1. III. Stoichiometry Stoy – kee – ahm –eh - tree
Chapter 12

2. Sections Click the section to jump to the slides
Mole Ratios
Mole-to-Mole Calculations
Mole-to-Mass Calculations
Particle Calculations
Molar Volume Calculations
Limiting Reactants
Percent Yields

3. Things you should remember
From the Moles Unit:
Identify particles as atoms, molecules (mc), and formula units (fun)
1 mole = 6.02 x 1023 atoms, molecules, or formula units
1 mole substance = mass (in grams) from the periodic table
From the Naming & Formulas Unit:
How to write a formula given a chemical name
From the Chemical Reactions Unit:
How to write a chemical equation given words
Balancing equations

4. I. Stoichiometry stoikheion, meaning element metron, meaning measure
Thus Stoichiometry- measuring elements!

5. III. Stoichiometry
Stoichiometry (sometimes called reaction stoichiometry to specify its use in analyzing a chemical reaction) is the calculation of quantitative (numbers) relationships between the reactants and products in a balanced chemical reaction.

6. A. Basis for Calculations
The basis for properly working stoichiometry problems is the Balanced Chemical Equation and the Mole Ratio. These are VITAL TO YOUR SURVIVAL IN STOICHIOMETRY!!!

7. 1. Review of a Balanced Chemical Equation
You must always check to ensure you have a proper chemical equation. It is not correctly balanced you can not use it for any calculations!!!!

8. Ex: Properly Balance the following equation2
__H2SO4(aq) + __NaHCO3(s) 
__Na2SO4(aq) + __H2O(l) + __CO2(g) 2 2

9. 2. Molar Ratio
Once the equation is properly balanced, there are relationships between all of the compounds involved. These are called MOLE RATIOS.
any two compounds can be written as a relationship in terms of moles

10. For the Reaction: H2SO4 + 2NaHCO3 Na2SO4 + 2H2O + 2CO2
What’s the relationship between NaHCO3 and Na2SO4?
2 mole NaHCO3 = 1 mol Na2SO4
(it’s just the coefficients!!)
What’s the relationship between H2SO4 and CO2?
1 mol H2SO4 = 2 mol CO2

11. For the Reaction: H2SO4 + 2NaHCO3 Na2SO4 + 2H2O + 2CO2
1 mol H2SO4 = 2 mol NaHCO3 or
1 mc H2SO4 = 2 mc NaHCO3
1 mol/mc H2SO4
2 mol/mc NaHCO3
1 mole/mc H2SO4

12. For the Reaction: H2SO4 + 2NaHCO3 Na2SO4 + 2H2O + 2CO2
1 mol H2SO4 = 2 mol CO2 or
1 mc H2SO4 = 2 mc CO2
1 mol/mc H2SO4
2 mol/mc CO2
1 mole/mc H2SO4

13. Example: Iron reacts with oxygen to create iron(III) oxide.
After writing a balanced equation, write down 3 possible relationships. (hint: just look at the coefficients.)
Skeletal equation:
Fe + O2  Fe2O3
Balanced equation:
4Fe + 3O2  2Fe2O3
Relationships:
4 mol Fe = 3 mol O2
4 mol Fe = 2 mol Fe2O3
3 mol O2 = 2 mol Fe2O3

14. Practice 6 2 3 Al2S3 + H2O  Al(OH)3 + H2S 1 mol Al2S3 = 6 mol H2O
1. Write three mole ratios (relationships) from the reaction below:
Al2S H2O  Al(OH) H2S 6 2 3
1 mol Al2S3 = 6 mol H2O
1 mol Al2S3 = 2 mol Al(OH)3
1 mol Al2S3 = 3 mol H2S
You should have 3
6 mol H2O = 2 mol Al(OH)3
6 mol H2O = 3 mol H2S
2 mol Al(OH)3 = 3 mol H2S

15. Practice
1. Write three mole ratios (relationships) from the reaction below:
Al2S H2O  Al(OH) H2S 6 2 3
1 mol Al2S3 = 6 mol H2O
1 mol Al2S3 = 2 mol Al(OH)3
1 mol Al2S3 = 3 mol H2S
6 mol H2O = 2 mol Al(OH)3
6 mol H2O = 3 mol H2S

