1. It is a very mathematical part of chemistry.
STOICHIOMETRY
From two Greek words: stoicheion (meaning "element") and metron (meaning "measure").
Stoichiometry deals with quantitative calculations about the masses (sometimes volumes) of reactants and products involved in a chemical reaction.
It is a very mathematical part of chemistry.

2. Branches of Stoichiometry
We have just finished studying mass relationships of elements in compounds (molar mass, converting grams to moles, moles to grams, moles to number of molecules). This branch of stoichiometry is called composition stoichiometry.
Now we are going to learn about reaction stoichiometry which involves the mass relationships between reactants and products in a chemical reaction.

3. Reaction Stoichiometry
Reaction stoichiometry is based upon chemical reactions and the law of conservation of mass.
All reaction stoichiometry calculations start with a BALANCED CHEMICAL EQUATION which gives the relative numbers of moles of reactants and products.

4. Mole Ratio
Solving any reaction stoichiometry problem requires the use of a mole ratio to convert from moles to grams or grams of one substance to moles or grams of another substance in a reaction. A mole ratio is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction. This information is gotten directly from the coefficients of the balanced chemical equation involved.

5. 2 𝑚𝑜𝑙 𝐴𝑙 2 𝑂 3 4 𝑚𝑜𝑙 𝐴𝑙 2 𝑚𝑜𝑙 𝐴𝑙 2 𝑂 3 3 𝑚𝑜𝑙 𝑂 2 4 𝑚𝑜𝑙 𝐴𝑙 3 𝑚𝑜𝑙 𝑂 2
Mole Ratio
Consider the balanced equation for the electrolysis of melted aluminum oxide to produce aluminum metal and oxygen gas.
2 𝐴𝑙 2 𝑂 3 𝑙 →4𝐴𝑙 𝑠 +3 𝑂 2 (𝑔)
Recall that the coefficients represent the relative amounts of moles of reactants and products. These relationships can be expressed in the following mole ratios (or their reciprocals):
2 𝑚𝑜𝑙 𝐴𝑙 2 𝑂 3 4 𝑚𝑜𝑙 𝐴𝑙 𝑚𝑜𝑙 𝐴𝑙 2 𝑂 3 3 𝑚𝑜𝑙 𝑂 𝑚𝑜𝑙 𝐴𝑙 3 𝑚𝑜𝑙 𝑂 2

6. 13.0 𝑚𝑜𝑙 𝐴𝑙 2 𝑂 3 × 4 𝑚𝑜𝑙 𝐴𝑙 (𝑠) 2 𝑚𝑜𝑙 𝐴𝑙 2 𝑂 3 =26.0 𝑚𝑜𝑙 𝐴𝑙 (𝑠)
Example
To solve relevant problems, you will need to choose the appropriate mole ratio for your problem.
For example, to find the amount in moles of aluminum metal that can be produced from 13.0 mol of aluminum oxide, we choose the mole ratio that relates Al (s) to Al2O3.
2 𝐴𝑙 2 𝑂 3 𝑙 →4𝐴𝑙 𝑠 +3 𝑂 2 (𝑔)
13.0 𝑚𝑜𝑙 𝐴𝑙 2 𝑂 3 × 4 𝑚𝑜𝑙 𝐴𝑙 (𝑠) 2 𝑚𝑜𝑙 𝐴𝑙 2 𝑂 3 =26.0 𝑚𝑜𝑙 𝐴𝑙 (𝑠)

7. Ideal Stoichiometric Calculations
Balanced chemical equations let us make predictions about chemical reactions without having to run the reactions in the laboratory.
Keep in mind the reaction stoichiometry calculations are for theoretical reactions that have been run under ideal conditions in which all the reactants have been converted into products.

8. Sample Problem
Ammonia, NH3, is widely used as a fertilizer and in many household cleaners.
How many moles of ammonia are produced when 6 mol of hydrogen gas react with an excess of nitrogen gas? (This means you have as much nitrogen gas as you need to use up all the hydrogen.)

9. Practice Problem Using Cards
Analyze: Given: 6 mol hydrogen gas (reactant) Unknown: moles of ammonia (product) Plan: First you’ll need the balanced chemical equation that relates ammonia (product) to hydrogen and nitrogen (reactants). __𝐻 2(𝑔) + __𝑁 2(𝑔) →__ 𝑁𝐻 3 (s) Next you’ll need the mole ratio (conversion factor) that relates hydrogen gas to ammonia so that you can convert moles of hydrogen gas (given) to moles of ammonia (unknown).

