2. Stoichiometry? According to Webster’s… “Pronunciation: `stoykee'âmitree WordNet Dictionary Definition: [n] (chemistry) the relation between the quantities of substances that take part in a reaction…”

3. How does it relate to past skills? Stoichiometry is a combination of the last 3 skills you have learned. 1) Predicting the outcome of a chemical reaction 2) Balancing 3) Converting chemical quantities

4. Why is it important? With Stoichiometry- you can predict the AMOUNT of any chemical that will be made in the reaction. So not only can you predict WHAT will be made (predicting skill), … You can state HOW MUCH of each chemical should be produced or how much of a chemical is needed. (Converting Chemical quantities skill).

5. One Important New Idea: The mole bridge During stoichiometry, you will, at some point, change from one chemical to another. You may only change chemicals when using mole to mole equivalents. This happens when moles of one chemical is on top and moles of a different chemical is on the bottom. ***Use your balancing coefficients as your numerical equivalents when going from mole to mole***

6. Some other new equivalents 1 g = 1000 mg 1 kg = 1000 g

7. What it looks like: problem & setup: 2.6 g FeCl 3 X g FeCl 3 mole FeCl 3 234.34 1 X mol KCl 1 3 mol FeCl 3 mol KCl g KCl 1 34.342 X 1 g KCl X mg KCl 1000 When 2.6 grams of iron III chloride reacts with an excess of potassium iodide, how many milligrams of potassium chloride will be made? FeCl 3 + KI  Fel 3 + KCl Balance… 33 = Can you find the mole to mole “bridge”? Note that at this point one chemical changes to another Where did the 1 and 3 equivalent come from? Don’t freak!

8. Your First Stoichiometry Problem A) What is your starting and finishing point? When 4.0 moles of aluminum bromide decomposes, how many moles of bromine will form? 4.0 mole AlBr 3 X mol Br 2 2 3 mole AlBr 3 THE REACTION: AlBr +3 AlBr 3 Al+Br 2 232 = 4.0 mol Start with the given B) What are we looking for? ? mol To summarize start/finish… THE PROBLEM: Use the balancing coefficients for mole to mole equivalents! Cal = 6 R=6.0 SSN= 6.0 x10 0 mol Br 2 But you can do it!

9. Stoichiometry problem 2 What was given? When 2.6 grams of hydrochloric acid reacts with an excess of barium hydroxide, how many moles of water will be made? 2.6 g HCl X g HCl mole HCl 36.46 1 X mol H 2 0 2 2 mol HCl Cal=.0713110258 R=.071 SSN= 7.1 x10 -2 mol H 2 0 2HCl + Ba(OH) 2  2H 2 O + BaCl 2 =

10. Stoichiometry problem 3 When 3.7 Liters of fluorine gas reacts with an excess of sodium (at STP), how many moles of product are made? 3.7 L F 2 X L F 2 mole F 2 22.4 1 X mol NaF 1 2 mol F 2 Cal=.3303571429 R=.33 SSN= 3.3 x10 -1 mol NaF F 2 + 2Na  2NaF =

11. Stoichiometry problem 4 When 2.6 grams of hydrochloric acid reacts with an excess of barium hydroxide, how many grams of barium hydroxide are needed? 2.6 g HCl X g HCl mole HCl 36.46 1 X mol Ba(OH) 2 2 1 mol HCl mol Ba(OH) 2 g Ba(OH) 2 1 X = 171.35 Cal= 6.109572134 R= 6.1 SSN= 6.1 x10 0 g Ba(OH) 2 2HCl + Ba(OH) 2  2H 2 O + BaCl 2

12. Stoichiometry problem 5 *Iron III chloride reacts with magnesium. If 555 grams of magnesium chloride is made, with how many grams of iron III chloride did you begin? 555 g MgCl 2 X g MgCl 2 mole MgCl 2 95.21 1 X mol FeCl 3 3 2 mol MgCl 2 mol FeCl 3 g FeCl 3 1 X = 162.20 Cal= 630.3329482 R= 6.30 x 10 2 SSN= 6.30 x10 2 g FeCl 3 2FeCl 3 + 3Mg  3MgCl 2 + 2Fe