16. Practice 2Al2O3  4Al + 3O2 2 mol Al2O3 = 4 mol Al
2. Aluminum is produced by decomposing aluminum oxide into aluminum and oxygen.
a. Write a balanced equation.
b. Write all the molar ratios that can be derived from this equation.
2Al2O3  4Al + 3O2
2 mol Al2O3 = 4 mol Al
2 mol Al2O3 = 3 mol O2
4 mol Al = 3 mol O2

17. NOTE:
NOTE: Every time you do a stoichiometric calculation you MUST use a mole ratio. The mole ratio allows you to compare one compound in an equation with another. Don’t forget to balance your chemical equations!!!

18. B. Calculating Problems

19. B. Calculating Problems - Stoich it up!
Before any stoich problem you have to set it up. Consider this the pre-game warm-up. This should become second nature to you.

20. Pre-game Warm-up: 1. Write a balanced reaction.
2. Determine your given & want
3. Determine your relationships. If you see…
2 different substances, determine their mole ratio
mass (g, mg, kg), calculate the molar mass of that substance
atoms, molecules, or fun, 1 mol = 6.02 x 1023 of that type
extras like mg or kg, you know what to do
Now you are ready to solve- IT’S GAME TIME.

21. Game Time: put your GIVEN OVER 1
place your relationships where the units cancel out diagonally
everything equal to each other goes above and below each other
cancel out your units until you are left over with your wanted

22. Here is a flow chart that we will dissect this unit to do our problems

23. For most of the examples we will be using this equation:
Haber Process:
N2 (g) H2 (g)  2NH3 (g)
Haber Process: an industrial process for producing ammonia from nitrogen and hydrogen by combining them under high pressure in the present of an iron catalyst
source: worldnet.princeton.edu

24. 1. Moles to Moles

25. 4.00 mol H2 2 mol NH3 = mol NH3 2.67 3 mol H2 1 Balanced Equation:
Ex 1: How many moles of ammonia are produced from 4.00 moles of hydrogen gas in the presence of excess nitrogen?
Balanced Equation:
N2 (g) H2 (g)  2NH3 (g)
No grams, No molar mass!!
Given :
Want :
Relationships:
4.00 mol H2
No fun, mc, No 6.02 x 1023!!
? mol NH3
2 mol NH3 = 3 mol H2
4.00 mol H2
2 mol NH3
= mol NH3
2.67 x
3 mol H2 1

26. 7.8 mol NH3 1 mol N2 = mol N2 3.9 2 mol NH3 1 Balanced Equation:
Ex 2: How many moles of nitrogen were used if 7.8 moles of ammonia were made in excess hydrogen gas?
Balanced Equation:
N2 (g) H2 (g)  2NH3 (g)
No grams, No molar mass!!
Given :
Want :
Relationships:
7.8 mol NH3
No fun, mc, No 6.02 x 1023!!
? mol N2
2 mol NH3 = 1 mol N2
7.8 mol NH3
1 mol N2
= mol N2
3.9 X
2 mol NH3 1

27. Ex 3: How many moles of hydrogen react with 13 moles of nitrogen?
Balanced Equation:
N2 (g) H2 (g)  2NH3 (g)
No grams, No molar mass!!
Given :
Want :
Relationships:
13 mol N2
No fun, mc, No 6.02 x 1023!!
? mol H2
1 mol N2 = 3 mol H2
13 mol N2
3 mol H2
= moles H2 39 x
1 mol N2 1

28. 2. Moles to Mass/ Mass to Moles

29. Steps to Success! 1. Write a balanced reaction.
2. Determine your given & want
3. Determine your relationships. If you see…
2 different substances, determine their mole ratio
mass (g, mg, kg), calculate the molar mass of that substance
atoms, molecules, or f.un, 1 mol = 6.02 x 1023 of that type
extras like mg or kg, you know what to do

30. Ex 1: How many grams of ammonia are produced from 4
Ex 1: How many grams of ammonia are produced from 4.00 moles of hydrogen gas in excess nitrogen?
Balanced Equation:
N2 (g) H2 (g)  2NH3 (g)
How do I know when to use mole ratios, molar mass or 6.02 x 1023?
Given :
Want :
Relationships:
4.00 mol H2
? g NH3
2 mol NH3 = 3 mol H2
1 mol NH3 = g NH3
Look at the “given” and “want” for clues
No fun, mc, No 6.02 x 1023!!

