1. Stoichiometric Calculations Stoichiometry

2. A. Proportional Relationships b I have 5 eggs. How many cookies can I make? 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar 5 eggs5 doz. 2 eggs = 12.5 dozen cookies Ratio of eggs to cookies

3. A. Proportional Relationships b Stoichiometry mass relationships between substances in a chemical reaction based on the mole ratio b Mole Ratio indicated by coefficients in a balanced equation 2 Mg + O 2  2 MgO

4. What can we do with Stoichiometry? b For the generic equation: R A +R B → P 1 + P 2 Given the……one can find the… Amount of R A (or R B )Amount of R A (or R B ) that is needed to react with it Amount of R A or R B Amount of P 1 or P 2 that will be produced Amount of P 1 or P 2 you need to produce Amount of R A &/or R B you must use

5. Given the equation… 2TiO 2 + 4Cl 2 + 3C → 2TiCl 4 + CO 2 + 2CO b How many mol chlorine will react with 4.55 mol carbon? b X mol Cl 2 = 4.55 mol C (4mol Cl 2 / 3mol C) b 6.07 mol Cl 2 will react with 4.55 mol carbon

6. Given the same equation… 2TiO 2 + 4Cl 2 + 3C → 2TiCl 4 + CO 2 + 2CO b What mass titanium (IV) oxide will react with 4.55 mol carbon? b X g TiO 2 = 4.55 mol C(2 mol TiO 2 /3 mol C)(79.9 g TiO 2 /1 molTiO 2 ) b 242 g TiO 2 will react with 4.55 mol carbon

7. And with the same equation… 2TiO 2 + 4Cl 2 + 3C → 2TiCl 4 + CO 2 + 2CO b How many molecules titanium (IV) chloride can be made from 115 g titanium (IV) oxide? b X m’c TiCl 4 = 115g TiO 2 (1mol TiO 2 /79.9g TiO 2 )(2 mol TiCl 4 /2 mol TiO 2 )(6.02 x 10 23 m’c TiCl 4 /1 mol TiCl 4 ) b 8.7 x 10 23 m’cules TiCl 4 can be made from 115 g TiO 2

8. B. 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 Core step in all stoichiometry problems!! Mole ratio - moles  moles 4. Check answer.

9. 1 mol of a gas=22.4 L at STP C. Molar Volume at STP S tandard T emperature & P ressure 0°C and 1 atm

10. C. Molar Volume at STP Molar Mass (g/mol) 6.02  10 23 particles/mol MASS IN GRAMS MOLES NUMBER OF PARTICLES LITERS OF SOLUTION Molar Volume (22.4 L/mol) LITERS OF GAS AT STP Molarity (mol/L)

11. D. Stoichiometry Problems b How many moles of KClO 3 must decompose in order to produce 9 moles of oxygen gas? 9 mol O 2 2 mol KClO 3 3 mol O 2 = 6 mol KClO 3 2KClO 3  2KCl + 3O 2 ? mol9 mol

12. b How many grams of KClO 3 are req’d to produce 9.00 L of O 2 at STP? 9.00 L O 2 1 mol O 2 22.4 L O 2 = 32.8 g KClO 3 2 mol KClO 3 3 mol O 2 122.55 g KClO 3 1 mol KClO 3 ? g9.00 L D. Stoichiometry Problems 2KClO 3  2KCl + 3O 2

13. D. Stoichiometry Problems b How many grams of silver will be formed from 12.0 g copper? 12.0 g Cu 1 mol Cu 63.55 g Cu = 40.7 g Ag Cu + 2AgNO 3  2Ag + Cu(NO 3 ) 2 2 mol Ag 1 mol Cu 107.87 g Ag 1 mol Ag 12.0 g? g

14. 63.55 g Cu 1 mol Cu D. Stoichiometry Problems b How many grams of Cu are required to react with 1.5 L of 0.10M AgNO 3 ? 1.5 L.10 mol AgNO 3 1 L = 4.8 g Cu Cu + 2AgNO 3  2Ag + Cu(NO 3 ) 2 1 mol Cu 2 mol AgNO 3 ? g 1.5L 0.10M

15. Stoichiometry “Pop” Quiz b Iron (III) reacts with water… Write the complete balanced equation How many liters of hydrogen (@ STP) will be produced from 8 moles of iron? How many grams of water are needed to produce 68 grams of Iron (III) oxide? 5.2 x 10 25 particles of iron will produce how many moles of hydrogen? b For the complete combustion of butane (C 4 H 10 )… Write the complete balanced equation In a room at 0°C and 1 atm pressure, how many m’cules oxygen react with 632 L butane? If 316 L liters of carbon dioxide gas are produced, how much butane (in moles) was combusted? b Silver nitrate reacts with sodium chloride… Write the complete balanced equation (including states of matter) How much precipitate (in g) is formed when 1.5 L of 0.10M silver nitrate is used? How many molecules of sodium chloride do you need to react with that amount of silver nitrate?

16. The Limiting Reactant A balanced equation for making a Big Mac® might be: 3 B + 2 M + EE  B3M2EE With……and… …one can make… 30 M excess B and excess EE 15 B 3 M 2 EE 30 B excess M and excess EE 10 B 3 M 2 EE 30 M 30 B and excess EE 10 B 3 M 2 EE

17. A balanced equation for making a tricycle might be: 3 W + 2 P + S + H + F  W3P2SHF With……and… …one can make… 50 P excess of all other reactants 25 W 3 P 2 SHF 50 S excess of all other reactants 50 W 3 P 2 SHF 50 P 50 S and excess of all other reactants 25 W 3 P 2 SHF

18. Solid aluminum reacts w/chlorine gas to yield solid aluminum chloride. 2 Al(s) + 3 Cl2(g)  2 AlCl3(s) If 125 g aluminum react w/excess chlorine, how many g aluminum chloride are made? = 618 g AlCl 3

19. If 125 g chlorine react w/excess aluminum, how many g aluminum chloride are made? = 157 g AlCl 3 b If 125 g aluminum react w/125 g chlorine, how many g aluminum chloride are made? 157 g AlCl 3 b We’re out of Cl 2.

20. b limiting reactant (LR): the reactant that runs out first Amount of product is based on LR. b Any reactant you don’t run out of is an excess reactant (ER).

21. Example Limiting Reactant Excess Reactant(s) Big Macsbunsmeat tricyclespedalsW, S, H, F Al / Cl 2 / AlCl 3 Cl 2 Al

22. How to Find the Limiting Reactant b For the generic reaction R A + R B  P b Assume that the amounts of R A and R B are given. b Should you use R A or R B in your calculations?

23. How to find LR steps… 1. Calc. # of mol of R A and R B you have. 2. Use the mole ratio from the balanced equation to calculate the required amount of the other reactant. 3. If resulting amount is less than what is given, that reactant is the LR.

24. % Yield molten + solid  molten + solid sodium aluminum aluminum sodium oxide oxide Al3+ O2– Na1+ O2– ___Na(l) + ___Al2O3(s)  ___Al(l) +___Na2O(s) 6 Na(l)+ 1 Al2O3(s)  2 Al(l) + 3 Na2O(s) Find mass of aluminum produced if you start w/575 g sodium and 357 g aluminum oxide. = 189 g Al

25. % Yield b This amount of product is the theoretical yield. amt. we get if reaction is perfect found by calculation b Now suppose that we perform this reaction in the lab and get only 172 grams of aluminum. Why? couldn’t collect all Al not all Na and Al2O3 reacted some reactant or product spilled and was lost

26. % Yield Calculation: % yield can never be > 100%

27. Find % Yield for the Previous Problem