13. Stoichiometry problem 6 When 444 g of lithium sulfide reacts with an excess of aluminum bromide, how many FU of lithium bromide will form? 444 g Li 2 S X g Li 2 S mole Li 2 S 45.95 1 X mol LiBr 3 6 mol Li 2 S mol LiBr FU LiBr 1 6.02 x 10 23 X Cal= 1.163386289 x10 25 R= 1.16 x10 25 SSN= 1.16 x10 25 FU LiBr = 3Li 2 S + 2AlBr 3  6LiBr + Al 2 S 3

14. Stoichiometry problem 7 When 4567.5 milligrams of potassium nitride reacts with an excess of chlorine gas, how many kiloliters of nitrogen gas will be made? 4567.5 mg K 3 N X mg K 3 N g K 3 N 1 X mol K 3 N 131.31 1 g K 3 N mol K 3 N mol N 2 2 1 X 1 X L N 2 22.4 1000L N 2 KL N 2 1 X Cal= 3.895819054 x 10 -4 R= 3.90 x 10 -4 SSN= 3.90 x 10 -4 KL N 2 1000 2K 3 N + 3Cl 2  6KCl + N 2 =

15. Bellwork: Date: Magnesium chlorate reacts with Lithium metal. If 5.58 x10 23 atoms are made, with how many milligrams of metal did you start? 1.29 x10 4 mg Li

16. Bell work Date: Limiting Reactant Warm-up 1) Chromic acid (aq) plus aluminum (s) produces... If 9.0 moles of hydrogen are produced, how many atoms of aluminum were in the reactants? 2) Ammonium sulfate (aq) is added to strontium hydroxide (aq)... If 8.1 X 10 31 FU of ammonium sulfate is present, how many milligrams of the other reactant is needed? 3.6 x 10 24 atoms Al 1.6 x10 13 mg of Sr(OH) 2

17. Bellwork: Date: Given that you have excess J, 10 Bread and 6 PB… 1) Calculate how much Bread you will need (if you start with 6 PB) Show setup. Include the word needed in your final answer. 2) Calculate how much PB you will need (if you have 10 bread. Show setup. include Needed in answer. 3) Look at your answer for #1 and #2. Det. the LR. 4) Use the LR to calculate how many PB and J Sandwiches can be made. 5) Compare how much ER you have, to how much you calculated you will need. How much will be left over? 2 1 1 Ratio:

18. The excess reactant was ________________. I had this amount to start: _______________. I calculated that I needed/used this amount: ______________. How can I determine how much would be left over?

19. Limiting Reactants/Reagents PB and J Activity (It’s PB J Time!)

20. Limiting Reactants/Reagents When two chemicals are combined… The one that runs out first is the limiting reagent. Because it runs out, it limits the amount of product made. Calculate the answer 2 times, starting with each reactant, the correct answer will be the one that is the LESSER AMOUNT. Only the Lesser amount is a realistic answer.

21. Limiting Reactants/Reagents When two chemicals are combined… The one that runs out first is the limiting reagent. Because it runs out, it limits the amount of product made. You must determine which reactant is the limiting reactant before you calculate how much product is made.

22. The Strategy I. Part I. 1. Start with one of the two reactants, and calculate the answer to the question. 2) Start with the OTHER of the two reactants, and calculate the answer to the same question. 3) Circle the Correct answer of the two calculations (the lesser one) 4) Determine the limiting reactant (LR) and the excess reactant (ER) LR= ER=

23. The Strategy Part II. (Finding out how much of the ER is left over) 1. Start with the LR, and calculate how much of the ER is used up/needed in grams. 2) Subtract (honors only) ER ( Started with- convert it to grams! ) - ER (used up- (already in grams: see 1) ER left over (in grams)!

