1. Wednesday, March 27 Edmodo Assignment
Watch the “Expanded Mole Map” video
Draw the expanded map
Click TURN IN and leave your answer to the problem solved in the video as a comment.
Print the PowerPoint notes for the unit. [You’ll want the sample problem slides.]
If needed, review the polyatomic ions, writing equations and balancing equations.

2. Unit 7 Stoichiometry
Chapter 9

3. FYI
"Stoichiometry" is derived from the Greek words στοιχεῖον (stoicheion, meaning element]) and μέτρον (metron, meaning measure.)
Stoichiometry = measuring elements

4. Info in a Chemical Equation
2K + CuSO4  Cu + K2SO4
A balanced chemical equation shows the correct ratios required for a chemical reaction to occur.
Knowing the ratio allows us to provide the appropriate amount of reactants AND predict the amount of products.

5. Mole to Mole Relationships
Since chemical equations give appropriate relationships of moles of each compound, mole ratios can be written.
Mole ratio: uses the coefficients in the balanced equation to show how two compounds are related in a reaction

6. Example 9.2 Goal: Write a mole ratio & solve the problem.
What number of moles of O2 will be produced by the decomposition of 5.8 mol of water?
First, the balanced chemical equation must be written for the situation described in the prompt.
2H2O  2H2 + O2

7. Example 9.2 (continued)
What number of moles of O2 will be produced by the decomposition of 5.8 mol of water?
2H2O  2H2 + O2
Next, the reaction shows that oxygen and water have a quantitative relationship: a mole ratio.
1 mole O2 : 2 moles H2O
Where did the numbers (1 & 2) come from?
Can you write other mole ratios from this equation?

8. Example 9.2 (continued)
What number of moles of O2 will be produced by the decomposition of 5.8 mol of water?
2H2O  2H2 + O2
Finally, we can solve this problem using the mole ratio as a conversion factor in our dimensional analysis.
Start with the given.
Set up the chart to cancel units AND compounds.
Multiply or divide as usual.

9. Example 9.3 Goal: Solve the problem with a partner.
Calculate the number of moles of oxygen required to react exactly with 4.30 mol of propane, C3H8, in the reaction described by the following:
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)

10. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)
iRespond Question F
Multiple Choice
66AA0EEC-F292-3C43-A0D0-4398BD51E3B3
Calculate the number of moles of oxygen required to react exactly with 4.30 mol of propane, C3H8, in the reaction described by the following:
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)
A.) moles of O2
B.) 7.17 moles of O2
C.) 21.5 moles of O2
D.) none of these
E.)

11. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)
Calculate the number of moles of oxygen required to react exactly with 4.30 mol of propane, C3H8, in the reaction described by the following:
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)
A.) moles of O2
B.) 7.17 moles of O2
C.) 21.5 moles of O2
D.) none of these

12. Example 9.4 Is another example needed?
Ammonia is used in huge quantities as a fertilizer. It is manufactured by combining nitrogen and hydrogen according to the following equation:
N2(g) + 3H2(g) = 2NH3(g)
Calculate the number of moles of NH3 that can be made from 1.30 mol H2(g) reacting with excess N2(g).

13. Suggested Homework
Page and 5

14. Mega Mole Map Mole to mole relationship
Mole Ratio = conversion factor (bridge)
In our examples so far, we’ve been given moles of one compound and asked to convert to moles of another compound.
What if I told you that I won’t always start out with moles or ask for moles? Starting or ending with mass, liters, or particles is very common.
Work with your partner to expand on our current Mole Concept Map.

15. Mega Mole Map Mole to mole relationship
Mole Ratio = conversion factor (bridge)
Mole to mass relationship
Molar Mass = conversion factor (bridge)
Mole to particle relationship
Avogadro’s # = conversion factor (bridge)
Mole to volume relationship
Molar volume = conversion factor (bridge)

16. Example 9.5 Goal: Use new mole concepts to solve the problem.
Consider the reaction of powdered aluminum metal and finely ground iodine to produce aluminum iodide. The balanced equation for this vigorous chemical reaction is:
2Al(s) + 3I2(s) = 2AlI3(s)
Calculate the mass of I2(s) needed to just react with 35.0 g of Al(s).
First, write the given and set up your chart.

17. Example 9.5 (continued)
Consider the reaction of powdered aluminum metal and finely ground iodine to produce aluminum iodide. The balanced equation for this vigorous chemical reaction is:
2Al(s) + 3I2(s) = 2AlI3(s)
Calculate the mass of I2(s) needed to just react with 35.0 g of Al(s).
Next, consult your map to determine a problem-solving path. How many steps will this problem require?
3 steps

18. Example 9.5 (continued)
Consider the reaction of powdered aluminum metal and finely ground iodine to produce aluminum iodide. The balanced equation for this vigorous chemical reaction is:
2Al(s) + 3I2(s) = 2AlI3(s)
Calculate the mass of I2(s) needed to just react with 35.0 g of Al(s).
Finally, plug in your conversion factors. Multiply and divide as usual to give your final answer.
g I2

19. Section 9.2 Review Questions Goal: Solve the problem with a partner.
Solutions of sodium hydroxide cannot be kept for very long because they absorb carbon dioxide from the air, forming sodium carbonate. The unbalanced equation is:
NaOH(aq) + CO2(g)  Na2CO3(aq) + H2O(l)
Calculate the number of grams of carbon dioxide that can be absorbed by complete reaction with a solution that contains 5.00 g of NaOH.

