Chapter 5 Chemical Reactions

1. Chapter 5Chemical Reactions

2. The Mole

3. The Mole • 1 mole = 6.022 x 1023 things • Avogadro’s number • The number of carbon atoms in 12g of C-12 • Abbreviation: “mol” How much does this weigh?

4. H2 (g) + Cl2(g)  2 HCl (g) Chemical Equations Coefficient Physical State Subscript Reactants Product(s)

5. Balancing Chemical Equations • Write the unbalanced equation • Balance the atoms of one element • Choose another element and balance it • Continue until all elements have the same number of atoms on both sides of the equation • Check yourself Zn(s) + HCl(aq)  H2(g) + ZnCl2(aq)

6. Problems • __ N2(g) + __ H2(g)  __ NH3(g) • __ Fe(s) + __ Cl2(g)  __ FeCl3(s) • __NH3(g) + __O2(g)  __NO(g) + __H2O(g) • __C5H12(l) + __O2(g)  __CO2(g) + __H2O(g)

7. General Reactions • Combination rxns: 2 or more substances react to form a single product • 2 H2 + O2 2 H2O

8. Decomposition rxns: single substance decomposes into 2 or more products • opposite of combination rxns • 2 H2O  2 H2 + O2

10. Single replacement/displacement: one element reacts with a compound to form a new compound and release a new element • 2 Na + 2 H2O  2 NaOH + H2

11. Exchange or Double replacement/displacement: an interchange of partners between two compounds • Pb(NO3)2(aq) + K2CrO4(aq)  PbCrO4(s) + 2 KNO3(aq)

12. Combustion rxns: the burning of a compound, usually a hydrocarbon, in oxygen to form heat, carbon dioxide and water

13. 5.7 Mole Calculations in Chemical Equations A balanced chemical equation also tell us: • The number of moles of each reactant that combine • The number of moles of each product formed 1 N2(g) 1 O2(g) 2 NO(g) + 1 mole of N2 1 molecule N2 1 mole of O2 1 molecule O2 2 moles of NO 2 molecules NO (The coefficient “1” has been written for emphasis.)

14. 5.7 Mole Calculations in Chemical Equations Coefficients are used to form mole ratios, which can serve as conversion factors. N2(g) O2(g) 2 NO(g) + Mole ratios: 1 mol N2 2 mol NO 1 mol O2 2 mol NO 1 mol N2 1 mol O2

15. 5.7 Mole Calculations in Chemical Equations Use the mole ratios from the coefficients in the balanced equation to convert moles of one compound (A) into moles of another compound (B).

16. 5.7 Mole Calculations in Chemical Equations Sample Problem 5.11 Using the balanced chemical equation, how many moles of CO are produced from 3.5 moles of C2H6? 2 C2H6(g) + 5 O2(g) 4 CO(g) + 6 H2O(g) Step [1] Identify the original and desired quantities. 3.5 mol C2H6 original quantity ? mol CO desired quantity

17. 5.7 Mole Calculations in Chemical Equations Step [2] Write out the conversion factors. 2 mol C2H6 4 mol CO 4 mol CO 2 mol C2H6 or 2 C2H6(g) + 5 O2(g) 4 CO(g) + 6 H2O(g) Choose this one to cancel mol C2H6. Step [3] Set up and solve the problem. 4 mol CO 2 mol C2H6 3.5 mol C2H6 x = 7.0 mol CO Unwanted unit cancels.

18. 5.8 Mass Calculations in Chemical Equations HOW TO Convert Moles of Reactant to Grams of Product Using the balanced equation, how many grams of O3 are formed from 9.0 mol of O2. Example sunlight 3 O2(g) 2 O3(g) [1] [2] Grams of product Moles of product Moles of reactant mole–mole conversion factor molar mass conversion factor

19. 5.8 Mass Calculations in Chemical Equations HOW TO Convert Moles of Reactant to Grams of Product Convert the number of moles of reactant to the number of moles of product using a mole–mole conversion factor. Step [1] 3 mol O2 2 mol O3 2 mol O3 3 mol O2 sunlight 3 O2(g) 2 O3(g) or mole–mole conversion factor Cancel mol O2 in Step [1].

