1. Stoichiometric Calculations
Stoichiometry – Ch. 11
Stoichiometric Calculations

2. Background on things you NEED to know how to do:
Name/write correct chemical formula
Write chemical equations
Balance chemical equations
Predict Products
Mole/mass conversions

3. Stoichiometry 74 tires 1 bicycle = 37 bicycles 2 tires
Stoichiometry uses ratios to determine relative amounts of reactants or products.
For example If you were to make a bicycle, you would need one frame and two tires.
1 frame + 2 tires  1 bicycle
If I had 74 tires, what is the most # of bicycles I could make?
74 tires
1 bicycle
= 37 bicycles
2 tires

4. Proportional Relationships
2 1/4 c. flour
1 tsp. baking soda
1 tsp. salt
1 c. butter
3/4 c. sugar
3/4 c. brown sugar
1 tsp vanilla extract
2 eggs
2 c. chocolate chips
Makes 5 dozen cookies.
I have 5 eggs. How many cookies can I make?
Ratio of eggs to cookies
5 eggs
5 doz.
2 eggs
= 12.5 dozen cookies

5. Proportional Relationships
Stoichiometry
mass relationships between substances in a chemical reaction
based on the mole ratio
Mole Ratio
indicated by coefficients in a balanced equation
can be used to determine expected amounts of products given amounts of reactants.
2 Mg + O2  2 MgO

6. Stoichiometry Steps 1. Write a balanced equation.
2. Identify known & unknown.
3. Line up conversion factors.
Mole ratio - moles  moles
Molar mass - moles  grams
Molarity - moles  liters soln
Molar volume - moles  liters gas
Mole ratio - moles  moles
Core step in all stoichiometry problems!!
4. Check answer.

7. Conversions LITERS OF GAS AT STP Molar Volume MASS IN GRAMS MOLES
(22.4 L/mol)
MASS IN
GRAMS
MOLES
NUMBER OF
PARTICLES
Molar Mass
(g/mol)
6.02  1023
particles/mol
Molarity (mol/L)
LITERS OF
SOLUTION

8. Mole – Mole Conversions
The first type of problems we encounter will go between moles and moles. For this we need to use mole ratios.
Ex: Write and balance the reaction between lead (II) nitrate and potasium iodide.
Pb(NO3)2 + 2KI 
2 KNO3 + PbI2
Mole ratio of potasium iodide to lead (II) iodide:
2 moles KI
1 mole PbI2

9. Mole to Mole Problems 2KClO3  2KCl + 3O2 9 mol O2 2 mol KClO3
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

10. Mole to Mass We can also convert from moles to mass, and mass to moles
For Example:
4 Al + 3 O2  2Al2O3

11. Mass to Moles: 4 Al + 3 O2  2Al2O3
If the reaction starts with .84 moles of aluminum, how many grams of aluminum oxide are produced?
.84 mol Al
2 mol Al2O3
101.9 grams Al2O3
4 mol Al
1 mol Al2O3
= 42.8 grams Al2O3

12. Mass to Mass Cu + 2AgNO3  2Ag + Cu(NO3)2 12.0 g Cu 1 mol Cu 63.55
How many grams of silver will be formed from 12.0 g copper reacting with silver nitrate?
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

13. Stoichiometry Problems – Mole-Mole
N2H4 + N2O4  N2 + H2O
? moles N2O4 = 2.72 moles N2H4 2 3 4
2.72 mol
? mol
2.72 mol N2 H4
1 mol N2O4
2 mol N2 H4
= 1.36 mol
N2O4

14. Stoichiometry with Gases
If the pressure and temperature are constant, the ratio of moles in the balanced equation is the ratio of liters in an all gas reaction.
At STP 1 mole of gas occupies 22.4 L
1 mole = 22.4 L

