2. Stoichiometry: Limiting Reactants Chapter 9 Lesson 3

3. PRACTICE PROBLEM 1. How many grams of gas can be produced from 0.8876 moles of HgO? 2 HgO  2 Hg + O 2 2. How many moles of fluorine are required to produce 12.0 grams of KrF 6 ? Given the equation: Kr + 3 F 2  KrF 6 3. How many grams of Na 2 CO 3 will be produced from the thermal decomposition of 250.0 g of NaHCO 3 ? 4. How many grams of CO 2 can be produced by the reaction of 75.0 grams of C 2 H 2 with excess oxygen? 5. Cu + 2 AgNO 3  Cu(NO 3 ) 2 + 2 Ag. How many grams of silver is produced when 125.0 g of copper is reacted with excess silver nitrate solution? 14.20g 0.182 mol 157.7 g 254 g 424.9 g

4. Limiting Reactants Caution: this stuff is difficult to follow at first. Be patient.

5. + 8 car bodies 48 tires8 cars plus 16 tires excess Limiting Reactants CB + 4 T CT 4

6. plus 8 hydrogen molecules excess + 8 carbon atoms 24 hydrogen molecules 8 methane molecules plus 16 hydrogen atoms excess Limiting Reactants C + 2 H 2 CH 4 Methane, CH 4

7. Container 1 Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 269

8. Before and After Reaction 1 All the hydrogen and nitrogen atoms combine. Before the reaction After the reaction Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 269 N 2 + H 2 NH 3 3 2

9. Container 2 Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 270

10. Before and After Reaction 2 Before the reactionAfter the reaction Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 270 N 2 + H 2 NH 3 3 2 excess limiting LIMITING REACTANT DETERMINES AMOUNT OF PRODUCT

11. Real-World Stoichiometry: Limiting Reactants LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World, 1996, page 366 Ideal Stoichiometry Limiting Reactants Fe + S FeS S = Fe = excess

12. Grilled Cheese Sandwich Bread + Cheese  ‘Cheese Melt’ 2 B + C  B 2 C 100 bread 30 slices? sandwiches Multiple Guess: 130 sandwiches 100 sandwiches 90 sandwiches 60 sandwiches 30 sandwiches Not enough information given 30 sandwiches

13. Limiting Reactants Available IngredientsAvailable Ingredients –4 slices of bread –1 jar of peanut butter –1/2 jar of jelly Limiting ReactantLimiting Reactant –bread Excess ReactantsExcess Reactants –peanut butter and jelly Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

14. Limiting Reactants Limiting ReactantLimiting Reactant –used up in a reaction –determines the amount of product Excess ReactantExcess Reactant –added to ensure that the other reactant is completely used up –cheaper & easier to recycle Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

15. Limiting Reactants aluminum + chlorine gas  aluminum chloride Al(s) + Cl 2 (g)  AlCl 3 2 Al(s) + 3 Cl 2 (g)  2 AlCl 3 100 g 100 g ? g A. 200 gB. 125 gC. 667 gD. 494 g

16. Limiting Reactants aluminum + chlorine gas  aluminum chloride 2 Al(s) + 3 Cl 2 (g)  2 AlCl 3 100 g 100 g x g AlAlCl 3 x g AlCl 3 = 100 g Al 27 g Al = 494 g AlCl 3 1 mol Al 2 mol AlCl 3 2 mol Al 133.5 g AlCl 3 1 mol AlCl 3 How much product would be made if we begin with 100 g of aluminum? Cl 2 AlCl 3 x g AlCl 3 = 100 g Cl 2 71 g Cl 2 = 125 g AlCl 3 1 mol Cl 2 2 mol AlCl 3 3 mol Cl 2 133.5 g AlCl 3 1 mol AlCl 3 How much product would be made if we begin with 100 g of chlorine gas?

17. Limiting Reactants – Method 1 1. Write a balanced equation. 2. For each reactant, calculate the amount of product formed. 3. Smaller answer indicates: –limiting reactant –amount of product Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

18. Limiting Reactants – Method 2 Begin by writing a correctly balanced chemical equation Write down all quantitative values under equation (include units) Convert ALL reactants to units of moles Divide by the coefficient in front of each reactant The smallest value is the limiting reactant!

