1. 1 Chapter 8 Quantities in Chemical Reactions Tro, 2 nd ed.

2. 2 STOICHIOMETRY: SOME REVIEW The molar mass of an element is its atomic mass in grams. It contains 6.0221 x 10 23 atoms (Avogadro’s number) of the element. The molar mass of an element or compound is the sum of the atomic masses of all its atoms.

3. 3 Avogadro’s Number of Particles 6.0221 x 10 23 Particles Molar Mass 1 MOLE

4. 4 How many moles of NaCl are present in 292.215 grams of NaCl? The molar mass of NaCl =58.443 g.

5. 5 STOICHIOMETRY Consider the combustion of acetylene: 2 C 2 H 2(g) + 5 O 2(g)  4 CO 2(g) + 2 H 2 O (l) 2 molecules 5 molecules 4 molecules 2 molecules 200 “ 500 “ 400 “ 200 “ 12x10 23 31x10 23 24x10 23 12x10 23 2 moles 5 moles 4 moles 2 moles Consider each of the formulas as molecules or moles Coefficients represent number of moles “Recipe” just like a cooking recipe: 2 cups of acetylene requires 5 cups of oxygen, etc. When we use the coefficients to relate the chemicals in an equation, we are doing STOICHIOMETRY

6. 6 STOICHIOMETRY 2 C 2 H 2(g) + 5 O 2(g)  4 CO 2(g) + 2 H 2 O (l) EXAMPLE: If 6 moles of C 2 H 2 are burned, then 15 moles of O 2 are needed, 12 moles of CO 2 are made and 6 moles of H 2 O are made How did we know? We used ratios. Practice with 0.102 moles of C 2 H 2 : 0.102 mol * 5 mol O 2 /2 mol C 2 H 2 = 0.255 mol O 2 0.102 mol * 4 mol CO 2 /2 mol C 2 H 2 = 0.204 mol CO 2 0.102 mol * 2 mol H 2 O/2 mol C 2 H 2 = 0.102 mol H 2 O

7. 7 Introduction to Stoichiometry: The Mole-Ratio Method Stoichiometry: The area of chemistry that deals with the quantitative relationships between reactants and products. Mole Ratio: a ratio between the moles of any two substances involved in a chemical reaction. The coefficients used in mole ratio expressions are derived from the coefficients used in the balanced equation.

8. 8 N 2 + 3H 2  2NH 3 1 mol2 mol3 mol The Mole-Ratio Method

9. 9 1 mol2 mol3 mol N 2 + 3H 2  2NH 3 The Mole-Ratio Method

10. 10 The Mole-Ratio Method The mole ratio is used to convert the number of moles of one substance to the corresponding number of moles of another substance in a stoichiometry problem. The mole ratio is used in the solution of every type of stoichiometry problem.

11. 11 The Mole Ratio Method for Stoichiometry 1. Have a balanced chemical equation. 2. Convert the quantity of starting substance to moles (if it is not already moles) 3. Convert the moles of starting substance to moles of desired substance. 4. Convert the moles of desired substance to the units specified in the problem.

12. 12 In the following reaction how many moles of PbCl 2 are formed if 5.000 moles of NaCl react? 2NaCl(aq) + Pb(NO 3 ) 2 (aq)  PbCl 2 (s) + 2NaNO 3 (aq)

13. 13 Mole Ratio Calculate the number of moles of phosphoric acid (H 3 PO 4 ) formed by the reaction of 10 moles of sulfuric acid (H 2 SO 4 ). Ca 5 (PO 4 ) 3 F + 5H 2 SO 4  3H 3 PO 4 + HF + 5CaSO 4 Step 2 Moles starting substance: 10.0 mol H 2 SO 4 Step 3 The conversion needed is moles H 2 SO 4  moles H 3 PO 4 1 mol5 mol3 mol1 mol5 mol