20. 5.8 Mass Calculations in Chemical Equations HOW TO Convert Moles of Reactant to Grams of Product Convert the number of moles of product to the number of grams of product using the product’s molar mass. Step [2] MM O3 = 16.0 x 3 = 48.0 g/mol molar mass conversion factor 48.0 g O3 1 mol O3 1 mol O3 48.0 g O3 Cancel mol O3 in Step [2]. or

21. 5.8 Mass Calculations in Chemical Equations HOW TO Convert Moles of Reactant to Grams of Product • Set up and solve the conversion. mole–mole conversion factor molar mass conversion factor Grams of product Moles of reactant 2 mol O3 3 mol O2 48.0 g O3 1 mol O3 x x 9.0 mol O2 = 290 g O3 Mol O2 cancel. Mol O3 cancel.

22. 5.8 Mass Calculations in Chemical Equations HOW TO Convert Grams of Reactant to Grams of Product Ethanol (C2H6O, molar mass 46.1 g/mol) is synthesized by reacting ethylene (C2H4, molar mass 28.1 g/mol) with water. How many grams of ethanol are formed from 14 g of ethylene? Example C2H4 + H2O C2H6O

23. 5.8 Mass Calculations in Chemical Equations HOW TO Convert Grams of Reactant to Grams of Product mole–mole conversion factor [2] Moles of reactant Moles of product molar mass conversion factor molar mass conversion factor [3] [1] Grams of reactant Grams of product

24. 5.8 Mass Calculations in Chemical Equations HOW TO Convert Grams of Reactant to Grams of Product C2H4 + H2O C2H6O molar mass conversion factor mole–mole conversion factor molar mass conversion factor Grams of reactant 1 mol C2H4 28.1 g C2H4 1 mol C2H6O 1 mol C2H4 46.1 g C2H6O 1 mol C2H6O x x x 14 g C2H4 Grams C2H4 cancel. Moles C2H4 cancel. Moles C2H6O cancel. Grams of product = 23 g C2H6O

25. 5.9 Percent Yield • The theoretical yield is the amount of product expected from a given amount of reactant based on the coefficients in the balanced chemical equation. • Usually, however, the amount of product formed is less than the maximum amount of product predicted. • The actual yield is the amount of product isolated from a reaction.

26. 5.9 Percent Yield Sample Problem 5.14 If the reaction of ethylene with water to form ethanol has a calculated theoretical yield of 23 g of ethanol, what is the percent yield if only 15 g of ethanol are actually formed? actual yield (g) theoretical yield (g) x 100% = Percent yield 15 g 23 g = x 100% = 65%

27. Stoichiometry Problems • If you have 10.0g C4H10 (butane), how many grams of water can you make upon combustion? • How much O2 do you need in problem 1 if you’d like to produce 7.39g CO2? • How much CO2 is produced upon combustion of 4.3g of propane (C3H8)?

28. Using the equation below, calculate the amount of glucose you started with if you produced 12.76g CO2? • How much glucose was consumed in order to produce 100.0 mL of H2O? __C6H12O6(s) + __O2(g)  __CO2(g) + __H2O(l)

29. Limiting Reagent • Limiting Reagent/Reactant/Factor: the reactant/factor that determines the amount of product formed • Other reactants are “in excess” • Cheaper reactants are usually in excess

30. 5.10 Limiting Reactants • The limiting reactant is the reactant that is completely used up in a reaction.

31. 5.10 Limiting ReactantC. Determining the Limiting Reactant Using the Number of Grams Sample Problem 5.20 [1] Convert the number of grams of each reactant into moles using the molar masses.

32. 5.10 Limiting ReactantC. Determining the Limiting Reactant Using the Number of Grams Sample Problem 5.20 [2] Determine the limiting reactant by choosing N2 as the original quantity and converting to mol O2. mole–mole Conversion factor 1 mol O2 1 mol N2 = 0.357 mol O2 0.357 mol N2 x The amount of O2 we started with (0.313 mol) isless than the amount we would need (0.357 mol) so O2 is the limiting reagent.

33. Problems • You have 10.0 moles H2 and 1.00 mol O2. How much H2O can you make? • You combust 10.2 mol propane in 7.80 mol O2. How much CO2 can you produce?

34. If 2.3 mol carbon disulfide reacts with 5.4 mol oxygen to form carbon dioxide and sulfur dioxide, what mass of sulfur dioxide is formed? • 5.50 g silicon dioxide reacts with 4.71g Carbon to from silicon carbide and carbon monoxide. What mass of carbon monoxide is formed?

35. Problems • From the previous question, the theoretical yield of CO is 5.13 g. If you obtained 4.32g CO, what was your % yield? • You react 4.41 mol carbon monoxide with 8.39 mol hydrogen gas to get 122g methanol. What is your percent yield?