10. Solution to Practice Problem
How many moles of NH3 can be created from 6 moles of H2 gas?
3𝐻 2(𝑔) + 𝑁 2(𝑔) →2 𝑁𝐻 3 (s)
The needed mole ratio between ammonia and hydrogen gas is:
2 𝑚𝑜𝑙 𝑁 𝐻 3 3 𝑚𝑜𝑙 𝐻 2
Calculation:
6 𝑚𝑜𝑙 𝐻 2 × 2 𝑚𝑜𝑙 𝑁 𝐻 3 3 𝑚𝑜𝑙 𝐻 2 =4 𝑚𝑜𝑙 𝑁 𝐻 3
Arrange your card so that moles of the known substance is on the bottom and moles of the unknown is on the top
Think of this as your mole ratio card

11. Mole to Mole Practice Activity
With a partner, you will practice converting mole to mole stoichiometry problems. We need some manipulatives, so each person will make a set of cards. However, the LEFT half of the room will be GROUP A and the RIGHT half of the room will be GROUP B, and each group will be making different cards. When we do the activity, you will partner with someone from the other group.

12. Mole to Mole Practice Activity
Now you’re going to make a set of MOLAR RATIO cards for two different chemical equations. First, you must correctly balance each equation. Next, you will make two-sided cards like this: Say the chemical equation is w AB + x CD → y AD + z CB One card will look like this:
w moles AB
x moles CD
x moles CD
w moles AB
Side 1 Side 2

13. Mole to Mole Practice Activity
w AB + x CD → y AD + z CB
Another card will look like this:
You keep doing this until you have paired ALL compounds in the equation and have created two-sided cards.
(If you have 4 compounds, there will be six cards; if you have 3 compounds, there will be three cards.)
w moles AB
y moles AD
y moles AD
w moles AB

14. Mole to Mole Practice Activity
GROUP A __Al2O3 → __ Al + __O2 Balance the equation. Then create a set of 3 molar ratio cards with the SAME pair of elements/compounds on each card, but reverse their placement. Remember to use the correct coefficients for each compound.
GROUP B _Fe2O3 + _CO → _Fe + _CO2 Balance the equation. Then create a set of 6 molar ratio cards with the SAME pair of elements/compounds on each card, but reverse their placement. Remember to use the correct coefficients for each compound.

15. Mole to Mole Practice Activity
GROUP A 2 Al2O3 → 4 Al + 3 O2 Balance the equation. Then create a set of 3 molar ratio cards with the SAME pair of elements/compounds on each card, but reverse their placement. Remember to use the correct coefficients for each compound.
GROUP B 1Fe2O3 + 3 CO → 2 Fe + 3 CO2 Balance the equation. Then create a set of 6 molar ratio cards with the SAME pair of elements/compounds on each card, but reverse their placement. Remember to use the correct coefficients for each compound.

16. Mole to Mole Practice Activity
GROUP A
_AlF3 + _O2 → _Al2O3 + _F2
Balance the equation. Then create a set of 5 molar ratio cards with the SAME pair of elements/compounds on each card, but reverse their placement. (You already made the Al2O3/O2 card.)
Remember to use the correct coefficients for each compound.
(8 cards when finished)
GROUP B
_C3H8 + _O2 → _CO2 + _H2O
Balance the equation. Then create a set of 6 molar ratio cards with the SAME pair of elements/compounds on each card, but reverse their placement.
Remember to use the correct coefficients for each compound.
(12 cards when finished)

17. Mole to Mole Practice Activity
GROUP A
4 AlF3 + 3 O2 → 2 Al2O3 + 6 F2
Balance the equation. Then create a set of 5 molar ratio cards with the SAME pair of elements/compounds on each card, but reverse their placement. (You already made the Al2O3/O2 card.)
Remember to use the correct coefficients for each compound.
GROUP B
1C3H8 + 5 O2 → 3 CO2 + 4 H2O
Balance the equation. Then create a set of 6 molar ratio cards with the SAME pair of elements/compounds on each card, but reverse their placement. Remember to use the correct coefficients for each compound.