31. 4.00 mol H2 2 mol NH3 17.031g NH3 3 mol H2 1 mol NH3 1 = g NH3 45.4
Ex 1: How many grams of ammonia are produced from 4 moles of hydrogen gas in excess nitrogen?
Balanced Equation:
N2 (g) H2 (g)  2NH3 (g)
Given :
Want :
Relationships:
4.00 mol H2
? g NH3
2 mol NH3 = 3 mol H2
1 mol NH3 = g NH3
4.00 mol H2
2 mol NH3
17.031g NH3 x x
3 mol H2
1 mol NH3 1
= g NH3
45.4

32. Remember your Steps to Success!
Ex 2: How many moles of ammonia are produced from 4.00 grams of hydrogen gas in excess nitrogen?
Balanced Equation:
N2 (g) H2 (g)  2NH3 (g)
Given :
Want :
Relationships:
4.00 g H2
? mol NH3
2 mol NH3 = 3 mol H2
1 mol H2 = 2.016g H2
Remember your Steps to Success!
No fun, mc, No 6.02 x 1023!!

33. 4.00 g H2 1 mol H2 2 mol NH3 2.016g H2 3 mol H2 1 = mol NH3 1.32
Ex 2: How many moles of ammonia are produced from 4 grams of hydrogen gas in excess nitrogen?
Balanced Equation:
N2 (g) H2 (g)  2NH3 (g)
Given :
Want :
Relationships:
4.00 g H2
? mol NH3
2 mol NH3 = 3 mol H2
1 mol H2 = 2.016g H2
4.00 g H2
1 mol H2
2 mol NH3 x x
2.016g H2
3 mol H2 1
= mol NH3
1.32

34. Practice
1. How many grams of nitrogen will react with 3.40 moles of hydrogen to produce ammonia?
2. How many grams of hydrogen are required to make 54.0 moles of ammonia in excess nitrogen?
3. How many moles of nitrogen react completely with 3.70 moles of hydrogen?
31.7 g N2
163 g H2
1.23 mol N2

35. 3. Mass to Mass

36. Remember your Steps to Success!
Ex 1: How many grams of ammonia are produced from 4.00 grams of hydrogen gas in excess Nitrogen?
Balanced Equation:
N2 (g) H2 (g)  2NH3 (g)
Given :
Want :
Relationships:
4.00 g H2
? g NH3
2 mol NH3 = 3 mol H2
1 mol H2 = 2.016g H2
Remember your Steps to Success!
1 mol NH3 = g NH3

37. Ex 1: How many grams of ammonia are produced from 4 grams of hydrogen gas in excess Nitrogen?
Balanced Equation:
N2 (g) H2 (g)  2NH3 (g)
Given :
Want :
Relationships:
4.00 g H2
? g NH3
2 mol NH3 = 3 mol H2
1 mol H2 = 2.016g H2
1 mol NH3 = g NH3
4.00 g H2
1 mol H2
2 mol NH3
17.031g NH3 x x x
2.016g H2 1
3 mol H2
1 mol NH3
= g NH3
22.5

38. Ex 2: How many grams of Nitrogen react with 12.0 grams of Hydrogen?
Balanced Equation:
N2 (g) H2 (g)  2NH3 (g)
Given :
Want :
Relationships:
12.0 g H2
? g N2
1 mol H2 = 2.016g H2
Remember your Steps to Success!
1 mol N2 = g N2

39. Ex 2: How many grams of Nitrogen react with 12.0 grams of Hydrogen?
Balanced Equation:
N2 (g) H2 (g)  2NH3 (g)
Given :
Want :
Relationships:
12.0 g H2
? g N2
1 mol N2 = 3 mol H2
1 mol H2 = 2.016g H2
1 mol N2 = g N2
12.0 g H2
1 mol H2
1 mol N2
28.014g N2 x x x
2.016g H2 1
3 mol H2
1 mol N2
= g N2
55.6