15. Molar Volume = 15.7 L H2 17.0 g Mg 1 mol Mg 24.31 g 1 mol H2 Mg
In the following reaction, if 17 g of Mg react, how many L of H2 forms?
Mg (s) + HCl (aq)  MgCl2 (aq) + H2 (g)
17.0 g Mg
1 mol Mg
24.31 g
1 mol H2 Mg
22.4 L H2
1 mol H2
= 15.7 L H2

16. Molar Volume Problems 2KClO3  2KCl + 3O2 ? g 9.00 L 9.00 L O2 1 mol
How many grams of KClO3 are req’d to produce 9.00 L of O2 at STP?
2KClO3  2KCl + 3O2
? g
9.00 L
9.00 L O2
1 mol O2
22.4 L
2 mol
KClO3
3 mol O2
122.55
g KClO3
1 mol
KClO3
= 32.8 g
KClO3

17. Molarity
Molarity is the number of moles of solute dissolved in one liter of solution.
Units are moles per liter or moles of solute per liter of solution.
Molarity abbreviated by a capital M
Molarity = moles of solute liter of solution

18. Molarity Problems = .43 mol NH4Cl = .145 L = 2.97 M NH4Cl .145 L
As an example, suppose we dissolve 23 g of ammonium chloride (NH4Cl) in enough water to make 145 mL of solution. What is the molarity of ammonium chloride in this solution?
23 g NH4Cl
1 mole NH4Cl
= .43 mol NH4Cl
53.5 g NH4Cl
145 mL
1 L
= .145 L
1000 mL
.43 mol NH4Cl
= 2.97 M NH4Cl
.145 L

19. Molarity Problems
Now, suppose we have a beaker with 175 mL of a 5.5 M HCl solution. How many moles of HCl is in this beaker?
175 mL
1 L
5.5 mol HCl
= .96 mol HCl
1000 mL
1 L

20. Molar Volume Problems Cu + 2AgNO3  2Ag + Cu(NO3)2 1.5 L .10 mol AgNO3
How many grams of Cu are required to react with 1.5 L of 0.10M AgNO3?
Cu + 2AgNO3  2Ag + Cu(NO3)2
? g
1.5L
0.10M
1.5 L
.10 mol
AgNO3
1 L
1 mol Cu
2 mol
AgNO3
63.55
g Cu
1 mol Cu
= 4.8 g Cu

21. Energy & Stoichiometry

22. Exothermic and Endothermic
Exothermic process – heat is released into the surroundings
Exo = Exit
Endothermic Process – heat is absorbed from the surroundings
Endo = Into
HEAT
HEAT

23. Thermochemical Equations
In a thermochemical equation, the energy of change for the reaction can be written as either a reactant or a product
Enthalpy: the heat content of a system at constant pressure (ΔH)
Endothermic (positive ΔH)
2NaHCO kJ Na2CO3 + H2O + CO2
Exothermic (negative ΔH)
CaO + H2O Ca(OH) kJ

24. Write the thermochemical equation for the oxidation of Iron (III) if its ΔH= kJ Fe(s) + O2(g)→ Fe2O3(s) kJ How much heat is evolved when 10.00g of Iron is reacted with excess oxygen?
Exo 4 3 2
10.00g Fe
1 mol
1652 kJ
=73.97 kJ of heat
55.85g Fe
4 mol Fe

25. Write the thermochemical equation for the decomposition of sodium bicarbonate, with a ΔH = kJ: 2 NaHCO kJ → Na2CO3(s) + H2O + CO2 How much heat is required to break down 50.0g of sodium bicarbonate?
Endo
1 mol NaHCO3
50.0 g NaHCO3
129 kJ
83.9 g NaHCO3
2 mol NaHCO3
=38.4 kJ of heat

26. Write the thermochemical equation for a single replacement of calcium oxide and water with a ΔH= kJ: CaO + H2O → Ca(OH) kJ How much energy is released when 100 g of calcium oxide reacts?
Exo
100 g CaO
1 mol CaO
65.2 kJ
56.07 g CaO
1 mol CaO
=116 kJ of heat