18. Practice Problem Using Cards
Analyze: Given: 6 mol hydrogen gas (reactant) Unknown: moles of ammonia (product) Plan: First you’ll need the balanced chemical equation that relates ammonia (product) to hydrogen and nitrogen (reactants). __𝐻 2(𝑔) + __𝑁 2(𝑔) →__ 𝑁𝐻 3 (s) Next you’ll need the mole ratio (conversion factor) that relates hydrogen gas to ammonia so that you can convert moles of hydrogen gas (given) to moles of ammonia (unknown).

19. Solution to Practice Problem
How many moles of NH3 can be created from 6 moles of H2 gas?
3𝐻 2(𝑔) + 𝑁 2(𝑔) →2 𝑁𝐻 3 (s)
The needed mole ratio between ammonia and hydrogen gas is:
2 𝑚𝑜𝑙 𝑁 𝐻 3 3 𝑚𝑜𝑙 𝐻 2
Calculation:
6 𝑚𝑜𝑙 𝐻 2 × 2 𝑚𝑜𝑙 𝑁 𝐻 3 3 𝑚𝑜𝑙 𝐻 2 =4 𝑚𝑜𝑙 𝑁 𝐻 3
Arrange your card so that moles of the known substance is on the bottom and moles of the unknown is on the top
Think of this as your mole ratio card

20. Mole to Mole Practice Activity
Each pair of students will have one paper, a dry erase marker, and the two sets of cards you’ve made.
Taking turns being the solver and the coach, you will answer the questions for two equations.
You will need to fill in the blanks on the paper and write your final answers in the provided space.
Turn in your paper with both names.

21. Mole to Mole Practice Activity
Balanced chemical equation #1:
__Al2O3 → __ Al + __O2
Problem #1: How many moles of oxygen can be created from 8 moles of Al2O3?
Problem #2: How many moles of aluminum can be created from 5 moles of Al2O3?
Problem #3: How many moles of Al2O3 are needed to create 9 moles of oxygen?

22. Mole to Mole Practice Activity
PRACTICE A SOLUTION
Balanced chemical equation #1:
2 Al2O3 → 4 Al + 3 O2
Problem #1: How many moles of oxygen can be created from 8 moles of Al2O3? 12 mol O2
Problem #2: How many moles of aluminum can be created from 5 moles of Al2O3? 10 mol Al
Problem #3: How many moles of Al2O3 are needed to create 9 moles of oxygen? 6 mol Al2O3

23. Mole to Mole Practice Activity
Balanced chemical equation #2:
__Fe2O3 + __CO → __Fe + __CO2
Problem #4: How many moles of iron can be created from 0.6 moles of CO?
Problem #5: How many moles of carbon dioxide can be created from 1.7 moles of Fe2O3?
Problem #6: How many moles of Fe2O3 are needed to create 7.6 moles of iron?

24. Mole to Mole Practice Activity
PRACTICE A SOLUTION
Balanced chemical equation #2:
1 Fe2O3 + 3 CO → 2 Fe + 3 CO2
Problem #4: How many moles of iron can be created from 0.6 moles of CO? 0.4 mol Fe
Problem #5: How many moles of carbon dioxide can be created from 1.7 moles of Fe2O3? 5.1 mol
Problem #6: How many moles of Fe2O3 are needed to create 7.6 moles of iron? 3.8 mol

25. Mole to Mole Practice Activity
Balanced chemical equation #3:
__AlF3 + __O2 → __Al2O3 + __F2
Problem #7: How many moles of oxygen are needed to react with 6 moles of AlF3?
Problem #8: How many moles of AlF3 are needed to create 5.3 moles of Al2O3?
Problem #9: How many moles of oxygen gas are needed to create 11 moles of fluorine gas?