40. Practice Determine the mass of NH3 produced from 280 g of N2.
What mass of nitrogen is needed to produce 100. kg of ammonia?
340 g NH3
8.22 x 104 g N2

41. 3. MC/F.UN/ Atoms to Mass

42. Ex 1: How many grams of ammonia are produced from 2
Ex 1: How many grams of ammonia are produced from 2.09 x 1014 mc of hydrogen gas?
Balanced Equation:
N2 (g) H2 (g)  2NH3 (g)
Given :
Want :
Relationships:
2.09 x 1014 mc H2
? g NH3
2 mol NH3 = 3 mol H2
1 mol NH3 = g NH3
1 mol H2 = 6.02 x 1023 mc H2

43. Ex 1: How many grams of ammonia are produced from 2
Ex 1: How many grams of ammonia are produced from 2.09 x 1014 mc of hydrogen gas?
Balanced Equation:
N2 (g) H2 (g)  2NH3 (g)
Given :
Want :
Relationships:
2.09 x 1014 mc H2
? g NH3
2 mol NH3 = 3 mol H2
1 mol H2 = 6.02 x 1023 mc H2
1 mol NH3 = g NH3
1 mol H2
2 mol NH3
2.09 x 1014 mc H2
17.031g NH3 x x x
3 mol H2 1
6.02 x 1023 mc H2
1 mol NH3
= g NH3
3.94 x 10-9

44. Ex 2: How many molecules of ammonia are produced from 40
Ex 2: How many molecules of ammonia are produced from 40.2 g of H2 in the presence of excess N2?
Balanced Equation:
N2 (g) H2 (g)  2NH3 (g)
Given :
Want :
Relationships:
40.2g H2
? mc NH3
2 mol NH3 = 3 mol H2
1 mol H2 = 2.016g H2
1 mol NH3 = 6.02 x 1023 mc NH3

45. Ex 2: How many molecules of ammonia are produced from 40
Ex 2: How many molecules of ammonia are produced from 40.2 g of H2 in the presence of excess N2?
Balanced Equation:
N2 (g) H2 (g)  2NH3 (g)
Given :
Want :
Relationships:
40.2g H2
? mc NH3
2 mol NH3 = 3 mol H2
1 mol H2 = 2.016g H2
1 mol NH3 = 6.02 x 1023 mc NH3
40.2g H2
1 mol H2
2 mol NH3
6.02 x 1023 mc NH3 x x x
2.016g H2
3 mol H2 1
1 mol NH3
= mc NH3
8.00 x 1024

46. Practice
How many mc of nitrogen will react with exactly 7.04 grams of hydrogen?
How many mc of ammonia will x 1021 mc of N2 produce in excess hydrogen?
7.01 x 1023 mc N2
4.18 x 1021 mc NH3

47. 1) How many mc of nitrogen will react with exactly 7
1) How many mc of nitrogen will react with exactly 7.04 grams of hydrogen?
Balanced Equation:
N2 (g) H2 (g)  2NH3 (g)
Given :
Want :
Relationships:
7.04 g H2
? mc N2
1 mol N2 = 3 mol H2
1 mol H2 = 2.016g H2
1 mol N2 = 6.02 x 1023 mc N2
7.04 g H2
1 mol H2
1 mol N2
6.02 x 1023 mc N2 x x x
2.016g H2
3 mol H2 1
1 mol N2
= mc N2
7.01 x 1023

48. How many mc of ammonia will 2
How many mc of ammonia will x 1021 mc of N2 produce in excess hydrogen?
Balanced Equation:
N2 (g) H2 (g)  2NH3 (g)
Given :
Want :
Relationships:
2.09 x 1013 mc N2
? mc NH3
2 mol NH3 = 1 mol N2
1 mol N2 = 6.02 x 1023 mc N2
1 mol NH3 = 6.02 x 1023 NH3
2.09 x 1013 mc N2
1 mol N2
2 mol NH3
6.02 x 1023 mc NH3 x x x 1
1 mol N2
6.02 x 1023 mc N2
1 mol NH3
= mc NH3
4.18 x 1021

49. 4. Molar Volume of Gases (Liters)

50. 4. Molar Volume of Gases (liters)
There is one new conversion: 1 mol of an ideal gas = 22.4 L of an ideal gas. For now we will assume every gas is an ideal gas at STP. You follow the same steps we have been following all along to solve this problem. In a reaction involving gases, Avogadro found that their volumes combine in whole number ratios. In other words, the coefficients in the chemical reaction also represent volume.