24. Limiting reagents problem When 2.6 grams of hydrochloric acid reacts with 7.44 grams of barium hydroxide, A) how many grams of water will be made ? B) How much excess reactant is left over? 2.6 g HCl X g HCl mole HCl… 36.46 1 Cal= 1.285024685 R= 1.3 g H 2 0 made 7.44 g Ba(OH) 2 X g Ba(OH) 2 mole Ba(OH) 2 … 171.35 1 = Cal= 1.564853224 R=1.56 g H 2 0 2HCl + Ba(OH) 2  2H 2 O + BaCl 2 LR= HClER= Ba(OH) 2

25. The Strategy II. 1. Determine how many moles of each reactant you have. (Convert both chemicals given to moles.) 2. Choose either reactant to begin with, …calculate how much of the other reactant is needed. 3. Compare the needed amount you calculated to what you actually have present of the other reactant- Determine which one will run out first. 4. Do all calculations starting with the limiting reactant!

26. When 2.6 grams of hydrochloric acid reacts with 7.44 grams of barium hydroxide, A) how many grams of water will be made ? B) How many grams of (excess) reactant will be left?

27. Limiting reagents problem When 2.6 grams of hydrochloric acid reacts with 7.44 grams of barium hydroxide, A) how many grams of water will be made ? B) How much excess reactant is left over in grams? 2.6 g HCl X g HCl mole HCl 36.46 1 = Cal=.071311026 R=.071 mol HCl present 7.44 g Ba(OH) 2 X g Ba(OH) 2 mole Ba(OH) 2 171.35 1 = Cal=.043419901 R=.0434 mol Ba(OH) 2 2HCl + Ba(OH) 2  2H 2 O + BaCl 2 1. Convert both reactants to moles… present

28. Limiting reagents problem 1.071 mol HClX mol Ba(OH) 2 2 1 mol HCl = Cal=.0355 R=.036 mol Ba(OH) 2 needed 2. Start with either reactant… … calculate how much of the other will be needed. R=.071 mol HCl present R=.0434 mol Ba(OH) 2 present Summary of reactants 3. Compare how much you have to how much you calculated you need… which reactant will run out first? R=.0434 mol Ba(OH) 2 present R=.036 mol Ba(OH) 2 needed More than enough is present 2HCl + Ba(OH) 2  2H 2 O + BaCl 2

29. The Limiting reactant is… Do all calculations starting with this number! Why? This chemical will run out first!... And therefore limits/determines how much can be made..071 mol HCl Now, on your own, start with the OTHER (the one we didn’t choose) and calculated how much of the other one will be needed. Do you get the same conclusion?

30. LR problem 1 cont. When 2.6 grams of hydrochloric acid reacts with 7.44 grams of barium hydroxide, how many grams of water will be made ? 2HCl + Ba(OH) 2  2H 2 O + BaCl 2.071 mol HCl X mol HCl mole H 2 0 2 X g H20 1 18.02 mol H 2 0 = Cal= 1.27942 R= 1.3 SSN= 1.3 x10 0 g H 2 O made 2 Start with LR If you accidentally start with the excess reagent.. Why would the answer be wrong?

31. LR problem 1 cont. Lastly, calculate how much excess reactant will be left over. The excess reactant was ________________. I had this amount to start: _______________. I Cal, that I needed to use this amount: ______________. How can I determine how much would be left over? Show your work. Round using fewest past the decimal (NOT fewest sig figs.) It’s different because you are subtracting- not multiplying or dividing.

32. Limiting reagents problem When 2.6 grams of hydrochloric acid reacts with 7.44 grams of barium hydroxide, A) how many grams of water will be made ? B) How much excess reactant is left over? 2.6 g HCl X g HCl mole HCl…go to Grams of Ba(OH) 2 36.46 1 2HCl + Ba(OH) 2  2H 2 O + BaCl 2 2) Subtract p 7.44 g Ba(OH) 2 ER ( Started with ) - 6.1g ER Ba(OH) 2 (used up) 1.34 g Ba(OH) 2 ER left over!

33. R=.071 mol HCl present R=.0868 mol HCl needed Not enough HCl is present HCl is the Limiting reactant.

34. Solubility of ionic compounds Very soluble compounds “dissolve” easily in water Insoluble compounds do not. How well a compound falls apart (into ions) in water This process is called dissociation

35. LR warm up 2.0 moles of aluminum choride is combined with 138.8 g of sodium oxide… A)LR? B)How many grams of precipitate made? C) How many FU left over? Na 2 O B) 7.610 x 10 1 g Al 2 O 3 C) (.5 moles) 3 x 10 23 FU AlCl 3