26. Molar Mass
Often reaction stoichiometry problems will also require one or more molar masses as conversion factors because we’re converting moles to masses.
2 𝐴𝑙 2 𝑂 3 𝑙 →4𝐴𝑙 𝑠 +3 𝑂 2 (𝑔)
In the example above, the rounded molar masses for the reactants and products are:
1 mol Al2O3= g 1 mol Al=26.98 g 1 mol O2=32.00 g
These are the molar mass conversion factors:
𝑔 𝐴𝑙 2 𝑂 3 1 𝑚𝑜𝑙 𝐴𝑙 2 𝑂 𝑔 𝐴𝑙 1 𝑚𝑜𝑙 𝐴𝑙 𝑔 𝑂 2 1 𝑚𝑜𝑙 𝑂 2

27. Molar Mass Manipulatives
You are now going to create molar mass cards for the compounds and elements on your mole ratio cards.
For EACH compound, you will create a two-sided card that has the molar mass on both sides, like this:
(If you have 4 compounds, there will be 4 cards; if you have 3 compounds, there will be three cards.)
35.89 g AB
1 mol AB
1 mol AB
35.89 g AB

28. Molar Mass Manipulatives
GROUP A
You will create two-sided molar mass cards for these compounds and elements:
g Fe2O3 / 1 mol Fe2O3
28.01 g CO / 1 mol CO
55.85 g Fe / 1 mol Fe
44.01 g CO2 / 1 mol CO2
44.10 g C3H8 / 1 mol C3H8
32.00 g O2 / 1 mol O2
18.02 g H2O / 1 mol H2O
GROUP B
You will create two-sided molar mass cards for these compounds and elements:
g Al2O3 / 1 mol Al2O3
26.98 g Al / 1 mol Al
32.00 g O2 / 1 mol O2
83.98 g AlF3 / 1 mol AlF3
38.00 g F2 / 1 mol F2

29. Conversions from Moles to Mass
Going from an amount of a given reactant in moles to an unknown amount of product in grams, you need to use TWO conversion factors: the mole ratio that relates those two substances and the molar mass of the product (substance to be expressed in grams). Example: ___ NH3 + ___ O2 → ___N2 + ___H2O How many grams of N2 are produced if 3.5 moles of oxygen gas react?

30. Mole to Mass Example
How many grams of N2 is produced if 3.5 moles of oxygen gas react? ___ NH3 + ___ O2 → ___N2 + ___H2O 3.5 𝑚𝑜𝑙 𝑂 2 × 2 𝑚𝑜𝑙 𝑁 2 3 𝑚𝑜𝑙 𝑂 2 × 𝑔 𝑁 2 1 𝑚𝑜𝑙 𝑁 2 4 3 2 6
Mole ratio
Molar Mass
= g N2

31. Mole to Mass Practice Activity
Each pair of students will have one paper, a dry erase marker, and the two sets of cards you’ve made.
Taking turns being the solver and the coach, you will answer the questions for two equations.
You will need to fill in the blanks on the paper and write your final answers in the provided space.

32. Mole to Mass Practice Activity
PRACTICE B1
Balanced chemical equation #1:
__Al2O3 → __ Al + __O2
Problem #1: How many grams of oxygen can be created from 10 moles of Al2O3?
Problem #2: How many grams of aluminum can be created from 30 moles of Al2O3?
Problem #3: How many moles of Al2O3 are needed to create grams of oxygen?

33. Mole to Mass Practice Activity
PRACTICE B1 SOLUTIONS
Balanced chemical equation #1:
2 Al2O3 → 4 Al + 3 O2
Problem #1: How many grams of oxygen can be created from 10 moles of Al2O3? 480 g O2
Problem #2: How many grams of aluminum can be created from 30 moles of Al2O3? g Al
Problem #3: How many moles of Al2O3 are needed to create grams of oxygen? g Al2O3

34. Mole to Mass Practice Activity
PRACTICE B1
Balanced chemical equation #2:
__Fe2O3 + __CO → __Fe + __CO2
Problem #4: How many grams of iron can be created from 15.2 moles of Fe2O3?
Problem #5: How many grams of carbon dioxide can be created from 8.1 moles of Fe2O3?
Problem #6: How many moles of CO are needed to create grams of iron?

35. Mole to Mass Practice Activity
PRACTICE B1
Balanced chemical equation #2:
__Fe2O3 + __CO → __Fe + __CO2
Problem #7: How many moles of Fe2O3 are needed to create grams of iron?
NOW TURN IN THIS PAPER (both names).
SWITCH PARTNERS AND GET ANOTHER PAPER PRACTICE B.