51. For gases: When you see liters…
1 mol __ = 22.4 L __

52. Remember your Steps to Success!
Ex 1: How many moles of nitrogen gas are needed to react with 44.8 liters of hydrogen gas to produce ammonia gas?
Balanced Equation:
N2 (g) H2 (g)  2NH3 (g)
Given :
Want :
Relationships:
44.8 L H2
? mol N2
1 mol N2 = 3 mol H2
1 mol H2 = 22.4 L H2
Remember your Steps to Success!
No grams, No molar mass
No fun, mc, No 6.02 x 1023!!

53. 44.8 L H2 1 mol H2 1 mol N2 = mol N2 0.667 22.4 L H2 3 mol H2 1
Ex 1: How many moles of nitrogen gas are needed to react with 44.8 liters of hydrogen gas to produce ammonia gas?
Balanced Equation:
N2 (g) H2 (g)  2NH3 (g)
Given :
Want :
Relationships:
44.8 L H2
? mol N2
1 mol N2 = 3 mol H2
1 mol H2 = 22.4 L H2
44.8 L H2
1 mol H2
1 mol N2
= mol N2
0.667 x x
22.4 L H2
3 mol H2 1
Given over 1
1 mol = 22.4 L
Mole ratio

54. Remember your Steps to Success!
Ex 2: If 5.00 moles of H2 react with excess N2, how many liters of NH3 are produced?
N2 (g) H2 (g)  2NH3 (g)
Given :
Want :
Relationships:
5.00 mol H2
? L NH3
2 mol NH3 = 3 mol H2
1 mol NH3 = 22.4 L NH3
Remember your Steps to Success!
No grams, No molar mass
No fun, mc, No 6.02 x 1023!!

55. Ex 2: If 5.00 moles of H2 react with excess N2, how many liters of NH3 are produced?
N2 (g) H2 (g)  2NH3 (g)
Given :
Want :
Relationships:
5.00 mol H2
? L NH3
2 mol NH3 = 3 mol H2
1 mol NH3 = 22.4 L NH3
5.00 mol H2
2 mol NH3
22.4 L NH3
= L NH3
74.7 x x
3 mol H2
1 mol NH3 1
Given over 1
Mole ratio
1 mol = 22.4 L

56. II. Limiting Reactants and Excess Reactants

57. “which reactant are you going to run out of first?”
In reality, a scientist does not always add chemicals in perfect proportions. There is usually one reactant that is in “excess”.
The limiting reactant or limiting reagent is the reactant that limits the amounts of the other reactants that can combine and the amount of product that can form in a chemical reaction.
“which reactant are you going to run out of first?”

58. “which reactant do you have extra of?”
The substance that is not used up completely in a reaction is sometimes called the excess reactant (excess)
Once one of the reactants is used up, no more product can be formed
“which reactant do you have extra of?”

59. Which is the limiting reactant+ 4 4 13
Which is the limiting reactant
4 shells
1 car
= 4 cars can be made if 4 shells are available
1 shell
13 tires
1 car
= 3.25 cars can be made if 13 tires are available
4 tires
Less product can be produced, therefore tires must be the limiting reactant

60. Activity
Write a recipe for the perfect burger

61. Ex. Consider the following balanced equation: Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g) If 1.21 moles of zinc are added to 2.64 moles of HCl, which reactant is in excess, and which is the limiting reactant?
Available
Needed
1.21 mol Zn
2 mol HCl
= 2.42 mol HCl
1 mol Zn
2.64 mol HCl
1 mol Zn
= 1.32 mol Zn
2 mol HCl

62. < > Limiting reactant Excess reactant Available VS Needed
Ex. Consider the following balanced equation: Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g) If 1.21 moles of zinc are added to 2.64 moles of HCl, which reactant is in excess, and which is the limiting reactant?
Available VS Needed
<
1.21 mol Zn mol Zn
Limiting reactant
>
2.64 mol HCl mol HCl
Excess reactant

