2. Chapter 12 Stoichiometry

3. The study of the quantitative, or measurable, relationships that exist in chemical formulas and chemical reactions.Stoichiometry

4. 2 H 2 + O 2  2 H 2 O Interpreting Balanced Chemical Equations Based on the mole ratio

5. Mole Ratio indicated by coefficients in a balanced equation Molar ratio of H 2 to H 2 O is 2:2 (Simplify 1:1) Molar ratio of O 2 to H 2 O is 1:2 2 H 2 + O 2  2 H 2 O

6. 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water. 2 H 2 + O 2  2 H 2 O In terms of moles

7. 4 grams of hydrogen react with 32 grams of oxygen to produce 36 grams of water. 2 H 2 + O 2  2 H 2 O In terms of mass

8. 2 hydrogen molecules react with 1 oxygen molecule to produce 2 water molecules. Notice that the number of molecules is NOT the same on each side of the arrow. 2 H 2 + O 2  2 H 2 O In terms of molecules

9. 4 atoms of hydrogen react with 2 atoms of oxygen to produce 2 water molecules, which are 4 atoms of hydrogen and 2 atoms of oxygen. 2 H 2 + O 2  2 H 2 O In terms of atoms

10. 44.8 L of hydrogen gas react with 22.4 L of oxygen gas to produce 44.8 L of water vapor. Notice that the number of liters of gas is NOT the same on each side of the arrow. 2 H 2 + O 2  2 H 2 O In terms of volumes

11. Law of Conservation of Matter & Mass 2 moles H 2 react with 1 mole of O 2 to form 2 moles of H 2 O. 2 H 2 + O 2  2 H 2 O H2H2H2H2 O2O2O2O2 H2OH2OH2OH2O 2 mol 1 mol 2 mol 2 g 1 mol 32 g 1 mol 18 g 1 mol 4 g 32 g 36 g + = VERIFIED!VERIFIED!

12. Lead will react with hydrochloric acid to produce lead (II) chloride and hydrogen. How many moles of hydrochloric acid are needed to completely react with 0.36 moles of lead? Practice Problem #1 Oh, no … where do I start?

13. Lead will react with hydrochloric acid to produce lead chloride and hydrogen. Write an equation and balance it. Pb + HCl  PbCl 2 + H 2 Pb + HCl  PbCl 2 + H 2 2

14. How many moles of hydrochloric acid are needed to completely react with 0.36 moles of lead? Determine Mole Ratio Pb + HCl  PbCl 2 + H 2 Pb + HCl  PbCl 2 + H 2 2 hydrochloric acid lead Coefficient = 1 Coefficient = 2 “wanted” “given”

15. How many moles of hydrochloric acid are needed to completely react with 0.36 moles of lead? Set up mole ratio “wanted” to “given” Pb + HCl  PbCl 2 + H 2 Pb + HCl  PbCl 2 + H 2 2wantedlead 1 2 hydrochloric acid given = = 0.36 moles of lead X moles of HCl = 1 2

16. How many moles of hydrochloric acid are needed to completely react with 0.36 moles of lead? Solve for “wanted” Pb + HCl  PbCl 2 + H 2 Pb + HCl  PbCl 2 + H 2 2 0.36 moles of lead X moles of HCl = 1 2 X = 2 (0.36 moles) X = 0.72 moles HCl

17. Umm … It would require 0.72 moles of hydrochloric acid. Is that right? That’s Correct!

18. Hem, hem – you have 9 more problems … ready, go.

19. How to solve, given an amount of one substance to another substance. “Stoich” Problems

20. Stoichiometry Steps 1. Write a balanced equation. 2. Identify “wanted” & “given”. 3. Convert given information to moles. 4. Determine Mole Ratio. (Moles of “wanted” : Moles of “given”) 5. Calculate Moles of “wanted”. 6. Convert to required units. KEY step in all stoichiometry problems! 4.Determine Mole ratio ( moles of “wanted” to moles of “given”)

21. What mass of bromine is produced when fluorine reacts with 1.72 g of potassium bromide? Mass-Mass Problems Help … I really need to know this right now!