36. Mole to Mass Practice Activity
PRACTICE B2
Balanced chemical equation #3:
__AlF3 + __O2 → __Al2O3 + __F2
Problem #1: How many grams of fluorine gas can be created from 10 moles of AlF3?
Problem #2: How many grams of Al2O3 can be created from 0.90 moles of oxygen gas ?
Problem #3: How many moles of AlF3 are needed to create 200 grams of Al2O3?

37. Mole to Mass Practice Activity
PRACTICE B2 Balanced chemical equation #4:
__C3H8 + __O2 → __CO2+ __H2O
Problem #4: How many grams of water can be created from 4.4 moles of propane (C3H8)?
Problem #5: How many grams of carbon dioxide can be created from 13.0 moles of propane (C3H8)?
Problem #6: How many moles of O2 are needed to create 300 grams of water?

38. Mole to Mass Practice Activity
PRACTICE B2
Balanced chemical equation #4:
__C3H8 + __O2 → __CO2+ __H2O
Problem #7: How many moles of propane are needed to create 1 kg of carbon dioxide?
NOW TURN IN THIS PAPER (both names).

39. Conversions from Mass to Moles
Going from an amount of a given reactant in grams to an unknown amount of product in moles, you again need to use TWO conversion factors: the mole ratio that relates those two substances and the molar mass of the reactant. In a spacecraft, the carbon dioxide exhaled by astronauts can be removed by its reaction with lithium hydroxide: 𝐶 𝑂 2 𝑔 +2 𝐿𝑖𝑂𝐻 𝑠 → 𝐿𝑖 2 𝐶𝑂 3 𝑠 + 𝐻 2 𝑂(𝑙) If g of carbon dioxide is used, how many moles of lithium carbonate will be produced?

40. Mass to Mole conversion
𝐶 𝑂 2 𝑔 +2 𝐿𝑖𝑂𝐻 𝑠 → 𝐿𝑖 2 𝐶𝑂 3 𝑠 + 𝐻 2 𝑂(𝑙) If g of carbon dioxide is used, how many moles of lithium hydroxide is needed? 𝑔 𝐶 𝑂 2 × 1 𝑚𝑜𝑙 𝐶 𝑂 𝑔 𝐶 𝑂 2 × 1 𝑚𝑜𝑙 𝐿𝑖 2 𝐶𝑂 3 1 𝑚𝑜𝑙 𝐶 𝑂 2
Molar Mass
Mole ratio
= 10 mol LiOH

41. Mass to Mole Practice Activity
PRACTICE C
Balanced chemical equation #1:
__Al2O3 → __ Al + __O2
Problem #1: How many moles of aluminum can be created from 65 grams of Al2O3?
Problem #2: How many moles of oxygen gas can be created from 375 grams of Al2O3?
Problem #3: How many grams of Al2O3 are needed to create 28 moles of oxygen?

42. Mass to Mole Practice Activity
PRACTICE C
Balanced chemical equation #4:
__C3H8 + __O2 → __CO2+ __H2O
Problem #4: How many moles of water can be created from 800 grams of propane (C3H8)?
Problem #5: How many moles of carbon dioxide can be created from 40 grams of oxygen gas?
Problem #6: What mass of O2 is needed to create 1.50 moles of water?

43. Mass to Mole Practice Activity
PRACTICE C
Balanced chemical equation #4:
__C3H8 + __O2 → __CO2+ __H2O
Problem #7: What mass of propane is needed to create 7.6 moles of carbon dioxide?
NOW TURN IN THIS PAPER (both names).

44. Stoichiometry Railroad Tracks
Now I’m going to teach you a method of arranging your problems in stoichiometry in a way that will give you a high degree of certainty that you have the correct answer.
In this method, the UNITS are KEY and will point the way toward the next conversion factor needed.
This method is called RAILROAD TRACKS. You may already have used this method in other conversions such as converting 10 m/s to mph:
See how the units are arranged so that they cancel?
10 m
3600 s
100 cm
1 inch
1 foot
1 mile s
1 hour
1 m
2.54 cm
12 inch
5280 feet
10 m/s = 22 miles/hour or mph

45. ‘Railroad Track’ Basic Rules
Every track must contain a NUMERIC VALUE, even if it’s just a ‘1’, (except for the FIRST track which has the known in the upper track but the bottom track is often empty).
Every track must contain a UNIT, usually moles or grams. This is so you can make sure the units cancel properly.
Every track must contain an ELEMENT or COMPOUND.