19. 18 Example #1 2Ca + O 2 → 2CaO How many moles of calcium oxide can you form starting with 15 moles of calcium and 12 moles of oxygen? 1. If not already, balance the chemical equation. Balanced. 2. Convert information given into units of moles. Already in units of moles. 3. Divide the mole amount of each reactant by its stoichiometric ratio from the balanced chemical equation. 15 moles Ca = 7.5 12 moles O 2 = 12 2 moles Ca 1 mol O 2

20. 19 Example #1 (con’t) 2Ca + O 2 → 2CaO How many moles of calcium oxide can you form starting with 15 moles of calcium and 12 moles of oxygen? 4. The reactant with the smallest quotient from step #3 will be the limiting reagent. Calcium will be the limiting reagent in this problem. 5. Work as typical stoichiometry problem using the limiting reagent. 15mol Ca x 2 mol CaO = 15mol CaO 2 mol Ca 19

21. Example #2 What mass of CO 2 can be produced by the reaction of 8.0 grams of CH 4 with 48 grams of O 2 according to the equation below? CH 4 + 2O 2 → CO 2 + 2H 2 O 1. If not already, balance the chemical equation. Balanced. 2. Convert information given into units of moles. Information given is in units of grams. We need to use the molar mass of each reactant to find moles of each reactant. 8.0g CH 4 x 1mol CH 4 = 0.50mol CH 4 16.04g CH 4 48g O 2 x 1mol O 2 = 1.50mol O 2 32.0g O 2

22. Example #2 (con’t) What mass of CO 2 can be produced by the reaction of 8.0 grams of CH 4 with 48 grams of O 2 according to the equation below? CH 4 + 2O 2 → CO 2 + 2H 2 O 3. Divide the mole amount of each reactant by its stoichiometric ratio from the balanced chemical equation. 0.5mol CH 4 = 0.5 1.5mol O 2 = 0.75 1 mol CH 4 2mol O 2 4. The reactant with the smallest quotient from step #3 will be the limiting reagent. CH 4

23. Example #2 (con’t) What mass of CO 2 can be produced by the reaction of 8.0 grams of CH 4 with 48 grams of O 2 according to the equation below? CH 4 + 2O 2 → CO 2 + 2H 2 O 4. Work as typical stoichiometry problem using the limiting reagent. 0.5mol CH 4 x 1mol CO 2 x 44g CO 2 = 22g CO 2 1mol CH 4 1mol CO 2

24. Limiting Reagents: shortcut Do two separate calculations using both given quantities. The smaller answer is correct. Ex3. How many g NO are produced if 20 g NH 3 is burned in 30 g O 2 ? 4NH 3 + 5O 2  6H 2 O+ 4NO 4 mol NO 5 mol O 2 x 30 g O 2 22.5 g NO= 30.0 g NO 1 mol NO x 1 mol O 2 32.0 g O 2 x 4 mol NO 4 mol NH 3 x # g NO= 20 g NH 3 35.3 g NO= 30.0 g NO 1 mol NO x 1 mol NH 3 17.0 g NH 3 x

25. Class work 1.2Al + 6HCl  2AlCl 3 + 3H 2 If 25 g of aluminum was added to 90 g of HCl, what mass of H 2 will be produced (try this two ways – with the steps & using the shortcut)? 2.N 2 + 3H 2  2NH 3 : If you have 20 g of N 2 and 5.0 g of H 2, which is the limiting reagent? 3.What mass of aluminum oxide is formed when 10.0 g of Al is burned in 20.0 g of O 2 ? 4.When C 3 H 8 burns in oxygen, CO 2 and H 2 O are produced. If 15.0 g of C 3 H 8 reacts with 60.0 g of O 2, how much CO 2 is produced? 5.How can you tell if a question is a limiting reagent question vs. typical stoichiometry?

26. 1 1 mol Al 27.0 g Al x # mol Al =25 g Al = 0.926 mol 2mol # mol HCl =90 g HCl 1 mol HCl 36.5 g HCl x = 2.466 mol 6mol HCl is limiting. 3 mol H 2 6 mol HCl x # g H 2 = 90 g HCl 2.0 g H 2 1 mol H 2 x 1 mol HCl 36.5 g HCl x = 2.47 g H 2 = 0.463 =0.411

27. Question 1: shortcut 2Al + 6HCl  2AlCl 3 + 3H 2 If 25 g aluminum was added to 90 g HCl, what mass of H 2 will be produced? 3 mol H 2 2 mol Al x 25 g Al = 2.78 g H 2 2.0 g H 2 1 mol H 2 x 1 mol Al 27.0 g Al x 3 mol H 2 6 mol HCl x 90 g HCl = 2.47 g H 2 2.0 g H 2 1 mol H 2 x 1 mol HCl 36.5 g HCl x