63. To find the limiting reactant and the excess reactant:
1) Determine the balanced chemical equation
2) Convert the given/available amounts reactants to the number of moles of the other reactant(s)
3) Compare the available amounts to the needed amounts:
*If the available amount > needed amount, it is excess reactant
* If the available amount < needed amount, it is the limiting reactant
4) To calculate the amount of excess:
-Available Amount – Needed Amount = Amount of Excess
5) To calculate the amount of product produced:
-G: amount of available limiting reactant
-W: amount of product

64. Example
Some rocket engines use a mixture of hydrazine, N2H4, and hydrogen peroxide, as the propellant. The products formed are nitrogen gas and steam. Which is the limiting reactant when 0.75 mol of N2H4 is mixed with 0.50 mol of H2O2? How much excess reactant, in moles, remains unchanged? How much of each product, in moles, is formed?

65. Step 1: Write a balanced reaction
Some rocket engines use a mixture of hydrazine, N2H4, and hydrogen peroxide, as the propellant. The products formed are nitrogen gas and steam.
N2H4
+ H2O2 2 
N2(g)
+ H2O(g) 4

66. N2H4
+ H2O2 2 
N2(g)
+ H2O(g) 4
Which is the limiting reactant when mol of N2H4 is mixed with mol of H2O2?
Available
Needed
Limiting Reactant
0.75 mol N2H4
2 mol H2O2
= 1.50 mol H2O2
1 mol N2H4
Excess
0.50 mol H2O2
1 mol N2H4
= 0.25 mol N2H4
2 mol H2O2

67. To find the limiting reactant and the excess reactant:
4. To calculate the amount of excess:
-Available Amount – Needed Amount = Amount of Excess

68. *Available – Needed = Remaining excess
How much excess reactant, in moles, remains unchanged?
Available
Needed
0.75 mol N2H4
2 mol H2O =
1.50 mol H2O2
1 mol N2H4
0.50 mol H2O2
1 mol N2H =
0.25 mol N2H4
Excess
2 mol H2O2 -
= 0.50 mol N2H4
*Available – Needed = Remaining excess

69. To find the limiting reactant and the excess reactant:
5. To calculate the amount of product produced:
-G: amount of available limiting reactant
-W: amount of product

70. How many moles of product are formed?
N2H4
+ H2O2 2 
N2(g)
+ H2O(g) 4
G: W: R:
0.50 mol H2O2
G: amount of available limiting reactant
mole N2
N2 is the first product
2 mol H2O2 = 1 mol N2
0.50 mol H2O2
1 mol N2
= 0.25 mol N2
2 mol H2O2
G: W: R:
0.50 mol H2O2
G: amount of available limiting reactant
H2O is the second product
mole H2O
2 mol H2O2 = 4 mol H2O
0.50 mol H2O2
4 mol H2O
= 1.0 mol H2O
2 mol H2O2

71. Balanced Equation: 8 Zn + S8  8 ZnS
Ex. Zinc metal and solid sulfur (S8) react to form solid zinc sulfide. If 2.0 moles of zinc are heated with 2.0 mol S8, identify the limiting reactant. How many moles of excess reactant remain. How many moles of product are formed?
Balanced Equation: 8 Zn + S8  8 ZnS
Available
Needed
2.0 mol Zn
1 mol S8
= 0.25 mol S8
8 mol Zn
1.0 mol S8
8 mol Zn
= 8.0 mol Zn
1 mol S8

72. < > Limiting reactant Excess reactant Available VS Needed
Ex. Zinc metal and solid sulfur (S8) react to form solid zinc sulfide. If 2.0 moles of zinc are heated with 1.0 mol S8, identify the limiting reactant. How many moles of excess reactant remain. How many moles of product are formed?
Available VS Needed
<
2.0 mol Zn mol Zn
Limiting reactant
>
1.0 mol S mol S8
Excess reactant

73. How much excess is present?
Ex. Zinc metal and solid sulfur (S8) react to form solid zinc sulfide. If 2.0 moles of zinc are heated with 1.0 mol S8, identify the limiting reactant. How many moles of excess reactant remain. How many moles of product are formed?
How much excess is present?
(Available Amount – Needed Amount = Amount of Excess)
1.0 mol S8 – 0.25 mol S8 = 0.75 mol S8 in excess