46. Example (mole to mole)
Let’s use the last sample problem. We converted 6 mol of hydrogen gas into an unknown amount (mol) of ammonia, using only one conversion factor, the mole ratio between ammonia and hydrogen gas:
Mole ratio
6 mol H2
2 mol NH3
3 mol H2
= 4 moles of NH3
Notice how the units have cancelled, leaving the desired unit!

47. Mole to Mass Conversion
Using the sample problem where the balanced equation for the electrolysis of melted aluminum oxide to produce aluminum metal and oxygen gas: 2 𝐴𝑙 2 𝑂 3 𝑙 →4𝐴𝑙 𝑠 +3 𝑂 2 (𝑔) We found the amount in moles of aluminum metal that can be produced from 13.0 mol of aluminum oxide was 26 mol Al(s). = 26.0 mol Al(s)
13.0 mol Al2O3
4 mol Al(s)
2 mol Al2O3
Mole ratio

48. Mole to Mass Conversion (cont.)
What if we want to know how much mass this is? We can extend the railroad tracks to convert the number of moles we found to find the equivalent number of grams of aluminum by using the molar mass as a conversion factor: = 701 g Al Stoichiometry problems use mole ratios and/or molar masses as conversion factors.
13.0 mol Al2O3
4 mol Al
26.98 g Al
2 mol Al2O3
1 mol Al
Mole ratio Molar mass

49. Mass to Mole Conversions
This is just the reverse of the prior type of problem. Here you start with a given mass and convert it to moles of another substance.
Example: The industrial manufacture of nitric acid starts with this reaction for the oxidation of ammonia. The products are nitrogen monoxide and water. Starting with 824 g of ammonia and excess oxygen, how many moles of NO and H2O are formed?
4𝑁 𝐻 3 𝑔 +5 𝑂 2 𝑔 →4𝑁𝑂 𝑔 + 6𝐻 2 𝑂(𝑔)
Analyze: Given: Amount of NH3 = 824 g
Unknowns: amount of NO and H2O (mol) produced
Plan: Convert mass of NH3 to moles of NO and H2O
To calculate this, we need the mole ratios between NH3 and NO and between NH3 and H2O; we also need the molar mass of ammonia, which is g.

50. Mass to Mole Conversions
4𝑁 𝐻 3 𝑔 +5 𝑂 2 𝑔 →4𝑁𝑂 𝑔 + 6𝐻 2 𝑂(𝑔)
Calculate:
To find moles of NO formed starting with mass of NH3:
= 48.4 mol NO
Molar mass
Mole ratio
To find moles of H2O formed.
824 g NH3
1 mol NH3
6 mol H2O
17.04 g NH3
4 mol NH3
= 72.5 mol H2O

51. Mass to Mass Conversions
The most complicated stoichiometry calculation converts a mass of one substance to the mass of another. The mass of the first substance is converted to moles using its MOLAR MASS, then the MOLE RATIO between the two substances is used to find the moles of the second substance, and finally this is converted to mass of the second substance by using its MOLAR MASS. We will keep extending our ‘railroad tracks.’

52. Mass to Mass Conversion
Example: Tin (II) fluoride, SnF2, is used in some toothpastes. It is made by the reaction of tin with hydrogen fluoride: 𝑆𝑛 𝑠 +2 𝐻𝐹 𝑔 →𝑆𝑛 𝐹 2 𝑠 + 𝐻 2 𝑔 How many grams of SnF2 are produced from the reaction of g HF with Sn? Given: Amount of HF = g Unknown: Mass of SnF2 produced (grams) Plan: Convert mass of HF to moles of HF using its molar mass, use the mole ratio of HF to SnF2, then convert moles of SnF2 to grams of SnF2 using its molar mass.

53. Mass to Mass Conversion
𝑆𝑛 𝑠 +2 𝐻𝐹 𝑔 →𝑆𝑛 𝐹 2 + 𝐻 2 g Plan: Convert mass of HF to moles of HF using its molar mass, use the mole ratio of HF to SnF2, then convert moles of SnF2 to grams of SnF2 using its molar mass.
30.00 g HF
1 mol HF
1 mol SnF2
g SnF2
20.01 g HF
2 mol HF
1 mol SnF2
Molar mass HF
Mole ratio
Molar mass SnF2
= g SnF2

54. Mass to Mass Practice Activity
PRACTICE D
Balanced chemical equation #2:
__Fe2O3 + __CO → __Fe + __CO2
Problem #1: How many grams of carbon dioxide can be created from 625 grams of Fe2O3?
Problem #2: How many grams of iron can be created from 250 grams of carbon monoxide?
Problem #3: What mass of Fe2O3 is needed to create grams of iron?

