contrast, the amount of product that
actually forms when the reaction is
carried out in the laboratory is called the
actual yield.
Day 2 or 3

48. The percent yield is the ratio of the actual
yield to the theoretical yield expressed as
a percent.

51. You try

52. Try percent problems in packe
• # 4 is more challenging.

53. Quiz
• Complete the rest of the limiter
worksheet
• Mini-quiz

54. Stoich of Heat
• Sec 10.3 SEE PG 334 EX 10.5
• 29-31 pg 354

55. • Heat content or enthalpy is denoted
by ∆H. The ∆H associated with an
equation tells you the amount of heat
either absorbed or released when a
chemical reaction occurs.

56. • ∆H can either be negative or positive.
The sign on ∆H lets you know if the
reaction is endothermic or exothermic in
nature. Recall that endothermic
reactions are those that absorb energy
from the surroundings. Exothermic
reactions are those that give off energy
in the form of heat.

57. • Exothermic Reaction:
2 B (s) + 3 H2 (g) + 3 H2O (g) -->
B2H6 (g) + 3 H2O (l) ∆H = -96 kJ
• When the equation is written this way, it shows that
the reaction is exothermic as the sign on ∆H is
negative.
• Alternatively, you may see the equation written in this
manner:
2 B (s) + 3 H2 (g) + 3 H2O (g) -->
B2H6 (g) + 3 H2O (l) + 96 kJ

58. • Endothermic Reaction:
4 NO2 (g) + 6 H2O (l) 
4 NH3 (g) + 7 O2 (g) ∆H = 1396 kJ
• Since the sign on ∆H is positive, we can
tell we're looking at a reaction which is
endothermic. (we need to ADD this
amount of heat to make it go)

59. • Another way this can be represented is:
• 1396 kJ + 4 NO2 (g) + 6 H2O (l)  4 NH3 (g) + 7 O2 (g)
• In this case, the amount of energy is
embedded in the equation as a reactant,
indicating that energy must be absorbed or
taken in so that the reaction can occur.

60. 2 B (s) + 3 H2 (g) + 3 H2O (g) 
B2H6 (g) + 3 H2O (l) ∆H = -96 kJ
• Ex: How much energy is released when 73.2 g of
B react as illustrated above?
73.2gB x
1molB
10.8gB
x
96kJ
2molB
= 325kJ energy evolved

61. Let’s try a couple
• How much heat is given off when 450.0 grams of
Propane are burned completely as shown in this
reaction?
• C3H8 + O2  CO2 + H2O ΔH = -2,220.1 kJ/mol
• TRY THESE ON THE WORKSHEET (Answers provided)
• THEN # 29-31 pg 354

62. Try the Stoich WS
• 3 problems in class.
• THEN worksheet Some like it hot
(homework) 1-3.
• Review for a stoich test.

63. Challenge
• Heat of dissolving NaOH calculate the
final temperature that can be reached.

64. Stoich of Heat

65. Stoich of Heat
• Cheeto Lab
– Heat lost= heat gained.
– Calories are a unit of heat
– 1000 calories = 1 Kilocalorie (Calories)
– Calculate Calories

66. Be able to
• Convert to and from
• Mole ratio: mole-mole
• Mass: g-g
• Volume gas @stp
• Heat kJ
• % yield and limiters
• Test has 5 questions.