74. To find the limiting reactant and the excess reactant:
4. To calculate the amount of excess:
-Available Amount – Needed Amount = Amount of Excess
5. To calculate the amount of product produced:
-G: amount of available limiting reactant
-W: amount of product

75. Balanced Equation: 8 Zn + S8  8 ZnS How much product was formed?
Ex. Zinc metal and solid sulfur (S8) react to form solid zinc sulfide. If two moles of zinc are heated with 1.00 mol S8, identify the limiting reactant. How many moles of excess reactant remain. How many moles of product are formed?
Balanced Equation: 8 Zn + S8  8 ZnS
How much product was formed?
G: W: R:
2 mol Zn
mol ZnS
8 mol Zn = 8 mol ZnS
2 mol Zn
8 mol ZnS
= 2 mol ZnS formed
8 mol Zn

76. Balanced Equation: C + H2O  H2 + CO
Practice
1. Carbon reacts with steam at high temperatures to produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam, identify the limiting reactant. How many moles of each product are formed?
Balanced Equation: C + H2O  H2 + CO
Available
Needed
2.4 mol C
1 mol H2O
= 2.4 mol H2O
1 mol C
3.1 mol H2O
1 mol C
= 3.1 mol C
1 mol H2O

77. < > Limiting reactant Excess reactant Available VS Needed
Practice 1. Carbon reacts with steam at high temperatures to produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam, identify the limiting reactant. How many moles of each product are formed?
Available VS Needed
<
2.4 mol C mol C
Limiting reactant
>
3.1 mol H2O mol H2O
Excess reactant

78. To find the limiting reactant and the excess reactant:
4. To calculate the amount of excess:
-Available Amount – Needed Amount = Amount of Excess
5. To calculate the amount of product produced:
-G: amount of available limiting reactant
-W: amount of product

79. Balanced Equation: C + H2O  H2 + CO
Carbon reacts with steam at high temperatures to produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam, identify the limiting reactant. How many moles of each product are formed?
Balanced Equation: C + H2O  H2 + CO
G: W: R:
2.4 mol C
G: amount of available limiting reactant
mole H2
H2 is the first product
1 mol C = 1 mol H2
2.4 mol C
1 mol H2
= 2.4 mol H2
1 mol C
G: W: R:
2.4 mol C
G: amount of available limiting reactant
CO is the second product
mole CO
1 mol C = 1 mol CO
2.4 mol C
1 mol CO
= 2.4 mol CO
1 mol C

80. Set VI: Reactions 2) Zn + Pb(NO3)2  Pb + Zn(NO3)2
3) Fe + 2HCl  H2 + FeCl2

81. III. Theoretical yield, Actual yield and Percent yield

82. C. Theoretical yield, Actual yield and Percent yield
The yield of a chemical reaction is the quantity of product one obtains from a given ratio of reactants.
The actual yield of a chemical reaction is the mass of the compound that you actually recover when you are done with the reaction.
The actual yield is also referred to as the experimental yield
The theoretical yield is the mass of compound you should obtain (theoretically) if everything goes perfectly. In all of the examples above we have been pretending that everything is perfect. All of our wanteds have been theoretical.

83. Ex 1: What is the theoretical yield of Na2SO4 in grams if 35 moles of NaOH is reacted with sufficient H2SO4?
Yield means product
Theoretical yield means mathematically, what should you get? (this is why we do stoichiometry calculations)

84. Ex 1: What is the theoretical yield of Na2SO4 in grams if 3
Ex 1: What is the theoretical yield of Na2SO4 in grams if 3.50 moles of NaOH is reacted with sufficient H2SO4?
2NaOH + H2SO4  Na2SO4 + 2H2O G: W: R:
3.50 mol NaOH
2 different substances
? g Na2SO4
You see g Na2SO4
2 mol NaOH = 1 mol Na2SO4
1 mol Na2SO4 = g Na2SO4
3.50 mol NaOH
1 mol Na2SO4
g Na2SO4 x x
2 mol NaOH
1 mol Na2SO4 1
Given over 1
Mole ratio
1 mol = g
= g Na2SO4
249