55. Mass to Mass Practice Activity
PRACTICE D
Balanced chemical equation #3:
__AlF3 + __O2 → __Al2O3 + __F2
Problem #4: How many grams of fluorine gas can be created from 80 grams of AlF3?
Problem #5: How many grams of Al2O3 can be created from 520 grams of oxygen gas ?
Problem #6: What mass of AlF3 is needed to create 480 grams of Al2O3?

56. Mass to Mass Practice Activity
PRACTICE D
Balanced chemical equation #3:
__AlF3 + __O2 → __Al2O3 + __F2
Problem #7: What mass of O2 is needed to create 390 grams of F2?

57. Remember!
Conversions aren’t just from reactant to product and product to reactant—they can be from reactant to reactant and product to product as well! Here’s an example: In a spacecraft, the carbon dioxide exhaled by astronauts can be removed by its reaction with lithium hydroxide: 𝐶 𝑂 2 𝑔 +2 𝐿𝑖𝑂𝐻 𝑠 → 𝐿𝑖 2 𝐶𝑂 3 𝑠 + 𝐻 2 𝑂(𝑙) How many grams of lithium hydroxide are required to react with 20 mol of CO2, the average amount exhaled by a person each day?

58. Analyze: Given: Amount of CO2= 20 mol Unknown: grams of LiOH needed
𝐶 𝑂 2 𝑔 +2 𝐿𝑖𝑂𝐻 𝑠 → 𝐿𝑖 2 𝐶𝑂 3 𝑠 + 𝐻 2 𝑂(𝑙)
Analyze: Given: Amount of CO2= 20 mol
Unknown: grams of LiOH needed
Plan: Convert moles of CO2 to grams of LiOH
Is the equation balanced????
We need the mole ratio between CO2 and LiOH; and we also need the molar mass of LiOH
Calculate:
Molar mass LiOH: = g LiOH/1 mol LiOH
20 𝑚𝑜𝑙 𝐶𝑂 2 × 2 𝑚𝑜𝑙 𝐿𝑖𝑂𝐻 1 𝑚𝑜𝑙 𝐶𝑂 2 × 𝑔 𝐿𝑖𝑂𝐻 1 𝑚𝑜𝑙 𝐿𝑖𝑂𝐻 = 𝑔 𝐿𝑖𝑂𝐻
MOLE RATIO MOLAR MASS

59. Review: Types of Reaction Stoichiometry Problems
Given and unknown quantities are in moles
amount of amount of
GIVEN substance (mol) → UNKNOWN substance (mol)
Given amount is in moles and unknown amount is in mass (often grams)
amount of amount of mass of
given substance→ unknown substance→ unknown substance
(mol) (mol) (g)
Here we use the MOLE RATIO between the two substances as the conversion.
Here we use TWO conversions, a MOLE RATIO and then the MOLAR MASS.

60. Review: Types of Reaction Stoichiometry Problems
Given is a mass in grams and unknown is an amount in moles
mass of amount of amount of
given substance → given substance → unknown substance
(g) (mol) (mol)
Here use MOLAR MASS and MOLAR RATIO to convert.
Given is a mass in grams and unknown is a mass in grams
mass of amount of amount of mass of
given substance → given substance → unknown substance → unknown substance
(g) (mol) (mol) (g)
Here use MOLAR MASS, MOLE RATIO, then a SECOND MOLAR MASS to convert.

61. Limiting Reactants (or Reagents)
In chemical reaction in a laboratory, usually there is an excess of one or more of the reactants. This means that when one of the reactants has been used up, the reaction stops, and no more product can be formed. The limiting reactant is the reactant that has been completely used up (none remains), which limited the amount of product that can form in a chemical reaction. The excess reactant is the reactant that is not completely used up in a reaction (some remains at the end of the reaction).