27. Write the thermochemical equation for the decomposition of magnesium oxide with a ΔH= kJ: 2 MgO → 2 Mg + O2 How many grams of oxygen are produced when magnesium oxide is decomposed by adding 420 kJ of Energy?
Endo
420 kJ
1 mol O2
31.98 g O2
=218 g of O2
61.5 kJ
1 mol O2

28. Stoichiometry in the Real World
Stoichiometry – Ch. 11
Stoichiometry in the Real World

29. Limiting Reactants Available Ingredients 4 slices of bread
1 jar of peanut butter
1/2 jar of jelly
Limiting Reactant
bread
Excess Reactants
peanut butter and jelly

30. Limiting Reactants Available Ingredients 24 graham cracker squares
1 bag of marshmallows
12 pieces of chocolate
Limiting Reactant
chocolate
Excess Reactants
Marshmallows and graham crackers

31. Limiting Reactants Limiting Reactant one that is used up in a reaction
determines the amount of product that can be produced
Excess Reactant
added to ensure that the other reactant is completely used up
cheaper & easier to recycle

32. Limiting Reactant Steps
1. Write the balanced equation.
2. For each reactant, calculate the amount of product formed.
3. Smaller answer indicates:
limiting reactant
amount of product actually possible

33. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g ? g 68.1 g
79.1 g of zinc react with 68.1 g HCl. Identify the limiting and excess reactants. How many grams of hydrogen can be formed?
Zn HCl  ZnCl H2
79.1 g
68.1 g
? g

34. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 68.1 g ? g 79.1 g Zn
1 mol Zn
65.39
g Zn
1 mol H2 Zn
2.02 g H2
1 mol
= 2.44 g H2

35. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 68.1 g ? g 68.1 g HCl
1 mol
HCl
g HCl
1 mol H2
2 mol
HCl
2.02 g H2
1 mol H2
= 1.89 g H2

36. Limiting Reactants Zn: 2.44 g H2 HCl: 1.89 g H2 Limiting reactant: HCl
Excess reactant: Zn
Product Formed: 1.89 g H2
left over zinc

37. Limiting Reactants #2 2Mg + O2  2MgO 5.42 g 4.00 g ? g
5.42 g of magnesium ribbon react with 4.00 g of oxygen gas. Identify the limiting and excess reactants. How many grams of magnesium oxide are formed?
2Mg O2  2MgO
5.42 g
4.00 g
? g

38. Limiting Reactants #2 2Mg + O2  2MgO 5.42 g 4.00 g ? g 5.42 g Mg
1 mol Mg
24.31
g Mg
2 mol
MgO Mg
40.31 g
MgO
1 mol
= 8.99 g
MgO

39. Limiting Reactants #2 2Mg + O2  2MgO 5.42 g 4.00 g ? g 4.00 g O2
1 mol O2
g O2
2 mol
MgO
1 mol O2
40.31
g MgO
1 mol
MgO
= 10.1 g MgO

40. A. Limiting Reactants #2 Limiting reactant: Mg Excess reactant: O2
Mg: 8.99 g MgO O2: 10.1 g MgO
Limiting reactant: Mg
Excess reactant: O2
Product Formed: 8.99 g MgO
Excess oxygen

41. Limiting Reactants
What other information could you find in these problems?
How much of each reactant is used – in grams, liters, moles
How much of excess reactant is left over – in grams, liters, moles

42. Percent Yield
measured in lab
calculated on paper

43. Percent Yield 2 mol KCl 74.55 g KCl 1 mol 1 mol 45.8 g K2CO3 K2CO3
When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl.
K2CO HCl  KCl H2CO3
1 mol
K2CO3
138.2 g K2CO3
2 mol
KCl
1 mol
K2CO3
74.55 g
KCl
1 mol
45.8 g
K2CO3
= 49.4 grams KCl

44. 46.3 grams KCl x 100 49.4 grams KCl % yield = 93.7 % Percent Yield
When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl.
46.3 grams KCl
x 100
49.4 grams KCl
% yield = 93.7 %