28. Question 2: shortcut N 2 + 3H 2  2NH 3 If you have 20 g of N 2 and 5.0 g of H 2, which is the limiting reagent? 2 mol NH 3 1 mol N 2 x # g NH 3 = 20 g N 2 = 24.3 g H 2 17.0 g NH 3 1 mol NH 3 x 1 mol N 2 28.0 g N 2 x 2 mol NH 3 3 mol H 2 x # g NH 3 = 5.0 g H 2 = 28.3 g H 2 17.0 g NH 3 1 mol NH 3 x 1 mol H 2 2.0 g H 2 x N 2 is the limiting reagent

29. Question 3: shortcut 4Al + 3O 2  2 Al 2 O 3 What mass of aluminum oxide is formed when 10.0 g of Al is burned in 20.0 g of O 2 ? 2 mol Al 2 O 3 4 mol Al x # g Al 2 O 3 = 10.0 g Al = 18.9 g Al 2 O 3 102.0 g Al 2 O 3 1 mol H 2 x 1 mol Al 27.0 g Al x 2 mol Al 2 O 3 3 mol O 2 x # g Al 2 O 3 = 20.0 g O 2 = 42.5 g Al 2 O 3 102.0 g Al 2 O 3 1 mol H 2 x 1 mol O 2 32.0 g O 2 x

30. Question 4: shortcut C 3 H 8 + 5O 2  3CO 2 + 4H 2 O When C 3 H 8 burns in oxygen, CO 2 and H 2 O are produced. If 15.0 g of C 3 H 8 reacts with 60.0 g of O 2, how much CO 2 is produced? 3 mol CO 2 1 mol C 3 H 8 x # g CO 2 = 15.0 g C 3 H 8 = 45.0 g CO 2 44.0 g CO 2 1 mol CO 2 x 1 mol C 3 H 8 44.0 g C 3 H 8 x 3 mol CO 2 5 mol O 2 x # g CO 2 = 60.0 g O 2 = 49.5 g CO 2 44.0 g CO 2 1 mol CO 2 x 1 mol O 2 32.0 g O 2 x 5. Limiting reagent questions give values for two or more reagents (not just one)

31. 34g NH 3 19g H 2 excess 23.0L H 2 25g Mg excess 164.3g Ag 3 PO 4 135.6g Na 3 PO 4 30 Class work Con’t 6. How many grams of NH 3 can be produced from the reaction of 28g of N 2 and 25g of H 2 ? 7. How much of the excess reagent in Problem 6 is left over? 8. What volume of hydrogen at STP is produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl? 9. How much of the excess reagent in Problem 3 is left over? 10. Silver nitrate and sodium phosphate are reacted in equal amounts of 200g each. How many grams of silver phosphate are produced? 11. How much of the excess reagent in problem 5 is left?

32. 31 12. Iron reacts with sulfur to form iron(II) sulfide. You have 32.0g of sulfur and 100g of iron. Calculate the number of grams of iron(II) sulfide that will form. 13. Zinc reacts with sulfuric acid in a single replacement reaction. You have 40g of zinc and 57g of sulfuric acid. What volume of hydrogen gas will form? 14. Silver nitrate and sodium bromide will react in a double replacement reaction. You have 24g of silver nitrate and 39g of sodium bromide. How many grams of sodium nitrate will form? 15. When hydrochloric acid reacts with iron(II) sulfide, a double replacement reaction is to be expected. You have 67g of hydrochloric acid and 58 of iron(II) sulfide. How many grams of hydrogen sulfide will form? 16. When aluminum is heated in oxygen, aluminum oxide is formed. You have 384g of each reactant. How many moles of aluminum oxide will form? 31 87.9g FeS 13.0L H 2 11.9g NaNO 3 22.5g H 2 S 6.45mol Al 2 O 3

33. Theoretical Yield The theoretical yield is the amount of product that can be made –In other words it’s the amount of product possible as calculated through the stoichiometry problem This is different from the actual yield, the amount one actually produces and measures

34. Percent Yield A comparison of the amount actually obtained to the amount it was possible to make Actual Yield Theoretical Yield Percent Yield =x 100