20. NaOH(aq) + CO2(g)  Na2CO3(aq) + H2O(l)
iRespond Question F
Multiple Choice
F90ADAB5-58A0-9A4C D250DD813AE
NaOH(aq) + CO2(g)  Na2CO3(aq) + H2O(l)
Calculate the number of grams of carbon dioxide that can be absorbed by complete reaction with a solution that contains 5.00 g of NaOH.
A.) 2.75 g CO2
B.) 5.50 g CO2
C.) 9.09 g CO2
D.) none of these
E.)

21. NaOH(aq) + CO2(g)  Na2CO3(aq) + H2O(l)
Calculate the number of grams of carbon dioxide that can be absorbed by complete reaction with a solution that contains 5.00 g of NaOH.
A.) 2.75 g CO2
B.) 5.50 g CO2
C.) 9.09 g CO2
D.) none of these

22. Limiting Reactants
As you know, a balanced chemical equation gives the perfect ratio of reactants needed to perform the reaction.
In reality, we rarely have the “perfect” ratio.
With an imperfect ratio, one reactant will run out before the other. The reactant running out will stop the reaction...limit the products.
The reactant that runs out = Limiting reactant
The reactant that remains = Excess reactant

23. Limiting Reactant Calculations
Using stoichiometry, we can calculate the following predictions:
Identify the limiting reactant
Identify the excess reactant
Predict the amount of product to form
Predict the amount of excess reactant remaining

24. Practice Problem 9.5
Calculate the mass of AlI3(s) formed by the reaction of 35.0 g Al(s) with 495 g I2.

25. Example 9.7
Suppose that 25.0 kg of nitrogen gas and 5.00 kg of hydrogen gas are mixed and reacted to form ammonia. Calculate the mass of ammonia produced when this reaction is run to completion.
Two “givens” means two calculations
The unknown is the same in the two calculations
The smallest answer is the best prediction as it tells when the limiting reactant will run out.

26. Follow-up Questions Identify the limiting reactant.
Identify the excess reactant.
How much H2 is used? Left?
How much N2 is used? Left?

27. Limiting Reactant Practice
1. The reaction between solid white phosphorus and oxygen produces solid tetraphosphorus decaoxide (P4O10). This compound is often called phosphorus pentoxide because its empirical formula is P2O5.
P4 + 5O2  P4O10
Determine the mass of tetraphosphorus decaoxide formed if 25.0 g of phosphorus (P4) and 50.0 g of oxygen gas are combined.
Identify the limiting reactant.
How much of the excess reactant remains after the reaction stops?

28. Section 9.2 Review Questions
You react 10.0 g of nitrogen gas with hydrogen gas according to the following reaction.
N2(g) + 3H2(g) = 2NH3(g)
What mass of hydrogen is required to completely react with 10.0 g sample of nitrogen gas?
What mass of ammonia is produced from 10.0 g of nitrogen gas and sufficient hydrogen gas?

29. Practice Problem 9.8
Lithium nitride, an ionic compound containing the Li+ and N3- ions, is prepared by the reaction of lithium metal and nitrogen gas. Calculate the mass of lithium nitride formed from 56.0 g of nitrogen gas and 56.0 g of lithium.

30. Follow-up Questions Identify the limiting reactant.
Identify the excess reactant.
How much of each reactant is used? Left?

31. Percent Yield Yield means product.
Calculating % yield is calculating what % of the product your experiment actually produced.
You are comparing your prediction to your action.

32. % Yield (Actual yield / theoretical yield) 100
Actual yield = either given in the problem or carried out in the lab
Theoretical yield = ALWAYS calculated by stoichiometry

33. The Plan: Stoichiometry Quiz (30 minute time limit)
Stoichiometry Stumpers
The lab practical will be similar to these examples.
5 point assignment due at the end of the period
Limiting Reactant Problem-based Learning

34. Stoichiometry Test Tips
Know how to write and balance a combustion reaction. (See #1 on your Stoichiometry Quiz.)
Remember, “a liter” or “a single gram” means 1 L or 1 gram. The given can be written this way.
“How many grams of product are produced?” - refers to theoretical yield (smallest answer) in a limiting reactant problem

35. 4. Metal products are solids.
5. STP = Standard temperature and pressure (doesn’t affect the calculation)
6. % yield = (ACTUAL/THEORETICAL) 100
7. Excess reactant - Use dimensional analysis starting with given amount of LR to calculate the amount USED; Subtract from ORIGINAL amount to calculate excess LEFT

36. Look closely at the wording…
Labeling the equation will help!
If you are given info about BOTH reactants, then you have a limiting reactant problem.
Example: What mass of NaCl will be produced by the reaction of 58.7g of NaI with 29.4g of Cl2 gas if the products are NaCl and I2?
If you are given info about ONLY ONE substance, then you have a simple stoichiometry problem.

37. More wording advice… LABELS!
If you are given info about BOTH reactants AND a product, then you’ve probably got a percent yield problem to solve. The product info is the actual yield.
Determine the percent yield for a reaction between 6.92g K and 4.28g of O2 if 7.36g of K2O is produced.