1. Stoichiometry Stoichiometry CDO High School

2. Stoichiometry Consider the chemical equation: 4NH 3 + 5O 2  6H 2 O + 4NO There are several numbers involved. What do they all mean? “stochio” = Greek for element “metry” = measurement Stoichiometry is about measuring the amounts of elements and compounds involved in a reaction.

3. Stoichiometry With Stoichiometry we find out that 4 : 5 : 6 : 4 do more than just multiply atoms. 4NH 3 + 5O 2  6H 2 O + 4NO 4 : 5 : 6 : 4 Are what we call a mole ratio.

4. Stoichiometry 4NH 3 + 5O 2  6H 2 O + 4NO 4 : 5 : 6 : 4 Can mean either: 4 molecules of NH 3 react with 5 molecules of O 2 to produce 6 molecules of H 2 O and 4 molecules of NO OR 4 moles of NH 3 react with 5 moles of O 2 to produce 6 moles of H 2 O and 4 moles of NO

5. Stoichiometry Equations

6. Mole to Mole Moles of Known Coefficient Known Coefficient Unknown

7. Stoichiometry Question (1) 4NH 3 + 5O 2  6H 2 O + 4NO How many moles of H 2 O are produced if 2.00 moles of O 2 are used? 2.00 mol O 2 2.40 mol H 2 O = Notice that a correctly balanced equation is essential to get the right answer 6 mol H 2 O 5 mol O 2

8. 4 mol NO 6 mol H 2 O Stoichiometry Question (2) 4 NH 3 + 5 O 2  6 H 2 O + 4 NO How many moles of NO are produced in the reaction if 15 mol of H 2 O are also produced? 15 mol H 2 O 10. mol NO =

9. Mole to Mass Mole of Known Coefficient Unknown Coefficient Known Molar Mass Unknown

10. 18.02 g H 2 O 1 mol H 2 O 6 mol H 2 O 4 mol NH 3 Stoichiometry Question (3) 4 NH 3 + 5 O 2  6 H 2 O + 4NO How many grams of H 2 O are produced if 2.2 mol of NH 3 are combined with excess oxygen? 2.2 mol NH 3 59 g H 2 O =

11. Mass to Mole Coefficient Unknown Coefficient Known Molar Mass Known Mass Known

12. 1 mol H 2 O 5 mole O 2 6 mol H 2 O Stoichiometry Question (4) 4 NH 3 + 5 O 2  6 H 2 O + 4 NO How many moles of O 2 are required to produce 42.3g of H 2 O? 42.3 gH 2 O 1.96 mol O 2 = 18.02 g H 2 O

13. Mass to Mass Conversions Coefficient Unknown Coefficient Known Molar Mass Known Mass Known Molar Mass Unknown

14. Stoichiometry Question (5) 4 NH 3 + 5 O 2  6 H 2 O + 4 NO How many grams of NO is produced if 12 g of O 2 is combined with excess ammonia? 4 mol NO 5 mol O 2 12 g O 2 9.0 g NO = 30.01 g NO 1 mol NO 1 mol O 2 32 g O 2

15. 3CuSO 4 + 2Al  Al 2 (SO 4 ) 3 + 3Cu What is the mass of copper formed when 15.8g of Aluminum is used? 3 mol Cu 2 mol Al 15.8g Al 55.8 g Cu = 63.55 g Cu 1 mol Cu 1 mol Al 26.98 g Al

16. Have we learned it yet? 4 NH 3 + 5 O 2  6 H 2 O + 4 NO a) How many moles of H 2 O can be made using 1.6 mol NH 3 ? b) what mass of NH 3 is needed to make 0.75 mol NO? c) how many grams of NO can be made from 47 g of NH 3 ?

17. Answers 4 NH 3 + 5 O 2  6 H 2 O + 4 NO a) b) c) 6 mol H 2 O 4 mol NH 3 x 1.6 mol NH 3 2.4 mol H 2 O = 4 mol NH 3 4 mol NO x 0.75 mol NO 13 g NH 3 = 17.04 g NH 3 1 mol NH 3 x 4 mol NO 4 mol NH 3 x 47 g NH 3 83 g NO = 30.01 g NO 1 mol NO x 17.04 g/mol NH 3 ÷

18. Reactions Involving a LIMITING REACTANT

19. LIMITING REACTANTS eactants R eactantsProducts 2 NO(g) + O 2 (g) 2 NO 2 (g) Limiting reactant = ___________ Excess reactant = ____________

20. Limiting Reactants: An Analogy If you were going to make pb and j sandwiches, and you had a new loaf of bread and a large jar of pb and a large jar of jelly, how many sandwiches can you make? CDO IB Chemistry SL20

21. Example: Limiting Reactants Mix 5.40 g of Al with 8.10 g of Cl 2. What mass of AlCl 3 can form? Al(s) + Cl 2 (g)  AlCl 3 (s) CDO IB Chemistry SL21

22. Cl 2 was the limiting reactant. Therefore, Al was present in excess. But how much? First find how much Al was required. Then find how much Al is in excess. How much of which reactant will remain when reaction is complete?