14. 14 Mole-Mass Calculations The object of this type of problem is to calculate the mass of one substance that reacts with or is produced from a given number of moles of another substance in a chemical reaction. If the mass of the starting substance is given, we need to convert it to moles. (Step 2) We use the mole ratio to convert moles of starting substance to moles of desired substance. (Step 3) We can then change moles of desired substance to mass of desired substance if called for by the problem. (Step 4)

15. 15 Mole Ratio Calculate the number of moles of H 2 SO 4 necessary to yield 784 g of H 3 PO 4. Ca 5 (PO 4 ) 3 F+ 5H 2 SO 4  3H 3 PO 4 + HF + 5CaSO 4 Method 1 Step by Step Calculations Step 1 Balance reaction given, with 784 grams of H 3 PO 4. Step 2 Convert grams of H 3 PO 4 to moles of H 3 PO 4. Step 3 Convert moles of H 3 PO 4 to moles of H 2 SO 4 by the mole-ratio method.

16. 16 Mole Ratio Calculate the number of moles of H 2 SO 4 necessary to yield 784 g of H 3 PO 4 Ca 5 (PO 4 ) 3 F+ 5H 2 SO 4  3H 3 PO 4 + HF + 5CaSO 4 Method 2 Continuous Calculation grams H 3 PO 4  moles H 3 PO 4  moles H 2 SO 4 The conversion needed is

17. 17 Mass-Mass Calculations Solving mass-mass stoichiometry problems requires all four steps of the mole-ratio method. You list the four steps here:

18. 18 Calculate the number of grams of NH 3 formed by the reaction of 112 grams of H 2. N 2 + 3 H 2  2 NH 3 Method 1 Step by Step Calculations Step 2 Convert 112 g of H 2 to moles. grams  moles Step 3 Calculate the moles of NH 3 by the mole ratio method.

19. 19 Calculate the number of grams of NH 3 formed by the reaction of 112 grams of H 2. N 2 + 3 H 2  2 NH 3 Step 4 Convert moles NH 3 to grams NH 3. moles  grams

20. 20 Calculate the number of grams of NH 3 formed by the reaction of 112 grams of H 2. N 2 + 3 H 2  2 NH 3 grams H 2  moles H 2  moles NH 3  grams NH 3 Method 2 Continuous Calculation

21. 21 GROUP PRACTICE: First write the balanced chemical reaction: ammonium phosphate reacts with calcium hydroxide to produce calcium phosphate, ammonia and water. Then answer: a. What weight of calcium hydroxide is need to produce 155 g of calcium phosphate? b. How many molecules of ammonia gas are released? c. What volume of water will be produced, assuming normal conditions? 2 (NH 4 ) 3 PO 4(aq) + 3 Ca(OH) 2(aq)  Ca 3 (PO 4 ) 2(aq) + 6 NH 3(g) + 6 H 2 O (l)

22. 22 a. 155 g Ca 3 (PO 4 ) 2 * 1mol/310 g = 0.500 mol Ca 3 (PO 4 ) 2 0.500 mol Ca 3 (PO 4 ) 2 * 3Ca(OH) 2 /1 Ca 3 (PO 4 ) 2 = 1.50 mol Ca(OH) 2 1.50 mol Ca(OH) 2 * 74 g/mol = 111 g b.0.5 mol Ca 3 (PO 4 ) 2 * 6 H 2 O/1 Ca 3 (PO 4 ) 2 = 3.0 mol H 2 O 3.0 mol H 2 O * 18 g/mol * 1 mL/1 g = 54 mL water c.0.5 mol Ca 3 (PO 4 ) 2 * 6 NH 3 /1 Ca 3 (PO 4 ) 2 = 3.0 mol NH 3 3.0 mol NH 3 * 6.0221 x 10 23 molecules/mol = 1.8 x 10 24 molecules

23. 23 Limiting-Reactant and Yield Calculations The limiting reactant is one of the reactants in a chemical reaction. It is called the limiting reactant because the amount of it present is insufficient to react with the amounts of other reactants that are present. The limiting reactant limits the amount of product that can be formed.