85. If they give you the actual yield and you figure out the theoretical yield, you can find the percent yield.
Actual Yield
Theoretical Yield
X 100
= % Yield
Same as…
Actual Yield
Theoretical Yield
% Yield
100 =

86. Your percent yield should not be greater than 100 because the theoretical yield is the MAXIMUM yield you can have.

87. Example: From the examples above, if 221 grams of sodium sulfate were actually collected, what is the percent yield?
From the experiment
221 g NaSO4
249 g NaSO4
X 100
= 88.8%
From the calculation

88. Practice
A student conducts a single displacement reaction that produces 2.75 grams of copper. Mathematically he determines that grams of copper should have been produced. Calculate the student's percentage yield.
A student completely reacts 5.00 g of magnesium with an excess of oxygen to produce magnesium oxide. Analysis reveals 8.10 g of magnesium oxide. What is the student's percentage yield? (*Hint: you must first find the theoretical yield.)
87.3%
97.7%

89. Practice Actual yield: 93.7 g
3. Consider the reaction below of zinc and hydrochloric acid to produce zinc chloride and hydrogen gas.
Zn + 2HCl  H2 +ZnCl2
If a sufficient mass of zinc metal combines with 58.0 grams of HCl, what is the theoretical yield of ZnCl2? What is the actual yield in grams if the percent yield is 87%.
Actual yield: 93.7 g

90. Practice 1
A student conducts a single displacement reaction that produces 2.75 grams of copper. Mathematically he determines that 3.15 grams of copper should have been produced. Calculate the student's percentage yield.
Actual from the experiment
2.75 g Cu
X 100
= 87.3 %
3.15 g Cu
calculated

91. actual X 100 = % yield Theo. Practice 2
A student completely reacts 5.00g of magnesium with an excess of oxygen to produce magnesium oxide. Analysis reveals 8.10 g of magnesium oxide. What is the student's percentage yield? (*Hint: you must first find the theoretical yield.)
Actual yield
actual
X 100
= % yield
Theo.

92. To calculate the theoretical yield
2Mg + O2  2 MgO G: W: R:
5.00 g Mg
2 different substances
g MgO
You see g Mg and g MgO
2 mol Mg = 2 mol MgO
1 mol Mg = g Mg
1 mol MgO = g MgO
5.00 g Mg
1 mol Mg
2 mol MgO
g MgO x x x
g Mg
2 mol Mg 1
1 mol MgO
Theoretical yield= g MgO
8.29

93. Practice 2
A student completely reacts 5.00g of magnesium with an excess of oxygen to produce magnesium oxide. Analysis reveals 8.10 g of magnesium oxide. What is the student's percentage yield? (*Hint: you must first find the theoretical yield.)
Actual yield
8.10 g MgO
X 100
= 97.7%
8.29 g MgO

94. Practice 3 Actual yield: 93.7 g
3. Consider the reaction below of zinc and hydrochloric acid to produce zinc chloride and hydrogen gas.
Zn + 2HCl  H2 +ZnCl2
If a sufficient mass of zinc metal combines with 58.0 grams of HCl, what is the theoretical yield of ZnCl2? What is the actual yield in grams if the percent yield is 87%.
Actual yield: 93.7 g

95. G: 58.0 g HCl W: R: g ZnCl2 Zn + 2HCl  H2 +ZnCl2
If a sufficient mass of zinc metal combines with 58.0 grams of HCl, what is the theoretical yield of ZnCl2? What is the actual yield in grams if the percent yield is 87%.
Zn + 2HCl  H2 +ZnCl2 G: W: R:
58.0 g HCl
2 different substances
g ZnCl2
You see g HCl & g ZnCl2
2 mol HCl = 1 mol ZnCl2
1 mol HCl = g HCl
1 mol ZnCl2 = g ZnCl2
58.0 g HCl
1 mol HCl
1 mol ZnCl2
g ZnCl2 x x x
g HCl
2 mol HCl 1
1 mol ZnCl2
Theoretical yield= g ZnCl2
108

96. Practice 3
If a sufficient mass of zinc metal combines with 58.0 grams of HCl, what is the theoretical yield of ZnCl2? What is the actual yield in grams if the percent yield is 87%.
Actual from the experiment
93.7 g
actual
X 100
= 87 %
108 g
calculated