62. Definitions
STOICHIOMETRY—branch of chemistry that deals with calculations about the masses of reactants and products involved in a chemical reaction. Composition: mass relationships between elements in compounds; Reaction: mass relationships between reactants and products in a chemical reaction LIMITING REACTANT—the reactant that is completely used up, which limits the amount of product that can form in a chemical reaction EXCESS REACTANT—the reactant that isn’t completely used up and doesn’t limit the amount of product that can form in a chemical reaction

63. MOLE--6. 022 x 1023 units of a substance (atoms, molecules, ions, etc
MOLE x 1023 units of a substance (atoms, molecules, ions, etc.) MOLE RATIO—a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction SYNTHESIS REACTION—chemical reaction in which two simpler substances combine to form a more complex product COMBUSTION REACTION—chemical reaction in which a reactant burns with oxygen to produce carbon dioxide and water DIATOMIC MOLECULE—a molecule composed of two atoms, usually refers to gases such as hydrogen, oxygen, and chlorine

64. Which is the limiting reactant?
In a chemical equation, determining the limiting reactant is straightforward but involves two calculations: calculate how much product is produced by each of the given amounts of reactants. Whichever reactant gives the smaller amount of product is the limiting reactant!
Example: Calculate how much product can be produced from mol of sodium and from 7.05 mol of chlorine and compare.
Plan: Use mole ratios to find the number of moles of one of the products and compare.

65. The balanced chemical equation is 2𝑁𝑎+ 𝐶𝑙 2 →2𝑁𝑎𝐶𝑙
Calculate how much product can be produced from mol of sodium and from 7.05 mol of chlorine and compare.
The balanced chemical equation is
2𝑁𝑎+ 𝐶𝑙 2 →2𝑁𝑎𝐶𝑙
21.75 mol 𝑵𝒂 × 2 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙 2 𝑚𝑜𝑙 𝑁𝑎 = 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙
𝟕.𝟎𝟓 𝒎𝒐𝒍 𝑪𝒍 𝟐 × 2 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙 1 𝑚𝑜𝑙 𝐶𝑙 2 =14.10 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙
Chlorine is the limiting reactant because it makes less product.

66. Which is the limiting reactant?
Remember: Whichever reactant gives the smaller amount of product is the limiting reactant! Example: How many grams of sodium chloride can be produced from g each of sodium and chlorine? Plan: Calculate how much product (NaCl) can be produced from both g of sodium and from g of chlorine and compare.

67. The balanced chemical equation is 2𝑁𝑎+ 𝐶𝑙 2 →2𝑁𝑎𝐶𝑙
How many grams of sodium chloride can be produced from 500. g each of sodium and chlorine?
The balanced chemical equation is
2𝑁𝑎+ 𝐶𝑙 2 →2𝑁𝑎𝐶𝑙
𝟓𝟎𝟎. 𝟎𝒈 𝑵𝒂× 1 𝑚𝑜𝑙 𝑁𝑎 𝑔 𝑁𝑎 × 2 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙 2 𝑚𝑜𝑙 𝑁𝑎 × 𝑔 𝑁𝑎𝐶𝑙 1 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙 = 𝑔 𝑁𝑎𝐶𝑙
𝟓𝟎𝟎. 𝟎𝒈 𝑪𝒍 𝟐 × 1 𝑚𝑜𝑙 𝐶𝑙 𝑔 𝐶𝑙 2 × 2 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙 1 𝑚𝑜𝑙 𝐶𝑙 2 × 𝑔 𝑁𝑎𝐶𝑙 1 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙 = 𝑔 𝑁𝑎𝐶𝑙
Since g of chlorine produces less NaCl than g of sodium, chlorine is the limiting reactant!

68. Now, how much excess sodium will you have at the end of the reaction?
Start with the given amount of Cl2 (500.0 g) as the starting point for another mass to mass calculation to find the amount of sodium actually used.
500.0 𝒈 𝑪𝒍𝟐× 1 𝑚𝑜𝑙 𝐶𝑙 𝑔 𝐶𝑙2 × 2 𝑚𝑜𝑙 𝑁𝑎 1 𝑚𝑜𝑙 𝐶𝑙2 × 𝑔 𝑁𝑎 1 𝑚𝑜𝑙 𝑁𝑎 = g Na
Since we started with g Na, we subtract g to find there is g remaining at the end of the reaction.