24. 24 H 2 + Cl 2  2HCl  + 7 molecules H 2 can form 14 molecules HCl 4 molecules Cl 2 can form 8 molecules HCl 3 molecules of H 2 remain H 2 is in excess Cl 2 is the limiting reactant 9.3

25. 25 Steps Used to Determine the Limiting Reactant 1. Calculate the amount of product (moles or grams, as needed) formed from each reactant. 2. Determine which reactant is limiting. The reactant that gives the least amount of product is the limiting reactant; the other reactant is in excess. 3. Calculate the amount of the other reactant required to react with the limiting reactant, then subtract this amount from the starting quantity of the reactant. This gives the amount of the that substance that remains unreacted.

26. 26 How many moles of HCl can be produced by reacting 4.0 mol H 2 and 3.5 mol Cl 2 ? Which compound is the limiting reactant? Step 1 Calculate the moles of HCl that can form from each reactant. H 2 + Cl 2 → 2 HCl Step 2 Determine the limiting reactant. The limiting reactant is Cl 2 because it produces less HCl than H 2.

27. 27 You Practice: Work on Skillbuilders 8.4 & 8.5 with a partner. Turn in one piece of paper with both your names on it showing all work.

28. 28 How many moles of silver bromide (AgBr) can be formed when solutions containing 50.0 g of MgBr 2 and 100.0 g of AgNO 3 are mixed together? How many grams of the excess reactant remain unreacted? Step 1 Calculate the grams of AgBr that can form from each reactant. MgBr 2(aq) + 2 AgNO 3 (aq) → 2 AgBr (s) + Mg(NO 3 ) 2(aq) The conversion needed is g reactant → mol reactant → mol AgBr → g AgBr

29. 29 How many grams of the excess reactant (AgNO 3 ) remain unreacted? Step 3 Calculate the grams of unreacted AgNO 3. First calculate the number of grams of AgNO 3 that will react with 50 g of MgBr 2. The conversion needed is g MgBr 2 → mol MgBr 2 → mol AgNO 3 → g AgNO 3 The amount of AgNO 3 that remains is 100.0 g AgNO 3 -92.3 g AgNO 3 =7.7 g AgNO 3

30. 30 Theoretical Yield The quantities of products calculated from equations represent the maximum yield (100%) of product according to the reaction represented by the equation. Many reactions fail to give a 100% yield of product. This occurs because of side reactions and the fact that many reactions are reversible.

31. 31 Theoretical Yield & Percent Yield The theoretical yield of a reaction is the calculated amount of product that can be obtained from a given amount of reactant. The actual yield is the amount of product finally obtained from a given amount of reactant. The percent yield of a reaction is the ratio of the actual yield to the theoretical yield multiplied by 100. Actual yield * 100 = Percent Yield Theor yield

32. 32 Silver bromide was prepared by reacting 200.0 g of magnesium bromide and an adequate amount of silver nitrate. Calculate the percent yield if 375.0 g of silver bromide was obtained from the reaction: MgBr 2 (aq) + 2AgNO 3 (aq) → 2AgBr(s) + Mg(NO 3 ) 2 (aq) Step 1 Determine the theoretical yield by calculating the grams of AgBr that can be formed. The conversion needed is g MgBr 2 → mol MgBr 2 → mol AgBr → g AgBr

33. 33 Silver bromide was prepared by reacting 200.0 g of magnesium bromide and an adequate amount of silver nitrate. Calculate the percent yield if 375.0 g of silver bromide was obtained from the reaction: MgBr 2 (aq) + 2AgNO 3 (aq) → 2AgBr(s) + Mg(NO 3 ) 2 (aq) Step 2 Calculate the percent yield. must have same units

34. 34 You Practice: Work on Skillbuilder 8.6 and problem 60 on page 257.