22. What mass of bromine is produced when fluorine reacts with 1.72 g of potassium bromide? 1. Write a balanced equation. Help … I really need to know this right now! F 2 + 2KBr  2KF + Br 2

23. What mass of bromine is produced when fluorine reacts with 1.72 g of potassium bromide? 2. Identify “wanted” and “given” Help … I really need to know this right now! F 2 + 2KBr  2KF + Br 2 “wanted” “given”

24. 1.72 g of potassium bromide 3. Convert “given” information to moles Help … I really need to know this right now! Molar Mass of KBr = 39 + 80 = 119 g/mol 1.72 g 119 g 1 mol = 0.01445 mol KBr

25. 4. Determine Mole Ratio Help … I really need to know this right now! F 2 + 2KBr  2KF + Br 2 “wanted” “given” Coefficients tell Mole Ratio “wanted” to “given” 1 : 2

26. 5. Calculate Moles of “wanted” Help … I really need to know this right now! F 2 + 2KBr  2KF + Br 2 Moles KBr Moles Br 2 1 = 2 0.01445 mol KBr = X 1 2 2 X = 2 X = 0.01445 mol X = X = 0.007225 mol

27. 6. Convert to required units X = Br 2 X = 0.007225 mol Br 2 0.007225 mol Molar Mass Br 2 80 + 80 = 160 g/mol 160 g 1 mol = 1.16 g Br 2

28. Stoichiometry

29. Limiting Reactants Available Ingredients 4 slices of bread 1 jar of peanut butter 1/2 jar of jelly Limiting Reactant bread Excess Reactants peanut butter and jelly

30. Limiting Reactants Available Ingredients Copper Wire 0.5 g AgNO 3 Limiting Reactant 0.5 grams AgNO 3 Excess Reactants Copper Wire

31. Limiting Reactant The reactant that limits the amount of product that can be formed.

32. When quantities of reactants are available in the exact ratio described by the balanced equation, they are said to be in Stoichiometric proportions.

33. Limiting Reactants Limiting Reactant used up in a reaction determines the amount of all products formed Excess Reactant added to ensure that the other reactant is completely used up usually cheaper & easier to recycle

34. Solving Problems – Limiting Reactants 1. Write a balanced equation. 2. For each reactant, calculate the amount of product formed. 3. The reactant that produces the smaller amount of product is the limiting reactant. Very similar to mass-mass problems!

35. Step 1: Write a balanced equation. Identify the limiting reactant when 1.22 g of oxygen reacts with 1.05 g of hydrogen to produce water. O 2 + 2H 2  2 H 2 O

36. Step 2: Identify the limiting reactant when 1.22 g of oxygen reacts with 1.05 g of hydrogen to produce water. O 2 + 2H 2  2 H 2 O For each reactant, calculate the amount of product formed.

37. Step 2: 1.22 g oxygen O 2 + 2H 2  2 H 2 O wanted given wanted = X 0.038 mol 2 1 given X= 0.076 mol H 2 O 32 g 1 mole = 0.038 mol O 2

38. Step 2: 1.05 g H 2 O 2 + 2H 2  2 H 2 O wanted given wanted = X 0.525 mol 2 2 given X= 0.525 mol H 2 O 2 g 1 mole = 0.525 mol H 2

39. Step 3: 1.22 g of O 2 would produce 0.0763 mol H 2 O The one that produces the smallest amount is your limiting reactant. 1.05 g of H 2 would produce.525 mol H 2 O Oxygen is your limiting reactant!

40. Limiting Reactants Identify the limiting reactant when 1.7 g of sodium reacts with 2.6 L of chlorine gas at STP to produce sodium chloride.

41. Step 1: Write a balanced equation. Identify the limiting reactant when 1.7 g of sodium reacts with 2.6 L of chlorine gas at STP to produce sodium chloride. 2Na + Cl 2  2NaCl

42. Step 2: Identify the limiting reactant when 1.7 g of sodium reacts with 2.6 L of chlorine gas at STP to produce sodium chloride. 2Na + Cl 2  2NaCl For each reactant, calculate the amount of product formed.

43. Step 2: 1.7 g Na 2Na + Cl 2  2NaCl wanted given wanted = X 0.0739 mol 2 2 given X= 0.0739 mol NaCl 23 g 1 mole = 0.0739 mol Na

44. Step 2: 2.6 L Cl 2 2Na + Cl 2  2NaCl wanted given wanted = X 0.116 mol 2 1 given X= 0.232 mol NaCl 22.4 L 1 mole = 0.116 mol Cl 2

45. Step 3: 1.7 g Na would produce 0.0739 mol NaCl The one that produces the smallest amount is your limiting reactant. 2.6 L Cl 2 would produce 0.232 mol NaCl Sodium is your limiting reactant! Click on the real player file called Sodium_Chlorine_2 to see a demo of this reaction

46. Percent Yield calculated on paper measured in lab

47. Percent Yield When 45.8 g of K 2 CO 3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical yield and % yield of KCl. K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g? g actual: 46.3 g

48. Percent Yield 45.8 g K 2 CO 3 1 mol K 2 CO 3 138 g K 2 CO 3 = 49.1 g KCl 2 mol KCl 1 mol K 2 CO 3 74 g KCl 1 mol KCl K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g? g actual: 46.3 g Theoretical Yield:

49. Percent Yield Theoretical Yield = 49.1 g KCl % Yield = 46.3 g 49.1 g  100 = 94.3% K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g49.1 g actual: 46.3 g