23. 2 Al + 3 Cl 2 2 AlCl 3 Calculating Excess Al

24. White Boards In the reaction BaCO 3 + 2HNO 3  Ba(NO 3 ) 2 + CO 2 + H 2 O, what mass of Ba(NO 3 ) 2 can be formed by combining 55g BaCO 3 and 26g HNO 3

25. Percent Yield

26. 2As 2 O 3 + 3C  3CO 2 + 4As ◦ If 8.87g of As 2 O 3 is used in the reaction and 5.33 g of As is produced, what is the percent yield?

27. Gas Stoichiometry Many chemical reactions involve gases as a reactant or a product Gas Stoichiometry – the procedure for calculating the volume of gases as products or reactants Gases also have a molar volume (L/mol) rather than concentration. This is the conversion factor used to convert (liters of gas) to (moles of gas)

28. Molar Volume Molar volume is the same for all gases at the same temperature and pressure (remember, all gases have the same physical properties) ◦ At STP, molar volume = 22.4 L/mol  This can be used as a conversion factor just like molar mass! At STP, one mole of gas has a volume of 22.4 L, which is approximately the volume of 11 “empty” 2 L pop bottles. STP = 22.4L/mol

29. Volume of Gas to Volume of Gas Volume of Gas Known Coefficient Unknown Coefficient Known Volume of Gas to Mass Volume of Gas Known Coefficient Unknown Coefficient Known 22.4 L 1 mol Molar Mass Unknown Mass to Volume of Gas Mass Known Molar Mass Known Coefficient Known Coefficient Unknown 22.4 L 1 mol

30. Example #1 If 300. 0g of propane burns, what volume of oxygen measured at STP is required for the reaction? C 3 H 8(g) + 5O 2(g)  3CO 2(g) + 4H 2 O (g) 300.0 g C 3 H 8 44.11 g C 3 H 8 1 mol C 3 H 8 5 mol O 2 22.4 L 1 mol 1 mol C 3 H 8 = 761.7 L O 2

31. Example #2 Hydrogen gas is produced when sodium metal is added to water. What mass of sodium is necessary to produce 20.0L of hydrogen at STP? 2Na (s) + 2H 2 O (l)  2NaOH (aq) + H 2(g) 20.0 L H 2 2 mol Na 22.4 L 1 mol 22.99 g Na 1 mol H 2 1 mol Na = 41.1 g Na

32. SOLUTIONS

33. Solutions Solute – what is being dissolved, usually present in the smallest amount Solvent – what is doing the dissolving, usually present in the largest amount Solution – solute and solvent combined Aqueous solutions – solutions in which water is the solvent

34. Concentration Concentration – describes how much solute is dissolved in solvent Saturated solution – when the solute can no longer dissolve in the given solvent Concentration unit (mol/L) often referred to as Molarity (M) [Concentration] = moles of solute volume of solution (L) CDO IB Chemistry SL34

35. Example: Concentration A solution contains 4.10 g of NaCl in 1.00 L. What is the solutions concentration in mol/L.? CDO IB Chemistry SL35

37. Standard Solution CDO IB Chemistry SL37

38. Example: Making a Standard Solution Calculate the mass of NaOH need to make 250 mL of a 0.200 mol/L solution. CDO IB Chemistry SL38

40. TITRATIONS

41. Titrations Laboratory Technique – which uses a standard solution to find the concentration of another solution.

42. Titration Calculation n 2 M 1 V 1 = n 1 M 2 V 2 n= Coefficient of substance M = molarity of substance V = volume of substance

43. Example: Titration What volume of 0.100 mol/L NaOH is required to titrate 25 mL of 0.300 mol/L solution of HCl to produce a neutral solution? CDO IB Chemistry SL43

44. H 2 SO 4 + 2NaOH  Na 2 SO 4 + 2H 2 O What is the concentration of 25.85 mL of H 2 SO 4 that is reacted with 21.50 mL of 0.300 mol/L solution of NaOH to produce a neutral solution?